Ask the Physicist!

Here T is the temperature and C is some constant. I will not bore you with the algebra, but you can, from these three equations, deduce the functional relationship between the final pressure and the initial volume:

I have graphed this for P/p=½ and for v=1 ft³; so you can see how the final pressure would change from 1000 ft³ to 5000 ft³. Note that as the room gets bigger and bigger, the quantity P'v/(nC)→½. But, the smaller the room is, the higher the final pressure is. But the higher the pressure is, the larger the number of molecules inside the bottle is at the end. But, we want as many molecules as possible out of the bottle for maximum "fizz", so the larger the room, the "fizzier" the bottle. Be sure to note, though, that the change is very small, only less that 0.1% for your rooms.

QUESTION:
A question I have is if particles did not have charge, would there just be gas floating around everywhere? Would matter clump up into dust since it might not move?

ANSWER:
If there were no electric charge in the universe there would be no atoms as we know them; there would be no electromagnetic radiation (light); presumably these particles would have mass so there would be gravity. But to say anything at all about how this uinverse would be constructed would require specifying if there were any other interactions among these particles besides gravity.

QUESTION:
The US Naval Academy makes this statement. "Often, ship power is roughly proportional to the cube of its speed, so doubling (2x) the speed of a destroyer from 15 knots to 30 knots will require 2^3 = 8 times as much power." If you could imagine the destroyer steam powered where if the destroyer was travelling at 15 knots, then the power available to stop it is the same as the power to push it forward. The destroyer could fire up more boilers and increase its speed to 30 knots, so now the power available to stop it also equal to the power pushing it forward at 30 knots.

My question is regarding the stopping distance for the two speeds. I was thinking the stopping distance at 15 knots is either 2 times or 4 times the stopping distance at 30 knots. I'm not sure this question can be answered.

ANSWER:
I think you are assuming that the ship brakes by either reversing the spin of the propellers or pointing them astern rather than aft. And so you assume you will have the same force and therefore the time to stop should equal the time to accelerate. Unfortunately, it is not that easy. A propeller pulling backwards rather than pushing forward causes "cavitation" which reduces its force considerably. Furthermore, because the sources of drag on the ship change as the speed changes, and the cavitation would become less severe as the ship slows; therefore the ships at 15 and 30 knots would be very unlikely behave in the linear fashion you suggest. The velocity dependence of the drag forces on a ship is quite complicated as you can read in the Naval Academy text you quoted. There is a nice little comment on how ships actually brake which you might find interesting.

QUESTION:
I am working on a concept for a sustainable energy project I am working on. I would like to create an art piece that powers a lightbulb through the friction of coins going down a ramp. Is this even possible? I know that energy can be created through friction but I'm not sure how and if coins can even create that kind of energy.

ANSWER:
I cannot see how this could possibly work. When a coin slides down the ramp, the work done by friction is converted to heat (a tiny fraction to sound). But, how would you extract the heat from the coin and the ramp? And how would you convert it into electrical energy to power a light bulb? I did some very rough calculations for a quarter (mass about 6 grams) sliding with constant speed about 10 cm/s down a ramp inclined at about 20⁰ and about a meter long. The force of friction would be on the order of about 0.02 Newtons and the energy generated by friction would be about 0.02 Joules; if it took 10 seconds to slide down, the power (rate of energy) would be about 0.002 Watts!

QUESTION:
Two men pick up a heavy log and carry it along a are they doing work if the height of the log above the ground does not change why do they get tired?

ANSWER:
See an earlier answer.

QUESTION:
why does it take twice as much energy as that of a 70mph pitch to throw 90mph pitch? Shouldn't twice mean 140mph?

ANSWER:
If the pitcher throws a pitch with speed v, he has given the ball energy E_v=½mv² where m is the mass of the ball. So the factor by which he must increase his speed from 70 mph to 90 mph is E₉₀/E₇₀=½m(90²)/½m(70²)=(9/7)²=1.65. So it doesn't take twice the energy, only 1.65 times the energy.

QUESTION:
So the law of conservation of energy states that energy can not be created or destroyed right? So in that case How did the Big Bang aka the formation of universe take place? As energy cannot just form out of nothing so There had to be energy present to convert it into the dark energy and matter. How did it happen?

ANSWER:
Nobody knows. All we know is that for times after the big bang the total energy which the big bang brought seems to not change. Sorry if this seems like a cop out, but it's the best I can do.

QUESTION:
If we watch a spaceship leave Earth and speed away really fast, we should see the time slow down for the spaceship. When the spaceship turns around and comes back, why doesn't the return trip null out the difference in time?

Secondly, if two other spaceships left Earth 120 degrees apart from each other, traveled at the same speed to the same distance, would we see their time slow exactly the same regardless of how fast or slow the Earth is moving?

Finally, if we can see the time on the spaceship slow, does the spaceship see the time on Earth slow because if motion is relative, the Earth is speeding away from the spaceship exactly as fast as the spaceship is leaving the Earth. Why does time always slow for the spaceship if it is not moving any faster than the Earth?

ANSWER:
First, note that site groundrules stipulate "… single, concise, well-focused questions"; I will answer your questions this time, but please comply if you submit questions in the future.

There is a difference between how we "see" a clock running and how fast that clock is running. Clocks run at different rates in different frames. Take the earth as one frame and and the spaceship as another. The spaceship's clock runs slower in the earth's frame and the earth's clock runs slower in the spaceship frame. You should look at the faq which addresses this question. You will see that when the two frames are moving apart, the other's clock "appears" to run slowly and when they are moving toward each other, the other's clock "appears" to run faster. But neither of these is indicative of how the clocks actually run the other's frame. This addresses your third question as well as I can.

Your first question suggests that time does run faster in one direction than the other but that is wrong because the rate a clock runs depends only on the speed, not the direction of that speed.

Your second question is tricky because usually when we talk about such things the earth is assumed to be at rest. But, of course, it isn't. But being at rest is not really an issue because even if the earth were moving with some constant speed in a straight line, the spaceships would have that velocity added to their own and everything would look to be the same from the earth's perspective as if the earth were at rest. But the earth is not moving with constant velocity but travels in its orbit around the sun; as the earth's velocity changed direction, the velocities of the two ships relative to the earth would change depending on the direction in which they launched. But the earth's speed in its orbit is tiny compared to the speeds of the ship and can be neglected for almost all purposes.

QUESTION:
If wind from a fan is blowing on an object at a constant speed, does the wind supply any force on the object? Thinking in terms of Newton's Second Law, I am wondering since having no acceleration makes the force 0.

ANSWER:
Suppose that there is a book sitting on a table and you push on it with a horizontal force but not hard enough to get it moving. Would you conclude, since the book is not accelerating that there is no force on it? Of course not, because you know perfectly well that you are pushing on it. But, there is no net force on it, so there must be some other force pushing back on the book just exactly opposite but equal to your force; that force is the friction between the book and the table surface. (It is important, if you know some physics, that this friction is not a Newton's third law force. That force is the force which the book exerts back on you which is equal and opposite but not a force on the book.)

Suppose that there is a book sitting on a table and a wind pushes on it with a horizontal force but not hard enough to get it moving. Would you conclude, since the book is not accelerating that there is no force on it? Of course not, because a wind will always exert a force on something it strikes. But, there is no net force on it, so there must be some other force pushing back on the book just exactly opposite but equal to the wind's force; that force is the friction between the book and the table surface.

QUESTION:
My son in law have a bet. A million dollars. It has to do with G force and what the force is that pushes the occupant of a vehicle into their seat upon the acceleration of the vehicle. I told him that my take is that while the occupant (O) of the acceleration vehicle AV is is accelerating at the same rate as the AV O is out of sync with the AV. Inertia holds the O in place when the vehicle begins to move. It is that lag in beginning times of acceleration due to inertia that pushes O into the seat. (and here's where my son in law breaks in to laughter)I say that until the acceleration stops (not just the rate of acceleration) given any slice of time during that acceleration the O is going slightly slower than the AV. Where I get all my derisive laughter is when I say that the O is going slower that the AV. Further, Nate says that the effect stops when the rate of acceleration stabilizes and is no longer increasing. I don't buy that. AS long as there is acceleration the effect continues (well. Or when gravity is no longer a factor). It doesn't need to be an increasing rate of acceleration. I'm a musician, Nate builds houses and computers. Neither of us knows what we're talking about. Do you?

ANSWER:
Good heavens, almost nothing you write makes any sense. A little tutorial is necessary. Speed measures the rate of change of position. If your speed is constant and is 10 miles/hour, you go 10 miles in the first hour, 10 miles in the second hour, 10 miles in the third hour, and so on. If speed is changing you introduce a new quantity, acceleration, which measures the rate of change of speed. If your acceleration is constant and is 5 miles per hour per second and you start from rest, your speed will be 5 miles/hour after one second, 10 miles/hour after two seconds, 15 miles/hour after three seconds, and so on. For purposes of enlightening you here, we will not talk about changing acceleration, assume that the acceleration is constant like in my little example above. Now, if the car with the occupant goes by, what do you see? You see a man who has an acceleration of 5 miles/hour/second. Newton's second law says that if there is some mass accelerating, there must be some force causing that acceleration. What force is that here? The only force in the direction of the acceleration is the seat which is exerting a force equal to the man's mass times his acceleration. Suppose we have a 200 lb man; without boring you with the details (getting the units right is a pain in the neck), the seat back exerts a 4.58 lb force on the man. If the car stops accelerating, this force disappears.

Wait a minute! What about the force which presses the man back into the seat? There is, in fact, no such force in spite of the fact that you would swear are being pushed back. What is actually happening is that the seat is pushing you forward with a force of 4.58 lb and your brain interprets what it feels as being pushed backward with a force of 4.58 lb. From the point of view of the observer, he is at rest in his seat but there is an unbalanced force on him in the forward direction; how can that be if there is no acceleration (remember he is at rest in his frame)? Newton's laws are not true in accelerated frames. But you can force Newton's laws to be true by adding a 4.58 lb force in the backward direction; we call this a fictitious force (because it is) but it forces Newton's laws to be obeyed.

QUESTION:
Do quantum fluctuations violate Einstein's special relativity? Now, I'll be honest, I am a total newbie when it comes to Physics. I would really like to know if my question is even reasonable, because I am not sure that quantum physics and general relativity are that related..

ANSWER:
The theory of special relativity is a classical theory and says nothing about quantum mechanics. Sometimes we hybridize the two when we want to know what nonrelativistic quantum mechanics can be examined when speeds are high, but the proper way to incorporate special relativity is via the Dirac Equation.

General relativity (GR) is a completely different theory, it is our current best theory of gravity. Since GR is a field theory, it should be able to be quantized; nobody to date has successfully quantized GR.

QUESTION:
you are traveling with the speed of light and then you take out a battery to splash out a strong beam of light towards the direction you're going. In this case, speed of light (speed of light emitted from from that battery) doubles or what?

ANSWER:
First and foremost, neither you nor any object with mass can travel the speed of light. You could get an inkling of the answer to your question by traveling with a speed 99% the speed of light. One of the tenets of the special theory of relativity is that the speed of light is independent of the speed of either the observer (you in your question) or the source. Hence the speed of your flashlight beam would be measured by you to be the same speed (3x10⁸ m/s) regardless of how fast you were traveling. That is the answer to your question. To go a little further, the result of special relativity for the relative speed of two objects moving in the same direction with speeds u and v relative to a third observer is v'=(u+v)/(1+(uv/c²)) where c is the speed of light. So let v=c (flashlight beam seen by you) and u=0.99c (your speed) and then c'=(0.99c+c)/(1+0.99c/c²))=1.99c/(1.99c)=c which is the speed seen by the third observer. (Classical physics, which is good for speeds much less than c, has v'=(u+v). Everybody sees c!

QUESTION:
I make cold brew coffee. I keep the coffee and creamer in my fridge and have a hot water tap that produces water at 190 degrees. Since the coffee and creamer are cold they bring down the overall temperature of the final cup of coffee. Because of this, I mix the coffee and creamer first and microwave on high for 30 seconds, then add the 190 degree water for a hot cup of coffee. Will the final temperature be the same if I mix the cold coffee and creamer with the 190 degree water, then microwave all ingredients for 30 seconds on high. c=coffe; cc=creamer hw=hot water m30= 30sec.microwave does : (c+cc)*m30+hw =(c+cc+hw)*m30

ANSWER:
This is a conservation of energy problem and can be solved by inspection. The microwave adds a certain amount of energy Q to the system (consisting of all the ingredients at their original temperatures) regardless of when in the mixing process you add it, so the final temperature will be the same. One proviso is that if any step of the mixing and heating results in a temperature exceeding the boiling point, all bets are off. In your particular case, if the 30 seconds of microwaving the c+cc boils it, you cannot be sure that the two would be the same; since these start somewhere around 35 degrees, I don't think that would happen. If you are really interested I could do the calculations for you (assuming c and cc and hw all have the same specific heat capacities to keep the algebra simple).

REPORT:
Thanks a lot for your quick response. At no time did any of the liquids reach boiling point. I concluded my "extremely crude" test this morning (didn't want to waste extra coffee yesterday)

c+cc*m=116, (c+cc)*m+hw=159
c+cc+hw=124, (c+cc+hw)*m=156

Considering how crude my liquid and temperature measurements were that's pretty close.

THANKS:
Excellent! I really appreciate it when I get feedback. I understand your not wanting to waste coffee; when trying cold brewing for myself I came to realize how much more a cup costs than with more traditional brewing methods.

QUESTION:
So my 16 year old son keeps his hands down low (around belly button level) when he is about to swing his baseball bat. His hands go back, no up much, and then he swings. He hits homeruns off of live pitchers. However, he doesn't hit many during practice when a coach is closer "tossing" the ball to him. My question is why can he produce more power or make the ball travel further when a player pitches to him? The other part of the question is: Can you determine the force produced after contact (you can use any numbers you want as the givens) if a batter swings like my son with low hands at belly button level vs the traditional high hands approach? My thought is that due to the path to the baseball my son swings at produces a more inline result with the pitch (matches pitched ball path better) therefore more resorting in producing force

ANSWER:
I can do an approximate calculation which will explain why he hits the ball farther with a faster pitch. I will assume that the collision between the ball and the bat is perfectly elastic (no energy is lost) which, while not exactly true, should be good enough for our purposes. The equation to use is

v_ball=[u_ball(m_ball-m_bat)+2m_batu_bat]/(m_ball+m_bat).

Here, v is the velocity after the two collide and u is the velocity before the collision and m is the mass; I hope it is obvious what the subscripts denote. Now, I looked up reasonable values of masses, incoming velocities, and bat speeds for high school kids:

m_ball=0.145 kg

m_bat=1 kg

u_bat=9 m/s (about 31 mph)

u_ball=-30 m/s (about 70 mph) for the game pitcher

u_ball=-13 m/s (about 30 mph) for warmup tosses

Note that I have chosen the direction from the plate to the pitcher to be positive, hence the negative signs on the pitched velocities. Now, I find that v_ball=38.1 m/s for the pitched ball, v_ball=25.4 m/s for the tossed ball. All other things being equal, the faster the pitched ball, the faster the hit ball. All this is for a one-dimensional calculation (all velocities right before and right after the collision are along a single straight line) and ignore the fact that there is the force which the batter is exerting on the bat; so do not expect quantitative agreement with real-life situations. Nevertheless, if all the possible complications could be included, the qualitative result would be unchanged.

Your second question I cannot answer because body mechanics are far too complex to deal with using simple physics. This is the domain for a good, experienced hitting coach who knows proper batting techniques from years of experience. I couldn't help noticing, though, that Babe Ruth had a low-hands swing! One thing in your question bothers me, though: "…the force produced after contact…"? Apart from gravity and air drag, there are no forces on the ball after it leaves the bat. Your son affects the ball only during the time the ball and bat are in contact with each other.

FOLLOWUP QUESTION:
The second question, I dont think I did a very good job asking it, is demonstrated in the attached picture. I am just curious if by having low hands and the bat being more on the same plane of the pitched ball, if it hits the ball further than a bat coming more down on the pitched ball (high hands = equals downward plane that bottoms out right at or just before contact with baseball). I hope this makes some sense....kinda hard to put into words.

ANSWER:
I have watched several videos of major leaguers with high-hands swings. None of them have bat trajectories which resemble your green arrow but look a lot more like your red arrow. A low-hands swing would not have the downard part of the swing you show. It may be that the high-hands swing allows getting a larger bat speed. Again, I think that this is better decided by a capable batting coach. Since most major leaguers have the high-hands swing, I can't help thinking that it must be advantageous somehow.

QUESTION:
This question is actually a couple of questions regarding super collider designs. Why do we not study collisions in detectors in say, a triangular shape? I understand the importance of the circular design to increase the energy as beams lap around the collider, but do we not have them collide towards a wall in a triangular path? My main question is how does the design of the collider affect its experimental value to detection of nuclear interactions? Are we just not interested in collisions at various angles, is it inefficient?

ANSWER:
One of the main reasons colliders are designed for "head-on" collisions is that this brings the most available energy to the collision. The colliding protons have zero net linear momentum and so none of the kinetic energy is "wasted" in conserving momentum. Also, from the perspective of accelerator design, it would be ridiculous; one of the reasons that the collider is so huge is that at any given time the deflection of the beam is a small angle. To have to deflect a beam through an angle of 60⁰ would require magnetic fields much larger than available even with superconducting magnets.

QUESTION:
So when we talk about inertial frame of references, we are talking about the thing that isn't moving in relation to a specific perspective? so say I stand still but the earth is moving around me, then the earth would be the inertial frame of reference? but if I move, the earth is still spinning but so slow that I think I'm the only thing 'moving', would I still be the inertial frame of reference? no right, since in my perspective the earth is the thing that looks as if it's not 'moving'?

ANSWER:
You are making this way too hard. The definition of an inertial frame is that if you find that Newton's first law is true (an object with no forces on it remains in a state of rest or moves with a constant speed in a straight line) you are in one. The earth is not an inertial frame because the surface of the earth is accelerating because its velocity is constantly changing direction. (For many applications it is a good approximation to consider the earth as an inertial frame because the acceleration is very small; nevertheless, it is not an inertial frame.) Once you have found one inertial frame, any other frame which moves with constant velocity relative to yours is also an inertial frame.

QUESTION:
Imagine an air-tight, light-tight box lined 100% (all sides, top, bottom, left right, front, back) with inward-facing mirrors inside which a lighted candle or a match is placed before the box is sealed. Once sealed, neither air nor light can exit or enter. When the flame eventually consumes all of the oxygen inside of the box and extinguishes, will the box remain lighted in perpetuity because of the lights constant reflection off of the mirrors?I am curious to read your answer. I suspect that my lack of knowledge of the properties of light will come into play here.

ANSWER:
Well, the candle does not do what I think you want it to do. It does not "consume all the oxygen", it turns it into carbon-dioxide or -monoxide. Why not just have a box from which you can pump out all the air? And in that box is a light bulb you can simply turn off. The answer to your question is given in an earlier answer. As you will see, it is the properties of mirrors, not light, which holds the answer to your question.

QUESTION:
Hello, I am arguing with another person about the following scenario: Which person expends more energy? A 180 lb man and a 140 lb woman both run 100 yards in the same amount of time. Who expended more energy? My contention is that the man did because of his mass. The person arguing with me said "Men have longer strides and take less steps to run down the field then women do. More Steps = more muscle contraction/relaxation x energy (atp) = women expend more." Please tell me your credentials as he won't believe me.

ANSWER:
Assuming that their is negligible air drag and feet to not slip (in other words no energy lost during the run) and they both start at rest, the person with greater mass expends more energy, in this case the man. Each person starts with zero energy and ends with a kinetic energy of ½Mv², generated solely by the individual.

Regarding my credentials, click the button labelled the physicist.

QUESTION:
Need to know how to calculate weight of a 1.9kg item falling from a distance of 3 m what is the weight when it hits the floor. I am not at all scientific enclined pls..or if you can help with answer pls

ANSWER:
The weight W is exactly the same as anywhere else near the surface of the earth, W=mg where m is the mass 1.9 kg and g is the acceleration due to gravity 9.8 m/s². Many countries express the weight of something in kg, but kg·m/s² is the proper science way to express weight and is called a Newton (N). If I can get any answer to your question I will translate it to kilograms at the end. Now, what you really want is the force which the weight exerts on whatever it hits (floor, your foot, etc.), not the weight. What if the weight is just sitting on your foot? Then the force which it exerts down on your foot is just equal to its weight 1.9x9.8=18.62 N (or, if you prefer, 1.9 kg). Incidentally, your foot exerts an upward force on the weight which is what keeps it from falling down. But, if it is falling on your foot, some force greater than the weight must be exerted on it because that is the only way to stop it from falling; of course that force is what your foot feels. What determines the magnitude of this force is how fast (speed v) the mass is falling and how quickly it stops (time t), F=mg+mv/t. Without going into details, the speed it hit your foot would be about 7.7 m/s and, if it takes 0.0004 seconds to stop (which would correspond to stopping in a distance of 3 mm), the force would be about 37,000 N or about 3,700 kg. Of course, this average force only lasts for about 0.4 milliseconds but would certainly hurt; after that the force would again be 1.9 kg. The way to make the force smaller would be to make the time longer. For example, if the 1.9 kg object was very soft it would compress and take much longer to stop.

QUESTION:
I understand the laser interferometry used in the LIGO experiment shows that we had a ripple experienced in spacetime - but how does this tiny wobble prove that the cause of it was black holes colliding? It might have been caused by Godzilla in Japan jumping into the sea for all i know. Whats the proof that it was black holes colliding and how do they know how many solar masses the black holes were from that data?

ANSWER:
There is more than one detector, one in Louisiana and one in Washington state; more recently detectors have been added in Italy and India. With more than one detector you can get an idea of where the signals originated, and it was certainly not in Tokyo bay. And if these observations were seismic, why did none of the thousands of seismographs around the earth detect them? But to me the most impressive "proof" is that the signal shape is so well fitted by the calculations. Our best theory of gravity is general relativity, the theory which predicted the presence of gravitational waves in the first place. By adjusting parameters in a general relativistic calculation, the data are very well fitted and things like the masses of the colliding black holes, total gravitational wave energy radiated, the angular momentum of the system, etc. can be extracted. One can also tell the difference between collisions between two black holes, two neutron stars, and one neutron star with one black hole.

QUESTION:
Hi, I've just learned that color is just a result of human visual processes, but that it is not an intrinsic property to all things. So if color is actually just wavelengths of different light, are these wavelengths infinite? Say for example, you and I were standing on a completely flat plain in a void, is there a certain distance away from you I could stand to stop perceiving you in color? Where, physically, do wavelengths stop or how far can and do they extend, if at all?

ANSWER:
Yes, light has a wavelength; "wavelengths" never "stop", and without a wavelength you do not have a wave (no light). Our eyes (actually our brain) can distinguish between different wavelengths and so we give names to different perceived wavelengths of light—red, green, blue, etc. That which is "just a result of human visual processes" is the names we give to different colors, not the physical properties of the light like wavelength.

QUESTION:
Since the Photon is the force carrier for EM waves, does that mean that photons are leaving one pole of a magnet and entering the other pole?

ANSWER:
No, the photons are "virtual photons", they pop into and out of existence, the time of existence being in accordance with the Heisenberg uncertainty principle (HUP). A photon has energy E and an uncertainty ΔE, and therefore if it were a "real" photon, would violate energy conservation. But, the HUP says that energy conservation may be violated but only for a time t approximately as long as t<h/ΔE where h is Planck's constant. The whole notion of photons being the force carriers is essentially a cartoon to help us visualize the quantization of a force field.

QUESTION:
Is it possible for a deflated basketball (outdoor Great Dane toy!) to become inflated after being left outside in 32° weather all night? Thank you!

ANSWER:
What is possible is that the ideal gas formula PV=NRT will provide a good qualitative answer to your question. P is the absolute pressure, V the volume, N some measure of the amount of air, R a constant, and T the absolute pressure. You should read an earlier answer similar to your question. Essentially, as long as the amount of air remains constant (no leaks), if the temperature increases either pressure or volume or both will increase.

QUESTION:
Why must the international Space Station maintain only a mere 18,000 miles per hour for their velocity to continuously fall around the Earth as Compared to the Earth which maintains a velocity of 68,000 miles per hour to maintain it's perpetual falling state around the Sun?

ANSWER:
For a circular orbit of radius R of an object of mass m about an object of mass M, mv²/R=mMG/R² if M>>m. Solving for v, v=√(MG/R); apparently M/R for the space station's earth orbit is much smaller than for the earth's sun orbit.

QUESTION:
Even though earths mass is not evenly distributed, how earth is spinning at its axis with constant speed?

ANSWER:
Any rigid object, regardless of its shape or mass distribution, will rotate about an axis passing through its center of mass. Only an external torque about this axis will change the rotation rate.

QUESTION:
We have a rolling platform 10'x10'x27 height. It weighs 2400 lbs. rough estimate on push force? On Plywood, level ground. I worked it to over 10.0000 n is that ballpark? Reason being we are working for general contractor, im the supervisor. He built this for us where we are allowable push force legislation states 80lbs. If someone gets hurt because we did try to move this tank , i own that , here you can be charged with criminal neglect. This fuys teing to save money but thats irrelevant, his ego is irrelevant, but legislation is the law.

ANSWER:
There is not enough information to calculate this.

FOLLOWUP QUESTION:
Ok, best case scenario. Now its bigger. 3000 lbs on 10"x4" wide castors ( wheels) see im trying to see if im within our safe parameters for push force on 4 men. Im moving it 30', It feels heavy but i need something to rebuttle this site supervisor trying to save money.

ANSWER:
It would be pretty easy to estimate if I knew the specs of the castors, in particular the coefficient of rolling resistance, C_rr, of the wheels. If they are ball bearing casters and the wheels are hard rubber, for example, one could probably estimate the rolling friction force of hard rubber on plywood neglecting the force due to the bearings. The force F needed to move your platform having a weight W can be estimated as F=C_rrW, so if you want F to be less than or equal to 80 lb and W=3000 lb, C_rr≤80/3000=0.027. If you look at the values of several types of wheels listed as examples in Wikepedia, you will see that almost every example has a much smaller value of C_rrthan 0.027; only a stage coach on a dirt road (0.039-0.73) or a car on sand (0.3) have larger values. It would appear that you can move this with four men each pushing with a force of 20 lb. On the other hand, if you read a little further down on Wikepedia you can estimate C_rrif the amount of depression z in the "floor" is not very small relative to the wheel diameter d=10", C_rr≈√(z/d). So, suppose that the plywood depresses by half a millimeter; I calculate that C_rr≈0.044 in which case the force to move the platform would be about 0.044x3000=133 lb. Over all, if you read more of the Wikepedia article you will see that rolling friction is a quite complicated problem.

QUESTION:
If the distance between two points on the surface of a sphere (presumably our three dimensional universe) is increasing at a rate of 72,000 meters/second/3.0857e16 meters (Hubble constant), how fast is the radius of the sphere (distance from the Big Bang) increasing? The radius may be in the range of 1.30558080522e26 meters (age of the universe).

ANSWER:
As stated on the site, I do not do astronomy/astrophysics/cosmology. However, you are asking a simple geometry question: if an arc length on the surface of a sphere is increasing at some rate, what is the rate at which the radius of that sphere is increasing? If the arc length is S and subtends an angle it subtends is A (in radians), then S=RA where R is the radius of the sphere. If the rate of change of S is dS/dt, then dR/dt=(dS/dt)/A. I do not understand your specification of rate which seems to have the units of s^-1; but you say it is the rate of increase of the distance between two points which should have the units m/s.

QUESTION:
Why do we measure the length of a pendulum from the centre of the bob and not from the point of attachment to the bob?

ANSWER:
This might not necessarily be the best way to characterize a pendulum. If the mass of the bob is much greater than the mass of the stick or string to which it is attached and if the length of the stick or string is much greater than the size of the bob, the best way to make the system well-modeled (for small angles) as a simple pendulum is to treat the center of mass as a point mass. By "simple pendulum" I mean a point mass M on a massless string of length d; in that case, the period T is approximately given by T≈2π√(d/g). That means we should use (d+R) instead of just d to correctly predict the period where R is the distance to the center of mass of the bob from the point of attachment of the string.

So now I would like to show that this is the case and what its limitations are. That may be much more than you want, but I like to lay these things out with some rigor just to please myself. I will look at the special case of a disk of radius R and mass M attached to a massless string of length d. This is called a physical pendulum and we need to ask what the pendulum's moment of inertia is about the point of suspension P. The moment of inertial of the disk about its center is I_CM=½MR² and, using the parallel axis equation, its moment of inertial about P is I_P=½MR²+M(R+d)². Now we use the rotational form of Newton's second law*, τ=-Mg(R+d)sinθ≈-Mg(R+d)θ=I_Pα=I_Pd²θ/dt² or d²θ/dt²≈-(Mg/I_P)θ which is the simple harmonic oscillator equation with a period T≈2π√(I_P/Mg(R+d)). This is the correct period for any shape of bob. Applying it to the disk, T≈2π√[(½R²+(R+d)²)/(R+d)g]. Suppose that R=0, a truly simple pendulum: T≈2π√(d/g) as expected. But suppose that R is much less than d but not zero; then ½R² is much less than (R+d)² and so T≈2π√((d+R)/g). This demonstrates why you use d+R instead of d when you specify the "length" of a pendulum. But, consider the other extreme—d=0; in that case, T≈2π√(3R/(2g)) and not 2π√(R/g).

*Torque equals moment of inertia times angular acceleration.

QUESTION:
Imagine a ball is travelling at constant velocity (it should be noted that initially it accelerates as it comes to motion from the state of rest. Then after moves at a constant velocity) now imagine it collide with a drinking glass. As we know the force which the ball would exert on glass is ball's mass times it's acceleration, here acceleration is zero so F=0 so will the glass move?

(I have asked this question to my teachers they said that in the process of collision the velocity of glass would change, hence acceleration is not equals to 0 so glass will move. But I disagree, why would collision affect the velocity? Because it exerts a force as a "reaction" of the "action" produced by the ball. As we know, according to newton's third law of motion, the reaction is equals and opposite to the action here action is itself zero so how can reaction be non zero. It means this is not the correct explanation of the above stated problem.)

ANSWER:
Your misunderstanding of the physics of what is going on has some important lessons to be learned. First, physics is, at heart, an experimental science and one should think about what you are trying to explain. So I ask you to imagine throwing a ball at a glass and ask yourself if, after the ball has hit the glass, the glass will remain at rest and the ball will move along unaffected. If the ball is very heavy compared to the glass, it may appear to be unaffected by the collision but the glass will affected; and if the glass is very heavy compared to the ball, it may appear to be unaffected by the collision but the ball will affected; but in general, both target and projectile are affected by the collision. The second lesson here is to not use some formula you have memorized (F=ma in your case) if you do not understand what it is applicable to.

So, let's do the physics of this situation correctly. When two objects collide, each exerts a force on the other. These two forces are, as you note, are equal and opposite, so the force on the whole system is zero, but both the glass and the ball experience equal and opposite forces. So each experiences an acceleration during the time of the collision, usually a quite short time but not zero. So, if it is a "head-on" collision, the ball will slow down (or even bounce back) and the glass will move away in the direction the ball was originally moving. You are right that F=0 for the system consisting of the ball plus glass, but that doesn't mean that nothing is changed. What is not changed is the linear momentum; for example, suppose the ball has a mass 2 kg and has a speed of 10 m/s before the collision, so it has a linear momentum of 20 kg·m/s and the glass has none; if the glass has a mass of 1 kg and a speed of 5 m/s after the collision, the ball has a speed of (20-1x5)/2=7.5 m/s making the total linear momentum still 20 kg·m/s.

QUESTION:
If my car weighs 1700 lbs and I am traveling 45 mph and traveling into the wind blowing 20 mph on an asphalt road what is the rate of friction on my tires that are fully inflated?

ANSWER:
There is not enough information to calculate the rolling friction force from your tires. I could estimate the air drag on your car which is probably more important than the rolling friction. But then I do not know how much power is being provided to the car by the engine to keep it going at constant speed. I also do not know other sources of friction, like in wheel bearings. What I can do is make a very rough estimate air drag and rolling friction force to see how they compare.

There is a handy estimate for air drag force, F_D=¼Av², where A is the area presented to the onrushing air and v is the speed of the air, 65 mph in your case. There is a catch, this works only if you work in SI units, but I can calculate the drag and then convert the answer to pounds. v=65 mph=29 m/s and I will guess A=4 m² (about 6'x6'); so F_D=¼x4x29²=841 N=189 lb.

The rolling friction force of a car's tires is F_R=C_rrW where W is the weight and C_rr is called the rolling resistance coefficient which is roughly between 0.007 and 0.014; if I take C_rr=0.01 then F_D=0.01x1700=17 lb. So the drag from the air is about 10 times bigger than the rolling friction at this speed.

We could take this a little further by estimating the power the car must be delivering to maintain its speed of 45 mph in this wind (neglecting other sources of friction). The engine must provide a forward force of 206 lb over a distance d in a time t and the power is P=Fd/t. But d/t is the actual speed of the car, 45 mph=66 ft/s, so P=206x66=13,596 ft lb/s=25 horsepower.

QUESTION:
This has to do with my mavic pro drone. It had a faulty battery and disconnected. At the time it was traveling 29 mph horizontally, 10.5 mph vertically and was 75 ft in the air at this time. The drone weighs 1.62 lbs. How far wiuld it travel before hitting the ground? I have a map of the area also if needed. Sorry if this is a simple question for you but i appriciate your time.

ANSWER:
I can give an approximate answer to this question. It depends on air drag being negligible and the terrain being reasonably level. I will first convert speeds to ft/s: 29 mph=42.5 ft/s and 10.5 mph=15.4 ft/s. The equations describing the motion are:

y=75+15.4t-16t²

v_y=15.4-32t

x=42.5t

v_x=42.5

Here x and y are horizontal and vertical directions. We know that when it hits, y=0, so we have to solve the quadratic equation -16t²+15.4t+75. Solving, I find that the time is t=2.7 s. Therefore the distance it will have traveled is x=42.5x2.7=114.7 ft. Since air drag has been neglected, you should find the crash site inside of about 115 ft unless the terrain has a large downward slope under the line of flight.

QUESTION:
This may at first sound like a high school homework problem, but it's not. I'm in my 50s and am just trying to work this out so that I can make an animation look a little more realistic (just for fun). I've idealized the problem for simplicity. I do have a daughter in high school, but she isn't taking physics until next year :-), It's been too long since I've tried to solve such equations, so I'm a bit rusty. Anyway here's the problem: mass M, moving at constant velocity V (along a frictionless surface), encounters a spring (aligned in the direction of motion) with spring constant K. After relearning the equations, I can pretty easily calculate the distance that the spring is compressed before bouncing back X = (M V^2 / K) ^ 0.5 . With more difficulty, I am able to calculate (hopefully correctly) the velocity of the mass with respect to distance V(x) = (V^2 - (K x^2 / M)) ^ 0.5. However, what I really want is an equation for X as a function of time: X(t).

ANSWER:
I have deleted a large amount of your question because you were "barking up the wrong tree" in your attempt to solve it. You clearly know a fair amount of physics, so likely you will kick yourself when I tell you how to go about solving this. A mass on a spring is a simple harmonic oscillator, so x(t)=Asin(ωt)+Bcos(ωt) where the angular frequency is ω=√(K/M)=2πf=2π/T; here, f is the frequency and T is the period. In your case you are going from the equilibrium position to the maximum value of x, so the time to reach the place where it comes (momentarily) to rest is ¼T=½π/ω=½π√(M/K). Since x(0)=0, B=0 and x(t)=Asin(ωt) and A is the amplitude (which you called X)where v(t)=0.

QUESTION:
Need help Understanding the concept of Constant pressure(THIS IS NOT A HOMEWORK PROBLEM,). Lets say you have a piston that is compressing a gas inside a cylinder. As the gas pushes up and raises the piston "WHY IS THE PRESSURE STAYING CONSTANT AND NOT CHANGING AS THE PISTON MOVES FROM A THE BOTTOM TO THE TOP OF THE CYLINDER."(please answer in the most simplest way possible Thank You.)

ANSWER:
The ideal gas formula for pressure P is P=NRT/V where N is the amount of gas, R is a conatant, and T is the temperature of the gas. Since you say that the gas is being compressed, I will assume that N is constant. V is decreasing which would mean that P is increasing if T is constant. Therefore, if P is constant, the temperature of the gas must be decreasing at the same rate as the volume is decreasing.

QUESTION:
doesnt the speed of light increase if it is captured by a black hole...

ANSWER:
When a mass falls in a gravitational field it increases in energy; it does this by speeding up. Similarly, when light "falls" in a gravitational field it also increases in energy; however, it cannot speed up but increases its energy by changing to a higher frequency (shorter wave length). The energy E of a photon of frequency f is E=hf where h is Planck's constant.

QUESTION:
You're in an igloo made of ice on an extremely cold night. You have a fire inside to keep you warm. As the interior warms, the ice on the inside melts, thinning the walls. But the severe outside temperatures prevent the igloo from melting away. As the walls continue getting thinner, melting on one side and super-cold on the other, does the dome expand, shrink, or stay the same size?
[from a paramedic from Canada]

ANSWER:
From the wording of your question, I will assume that the temperature inside the igloo is increasing with time. If it is indeed warmer inside than outside, heat is flowing through the ice and there is a temperature profile for which the temperature increases linearly from the outside surface to the inside surface. As the temperature increases inside, the average temperature of the ice increases; in fact, at every depth into the ice the temperature is increasing. When ice warms, it expands. Therefore, I believe that the igloo will increase in size.

QUESTION:
It was -40 this morning so the kids boiled some water and threw it outside. Super cool effect. We wondered, "what would happen if we did this with coloured water." We put red food colouring (gel) in it and it dissolved. When it was thrown, it almost immediately separated from the water and fell downward while the plume of evaporation went up. (I wish I could attach the photos for you to see). Why would this be? Would the food colouring have a higher boiling point? Can you help me understand what happened here. Also, would the implications be that the water cycle has been cleaning the water for years, but with new chemicals with lower boiling points be entering our water system and unable to be removed?

ANSWER:
It is very hard to give a definitive answer because your description is pretty vague. My best guess would be that something in the food color made it remain liquid when the water froze so that it separated. For example, propylene glycol, which is used in many food colorings, has a melting point around -60°C so it would not freeze at -40°.

QUESTION:
Since white light is made up of all of the rainbow colors, when it enters a prism and the light bends, the white light spreads out into the different colors of the rainbow, which is of course known as dispersion. My confusion is this: When the light wave goes from one medium to another, the frequency of the wave remains UNCHANGED. However, the wavelength CHANGES, causing the wave to change speed as it moves from one medium to another. The larger wavelengths have a greater speed (red and orange), and the smaller wavelengths have a lower speed (blue and violet). Since they have different speeds through the prism, they bend by different amounts as they exit the prism. To sum up my question: Doesn't the SAME wave in the spectrum have to have the SAME FREQUENCY, and therefore the SAME WAVELENGTH? For example, Red has one single wavelength with its corresponding frequency. So how can you reconcile this with an UNCHANGED frequency but a CHANGED wavelength? That won't be Red anymore! Would it? What am I missing/misunderstanding here? Can't wrap my head around this at all!

ANSWER:
The wavelength λ of a wave of frequency f and speed v is λ=v/f. Since the speed of a wave inside the prism is smaller than outside, it does not have the "SAME WAVELENGTH". What is red outside is not the same wavelength inside but shifted toward the blue.

QUESTION:
My son and I wanted to design a snooker/pool ball rotation experiment. By wanting to study the reverse rotation of a ball when flip the ball with a finger, rotate it forward on the table with the technique of flip the ball to spin back (similar backspin /screw-back shot in snooker/pool).
1.How are the theories of physics explained?
2.How will we design this experiment and How are the mathematical models going?

ANSWER:
It is actually easier to do experiments which show the same thing with a ping pong ball or a hula hoop. I can explain to you the physics of what is going on. I cannot tell you how to design your experiment since I know neither your goals nor your son's age. If this is a science fair idea, it would be inappropriate for me to design your son's experiment because then it would not be his experiment. So, how does physics describe this reversal of motion? I will not get too quantitative. In the figure above, the ball is spinning with a counterclockwise rotation as shown, and it is a rapid enough rotation that the ball is sliding in the forward direction at the point where the ball touches the table top. The center of mass of the ball is moving forward with a velocity v. There are three forces in this problem and two of them are not important for us. One force, vertically down, is the weight W of the ball and the second is the force N which is the force which the table exerts on the ball, vertically up; since there is no motion in the vertical direction, these two forces must be equal and opposite and thus cancel out. (This is Newton's first law and not Newton's third law which is often called "action and reaction".) Finally, the important force for us is the frictional force f which the table exerts on the ball; because the ball is sliding forward, f must act backwards and be horizontal. Newton's second law says two things in this problem:

f will cause the center of mass of the ball to slow down and
the torque due to f about the center of mass will cause the ball's rotational velocity ω to slow down.

Now, three things can happen, depending on the values of v and ω:

If v is large and ω is small, the ball will stop slipping before it stops moving forward and from that point on it will simply keep rolling forward (not slipping) with constant speed because f will disappear if there is no slipping. (In the real world there is still some friction from other forces and the ball will eventually stop.)
If v is small and ω is large, the ball will stop moving forward before it stops rotating; but if it is still rotating f will still be there and the ball will start moving in the backward direction, speeding up. Eventually the torque from the friction will slow the rotation to the point that the ball is rolling without slipping but back toward you.
If v and ω are just right, the ball will stop moving forward at the instant that it stops rotating and just stop dead.

Hope that this answers your question about the "theories of physics" relevant to your experiment.

QUESTION:
How is it we are able to see matter if all the atoms and sub atomic particles that everything is made up of are invisible to the naked eye? It doesn't make sense that electrons, protons and quarks are point masses with zero dimensions but define the size of the atom by their separation. So if an atom is mostly empty space which light cannot reflect off. How is anything visible?

ANSWER:
Only the electron is considered to be a point mass, but the size of the elementary particles which make up visible objects is irrelevant. You see something by observing light and light interacts with objects you see because it interacts with all the tiny particles which are the constituents of macroscopic objects.

QUESTION:
A Thought experiment: A square box, a stick man in the bottom right corner, an apple in the top left corner. From the perspective of the stick man, does the apple fall from the top down? Or does the floor rise to the position of the apple? Are both paths equally valid? Can the stick man distinguish one way or the other?

ANSWER:
What you are describing is the equivalence principle. There is a story, probably apocryphal, about Albert Einstein, looking out the window of his office in the Swiss Patent Office where he got his first job. Across the square there was a painter on a ladder painting the building. The painter fell off the ladder and, as he was falling to the ground, Einstein thought "there is no experiment which he could do to distinguish whether he is in a gravitational field with a gravitational acceleration g or is in empty space with an acceleration g." If your man is in a box and sees the apple accelerating toward him with acceleration g, it could be either that the box is accelerating up with an acceleration g or that he is in a gravitational field the gravitational acceleration of which is g. There is no way, other than opening the box and looking around, that he could distinguish between the two possibilities. The equivalence principle is a keystone of the general theory of relativity.

QUESTION:
I'm not a scientist just a observer, tinkerer. I'm looking for the equation to figure out the amount of air volume and velocity of that given volume to generate 1lb of force. Google is being obscure about it. And again I'm probably just dumb. However I know someone who is better than me with math could give me the equation then I could use that to further my own work quicker. Also just to be clear we are talking like regular air outside. If you want the region for density calcs and stuff the American south east coast so high humidity air so maybe slightly denser than other areas though I'd prefer those calculations be left out.

ANSWER:
I think you want pressure, not force. A slight breeze will exert way more than 1 lb on a large sailboat sail but only an ounce or two on a pea. So, two things matter, the force F of the air on an object with area A. As you note, the force depends on the density of the air and other properties of the air. But, you need to appreciate that any calculation of air drag is going to be an approximation. There is a very handy little formula which makes it easy to estimate this for air near sea level pressure: P=F/A=¼v² where v is the speed of the air. The only catch is that this works only if you used SI units, F in N (newtons), A in m² (square meters), and v in m/s (meters per second). But, this is really no big deal, you just need to be able to go back and forth between SI and English system. There is an excellent little program, named Convert, which allows you to do this and it is easy to use. I will walk you through an example: Suppose you want to exert a force of 1 lb on an area of 1 ft²; convert lb to N and ft² to m²:

So you want a pressure of 4.45 N/0.0929 m²=47.9 N/m². Now, our formula says that this is 47.9=¼v². Solving, v=√(4x47.9)=13.8 m/s. Now, maybe you want this in miles per hour, so convert:

So, you need a wind velocity of 30.9 mph. As you can see, volume of air has nothing to do with it, just the area on which the wind is blowing.

QUESTION:
There is an often seen problem of finding the angle at which a particle sliding from the top of a smooth sphere leaves the sphere. However, I am trying to calculate the time it takes the particle to reach the point of departure. When I try to so the integration, I get an infinite time... . How do I resolve the problem?

ANSWER:
I will assume that you are familiar with the classic problem of finding the angle θ at which a point of mass m , starting from rest at θ=0 and sliding frictionlessy down the surface of a sphere or cylinder of radius R leaves the surface. The answer is

∫dt=√(R/(2g))∫[dθ/√(1-cosθ)]
=2√(R/(2g))sin(θ/2)[ln(sin(θ/4))-ln(cos(θ/4))]/√[1 - cos(θ)]+constant

The integral is shown below for R=1 m, g=9.8 m/s². (I did not do the integral off the top of my head! I used the wonderful resource Wolfram Alpha.)

Now, integrating from t=0 to t and from θ=0 to 0.841, one gets t=∞, just as you found. But, this is not unexpected because when θ=0, the mass is in equilibrium and will remain at rest until you nudge it. If you calculate the definite integral from some θ to 0.841, you get the other curve shown (red). This is instructive but probably not what you want because each point θ on the curve is the time it takes before exiting if it starts with the velocity it would have had if it fell from the top; for example, for θ=0.4, t=0.23 s is the time before leaving the surface if the mass started there with speed v=√[2Rg(1-cosθ)]=√[2x1x9.8(1-cos0.4)]=1.2 m/s. An example which more closely would be of interest to you would be the smallest angle I plot, θ=0.005=0.29⁰, where the time to launch is about 1.6 s.

To get a little more quantitative, we could do a small angle approximation, θ«1. Then, 1-cosθ≈θ²/2, so ∫dt≈√(R/g)∫[dθ/θ]=√(R/(g))lnθ, and the definite integral if you start at θ₀is t≈[√(R/g)]ln(0.841/θ₀). So if you start at θ=0.005, t≈1.64 s. It is quite remarkable how well the small angle approximation works as shown by the green curve in the picture.

This, I believe, answers the question. It might be interesting to calculate times and angles of launch as a function of starting angles θ>0 and the particle initially at rest; I will add this later if I get around to working it out.

ADDED NOTE:
I have calculated times for starting angles 0⁰<θ₀≤90⁰. The object starts at rest at θ₀ and leaves the surface at an angle θ_xafter a time t. The main difference between this and the earlier is that the starting potential energy is mgRcosθ_x; this changes the velocity to v=√[2Rg(cosθ₀-cosθ)] and θ_x=cos^-1(2cosθ₀/3). Everything else proceeds the same as above except the indefinite integral now has the form

Where F is an elliptic integral of the first kind. Plots of t and θ_x are plotted in the graph for 0⁰<θ₀≤90⁰. Also, I have added the t graph above (black curve) so it can be compared to the calculations done earlier in this question.

QUESTION:
The Energy principle or the law of conserving energy says that energy cannot be created or destroyed, just transformed from one form to another. So if I lift a 1kg weight 10 km I add some energy to it, then I release it and this will be transformed first into kinetic energy, then into heat as it hits the ground. Ok, fair enough. But suppose we have some rock that is drifting through space. Then the earth comes by and catches it in the gravititional field. It falls towards earth, slowly at first, then accelerating, gaining kinetic energy, which is then tranformed to heat maybe equalling many Hiroshimas when it hits. But where does this energy come from according to the Energyprinciple ? It is generated by the gravity field, the rock is accelerated to high speeds and energy is added to it, not converted from something else !? How does the law of conserving energy hold up in this case ? Where does the energy come from ?

ANSWER:
It's really no different from your dropping a rock. In your first example, you gave energy to the weight you lifted. But, if your space rock is some some distance from the earth, it still has potential energy even though you did not give that energy to it. That potential energy is what gets transformed into kinetic energy at the earth's surface.

QUESTION:
When a light beam originates from a laser or any other luminous object then it must first undergo acceleration though for a very short time but still it must or how can it go from rest to the speed of light without accelerating?

ANSWER:
Light does not start out at rest, it has its speed at the instant of creation.

QUESTION:
I am thinking that gravity is just a myth. Isn't gravity a direct contradiction of the first law of motion? Where is the force which causes objects to fall spontaneously? And how can gravity cause two tides, each tending in an opposite direction?

ANSWER:
I am not sure where you are coming from. If you say gravity violates the first law, you must be saying that gravity is not a force. And, if that is the case, you must know a little about general relativity (GR) which is our best theory of gravity. And you must therefore know that GR is most often presented as mass deforming space time, implying that gravity is geometry, not a force. However, two recent answers (1 and 2) addressed this issue. The takeaway from those is that, although one possible interpretation of GR is the deformed spacetime, GR is a field theory and therefore it can also be interpreted as there being forces associated with the fields. Regarding your final problem, tides, the reason is quite simple and is called the "tidal force"; the gravitational force from the moon gets smaller as distance increases. Therefore the force on the side closest to the moon is bigger than the force on the center which is bigger than the force on the far side. All three forces are toward the moon but, relative to the center of the earth, the near-side and far-side forces both point away from the center of the earth.

QUESTION:
I have a question about the speed of light. I understand nothing can travel that fast except light itself. I also understand that the closer you get to the speed of light the slower time goes for the person who is traveling that fast to the point where if you are actually at the speed of light time virtually stops. Does that mean that if you were able to travel faster than the speed of light, time would start to reverse and you would basically time travel? And is that the reason physicists theorize you can't travel back in time because you can't travel faster than that?

ANSWER:
Your first understanding is correct, your second is not. The time for the person traveling fast relative to you does not go slowly as observed by her—it is you who observes that her clock is slower than yours. You cannot go as fast as light because it would take an infinite amount of energy to get you there and there is not an infinite amount of energy in the universe.

QUESTION:
Ok a person I know and I are arguing over this haha. If 2 vehicles are traveling down a highway at 50 mph and one vehicle(sedan) hits the other vehicle (Snap on box truck) in a pit Maneuver position from 10 yards away (Vehicle swerved into the box truck moving forward hitting it were the rear wheels are) is it possible for the snap on truck to roll backwards on top of the car... rolling in the opposite direction of the way the momentum its going. Box trucks are approximately 3500 to 12000 pounds. I said for a car to hit a truck wile there both moving at 50 mph in the same direction and get it to roll backwards on top of the car it would have to be moving at a speed that's not possible to reach in that or probably any car manufactured. Can you tell me 1. How fast you have to be moving in a sedan to hit a Snap on box truck going 50mph to get it to roll on top of you ( in the opposite direction of momentum) 2.is it possible for the sedan to hit the truck from 10 yards away in a pit maneuver position and get it to roll backwards, towards you, on top of you. 3. If the sedan hits a Snap On box truck(brings tools and parts to advance auto) in a pit maneuver position right where the rear wheels are located, what scenario would it have to be for this to happen. I said only way is if the sedan pits the truck and kicks off to the right the person is gonna try to counter steer back to the left skidding out of control and the truck now caught up and rolled on top of it. The other person says the driver side was down, laying on the sedan but it pit the truck on the left (driver side) also.. I say it's not even possible to get 4000 pounds traveling at 50mph to roll the opposite way the momentum is. It had to of pit the truck from the driver side and kicked off out across the highway and came back to the left where the truck then hit it in a sliding position and rolled on it. I don't see how it could happen any other way.

ANSWER:
I had no idea what a pit maneuver was, but now I do. I am afraid that I do not find your description very clear but I do expect that it is not impossible for the truck to roll. If the car were hitting the truck from the right and the truck driver steered hard left to try to avoid the hit and the center of gravity of the truck was high, it could roll. However, trying to calculate the possibility of this happening is very difficult and depends on many details of the situation. You might be interested in an earlier answer which investigates rolling of a bus.

QUESTION:
I notice when I stir my hot chocolate in a cup and then tap on the side with the spoon, that the pitch of the sound that comes from my spoon hitting the side of the cup seems to go up, and then stabilize to just one pitch if you will when the liquid has stopped spinning. I was wondering if there is a physics term for this, And what that is.

ANSWER:
Think of your chocolate cup as a bell. A bell, being made of an elastic material*, will ring with a particular pitch; that is because, given its geometry and composition, there are only specific frequencies of vibration, known as overtones, which persist and do not quickly damp out. [If you actually analyzed the sound you would find many of the overtones present, but your brain causes you to deduce that the pitch you hear is the lowest which is called the fundamental.] Now, take an empty cup and strike it, then fill it and strike it; what do you hear? The movie above (tapping a cup while pouring in water) which I made shows that as the water gets deeper the pitch gets lower. I was not able to replicate your observation. When you get the cocoa spinning it tends to push out toward the inner surface of the cup because of the centrifugal force, so as it stops the level of the cocoa at the cup goes down and at the center of the fluid comes up. You noted that the pitch gets higher which is in accord with the observation that a full cup rings with a lower pitch than a less full cup; so if the level to which the water comes up on the surface of the cup is the important thing, we can at least qualitatively understand what is going on.

*By elastic I mean that if you deform it a small amount it will tend to go back to its original shape.

QUESTION:
Is the energy of a gravitational field really considered to be negative by the scientific community? If so, how big is the negative energy relative to the source? I mean: it is logical and i can find this on the internet but what is the value of that information? I also can find that the total energy of the gravitational field is equal but opposite to the source energy (E=mc^2) effectively saying that the total sum of energy in the universe is zero. Is this correct?

ANSWER:
I think that E=mc² has nothing to do with your question. It is not clear what you mean by the "energy of a gravitational field". It is shown in a NASA publication that the energy density of a point mass M at a distance r is u_G=g²/(8πG) where g=MG/r² and G is the universal gravitational constant. This quantity is definitely positive and so would be the total energy of this field in any volume in space. Maybe you are thinking of the gravitational potential energy of a point mass, U=-MG/r+MG/r₀ where r₀ is the position where we choose the potential energy to be zero. If we choose r₀=∞, U=-MG/r<0 everywhere. This is convenient because the total energy of a particle is E=U+K and if E<0, the particle is bound (orbit) and if E>0, it is unbound (scattering).

QUESTION:
My 9 year old son and I are working on his science fair project. He wanted to try to combine sledding and physics. His hypothesis is that adding weight, by combining tubes/ridders, would increase the speed. Our local sledding hill uses man made snow and only allows supplied tubes that are 40 inches in diameter. Our control was one rider/tube. By finding the mean time on runs, we concluded that 2 tubes/ridders was the fastest. Each additional ridder after 2 resulted in slower speeds. It seemed almost proportional that each additional ridder after 2 resulted in slower and slower times. I'm a 5th grade math teacher and could use some help explaining the science behind this.

ANSWER:
Your findings are not in accordance with simple physics. I will give you a quick discussion of how a block on an incline will accelerate if released at the top; your son will not be able to understand but you, being a math teacher, should and could convey the essense of it to him. The frictional force f is proportional to the force which presses the two together and that force is proportional to the weight W times the cosine of the incline angle θ; the proportionality constant μ is called the coefficient of kinetic friction and depends on the nature of the surfaces sliding on each other, snow and tube in your case. So we have f=-μWcosθ. (The minus sign means f acts up the hill.) But there is also the force of gravity down the hill which is F=Wsinθ, so the net force on the block will be F+f=W(sinθ-μcosθ). Now, this force (F+f) causes an acceleration which is proportional to the force divided by the weight; the proportionality in this case is 32 if you are measuring weight in pounds. So we have a=32(F+f)/W=32(sinθ-μcosθ). Since the weight cancels out, the motion down the hill should be identical no matter how much weight you have. For example if μ=0.1 and θ=30⁰, a=32(0.5-0.1x0.87)=13 ft/s², about 40% of gravitational acceleration (32 ft/s²), regardless of the weight.

I believe that your data do not show the expected behavior because snow is difficult to assign a coefficient of friction to. Snow can be wet, dry, loose, compacted, etc.; for example, μ for waxed wood (ski) on wet snow is about 0.1 but on dry snow about 0.4, a huge difference. Furthermore adding more weight is very likely to change the frictional force by causing more packing or by "plowing" up a little pile of snow in the front. In order to successfully make these measurements, you would have to take extraordinary pains to keep the conditions of the snow identical for the many runs. Overall, snow is not going to provide a really scientific study of friction.

Here is an important lesson to be learned from all this. You know Galileo's experiment demonstrating that heavy objects, when dropped, fall at the same rate as light objects (assuming that air drag is not important). The reason for this is just the same as the reason that the objects of different weights sliding down an incline should move down at the same rates. So you can demonstrate to him the overriding principle by dropping a baseball and a bowling ball from the same height simultaneously.

QUESTION:
When sunlight passes through a prism you could say it gets seperated by color frequencies. But what happens to the particle aspect of light..? Thank you...

ANSWER:
Photons get separated by their energies. The energy of a photon is proportional to the frequency of the corresponding light wave.

QUESTION:
Hey I'm really interested in time travel theory. Assuming to travel through time one needs to travel faster than the speed of light could one hitch a ride on an object that flies faster than the speed of light naturally like a comet and thereby travel through time? Has that ever been explored? Are there books or papers on that subject? Are there scientists that explore that idea?

ANSWER:
I am sorry, but almost everything you say in your question is wrong. First, nothing can travel faster than the speed of light. Comets travel at speeds far below the speed of light. According to the standard laws of physics, time travel is permitted into the future but not the past. This is achieved by traveling at speeds comparable to the speed of light; to get a feeling for how you can travel to the future, see my earlier discussion of the twin paradox. There has been some speculation about time travel by warping space-time to open "wormholes", but that is indeed speculative.

QUESTION:
A vulcan bomber with an approximate loaded with fuel weighs 70,000 pounds has 70,000 pounds of thrust per sq inch in engines combined. Does it rely on the lift of the wings and momentum to accelerate and go up?

ANSWER:
Whenever the thrust is larger than the weight, the craft can fly without wings. When you say the engines have 70,000 lb of thrust per square inch, if you want the thrust you need to know the area of the nozzles of the engines because lb/in² is pressure, not force. If they are, for example, 100 square inches the thrust would be 7,000,000 lb.

However, your numbers seem far off compared to data I found on Wikepedia. The Vulcan B.1 had a maximum takeoff weight of 170,000 lb and a thrust from its four engines of 4x11,000=44,000 lb. Hence thrust/weight=0.25. This plane cannot fly without wings! By the way, the wings do not provide acceleration, they provide lift.

QUESTION:
Is it true that in order to break through the Coulomb Barrier for Nuclear Fusion, would you have to exceed temperatures of 120 Million kelvin to do so?

ANSWER:
That is the right order of magnitude. It is interesting to note that this is about 8 times hotter than the core of the sun where most of the fusion takes place. One reason for this difference is that because of the huge gravitational force the core of the sun is compressed to a very large pressure so the nuclei are closer together to start with. The other reason is that in a fusion reactor, possible pressures require the fusion to between deuterium and tritium nuclei whereas the sun is deuterium-deuterium fusion; if we could achieve higher pressures we could feasibly do deuterium-deuterium fusion in a reactor at a lower temperature.

QUESTION:
I was wondering about something. I imagine an object floating on electromagnetic rails. That object could be an iron petanque ball. Pushing it along the rails demands somes effort. Then, I want to push another object on this rail. It is an object made of 1000 petanque balls that are welded together. It is much harder to accelerate that object along the levitating rails. Then I was wondering about the fact that, on the one hand, objects have difficulty to be accelerated but, on the other hand, they make other objects accelerate towards them, like the earth. Could it not be then that the difficulty to be accelerated for the 1 iron ball, the 1000 iron balls welded together and any other object, is due to the accelerating zone towards themselves of their own making? So any object would create a zone around it with an acceleration towards themselve. It affects not only objects at a distance, like the accelerating zone created by earth that makes objects fall towards it. But it also affects the very object provoquing that accelerating zone around it by making it difficult to accelerate that object.

ANSWER:
The fact that your objects are being levitated by the rails is not really relevant to your question. What it means, I would say, is that we can ignore any friction in this problem. Your first statement, that "[p]ushing it along the rails demands somes effort" is not true; Newton's first law says that an object which experiences no forces will move with constant velocity. The second statement, that "It is much harder to accelerate that object [1000 balls] along the levitating rails" is perfectly true but no mystery: that is just Newton's second law which states that the acceleration of an object is proportional to the force applied and inversely proportional to its mass. It has nothing whatever to do with the gravitational interactions among the balls.

QUESTION:
I am trying to understand Einsteins Theory of Relativity. The part I can not quite wrap my head around is that time moves slower as you move faster through space, while everything else continues as normal. So if this is true, then that makes me wonder, what two craft had a race from Earth to Pluto or something. One craft moves at 10,000 m/s, the other moves at 99% the speed of light. The time on near light speed craft slows drastically, while the other is negligibly changed. So then would the slower craft arrive first, because its not being held back by severe time dilation?

ANSWER:
For purposes of calculation I will use 5.9x10¹² m as an average earth-Pluto distance. On both ships time moves perfectly normally. But on the fast ship the distance to pluto is drastically reduced by length contraction, L'=L√(1-(v/c)²)=5.9x10¹²√(1-(0.99)²)=8.3x10¹¹ m. So the fast ship will take, by his clock, t_fast=8.3x10¹¹/(0.99x3x10⁸)=2739 s=46 min. For the slow ship, which experiences almost no length contraction, the time according to his ship's clock is t_slow=5.9x10¹²/10⁴=5.9x10⁸ s=19 yr.

QUESTION:
an open bucket filled with water is placed on the ground in direct sunlight . As water begins to evaporate, what effect, if any , will this have on the pressure the bucket is exerting on the ground

ANSWER:
The pressure on the ground is the weight of the bucket plus water divided by the area of the bucket's contact with the ground. As the water evaporates there is less water and the water in the bucket will weigh less; pressure gets smaller.

QUESTION:
Can the elements relative to our solar system truly be used to explain the universe? Or is it all speculation? Is it possible for galaxies to have a periodic table made up of elements we have yet to discover? Or maybe its even narrower than that, refereing to neighboring solar sytems within our own galaxy.

ANSWER:
As I clearly state on the site, I do not do astronomy/astrophysics/cosmology. In your case I would have to argue that no speculation is involved. No evidence has ever been found that the laws of physics are not constant all over the universe. Hence, the universe should be homogeneous regarding what it is made of. The way astronomers study the properties of the universe relies heavily on spectroscopy, the measurement of the wavelengths of light coming from distant stars. The spectrum of any given atom is unique to that particular atom; identical spectra to those on earth are found in distant stars.

QUESTION:
So, I understand that the moon travels in a straight line but that spacetime is bended due to the presence of the earth so the moon travels around the eart. If this is correct, what would happen if I was on the moon and directed a really powerfull laser beam in the same direction as the straight line motion of the moon? Will the light also go in a straight line in a bended spacetime and travel around the earth like the moon?

ANSWER:
Suppose that the moon were given a different speed, would it still follow the same nearly circular orbit? No, the orbit it moves in is determined by its speed. So light, having a much larger speed would feel the gravitational fields from the earth and moon and follow a "straight line" path but it would be very little curved compared compared to that of the moon. There is certainly no doubt that gravity can bend light, but a photon can never orbit the earth because its velocity is greater than the escape velocity for any altitude. However, for a big enough mass a photon can orbit; if you are a distance R=3MG/2 from a black hole of mass M, photons will move in a circular orbit. If you were this distance from a black hole and looked in a tangential direction you would see the back of your head! Elliptical orbits are not possible because the speed of a photon has a constant speed.

QUESTION:
What are the meanings of and are there any practical applications to higher-order derivatives of motion such as metres per second cubed and to the power of 4,5,6 7...?

ANSWER:
The third time derivative of position, j=d³r/dt³, is called jerk. It is also the time derivative of acceleration, j=da/dt, and aptly named because if acceleration is not smooth and constant, like when someone is learning how to drive a standard transmission car, motion is jerky. The time derivative of jerk, s=dj/dt, is called snap. After snap, the 5th and 6th derivatives are called crackle and pop, a tribute to Rice Crispies, I guess! If you want to read a whole article about this question, check out this link.

QUESTION:
what purpose does c² serve in E=mc² other than to make E a ridiculously large quantity of really small units instead of E=m where E is a 'normal' quantity of ridiculously large units?

ANSWER:
Purpose? An equation in physics cannot be changed randomly. Any particular term does not have a "purpose", the equation as a whole is the description of a fact. If an object has a mass m and you could figure out how, the amount of energy you could extract from that mass is, in fact, very huge. Most processes in nature do not result in mass just disappearing and some energy appearing, rather some small fraction of mass disappears and energy appears. That is how the sun works: when you create helium from hydrogen you end up with about 1% less mass than you started with. One example of complete annihilation mass is when an electron meets its antiparticle, the positron, they both disappear and electrictromagnetic radiation with energy precisely equal to the sum of their masses times c² appears. Finally, I should note that E=m is an impossible equation because mass and energy do not have the same units; for example, the equation 1 meter=4 kilograms make no sense.

FOLLUP QUESTION:
I think you missed my point. I was not randomly changing an equation in physics. We made up a unit of measure (E) which is some number (m - mass which is universal regardless of place) times some other ridiculously huge number (c² - where the speed of light, c, can be expressed in any number of different earth units - miles per second, kilometers per second, parsecs per century, feet per hour etc. etc.). But what it boils down to is that the energy in some quantity of matter is directly related to it's mass, whatever quantity of units you want to use to measure it.

ANSWER:
The most used system of units, the SI system (meter, kilogram, second, …) are based on historical decisions which fitted the magnitudes of things we normally encounter in science. A second was approximately the the time of human heartbeats, the meter was on the order of our own sizes, the kilogram was about the mass of something you might pick up at the supermarket, etc. But, in the late 19th and early 20th centuries when scientists started examining things going very fast, things being very small, etc., it was inevitable that very large or very small numbers would arise in our theories and calculations. In many cases, the mathematics emerging from theories could be much more concisely written by introducing different systems of units. One which you would like, I think, is the planck system where many fundamental constants are chosen to be equal to 1: the speed of light, c=1; rationalized Planck constant, ℏ=1; permittivity of free space, ε₀=1; permiability of free space, μ₀=1; etc. So, you see, E=m is the famous equation in natural units. If you want to look into this more deeply, see the Wikepedia article on natural units.

QUESTION:
How many miles of devastation would be caused by a 50 football field high tsunami. We live about 75 miles away from the ocean. They think it would affect our town, but I think this is wrong. I don’t think the amount of energy created could affect us from that far away.

ANSWER:
I don't know where you got the "50 football field" number, but it is unlikely to be close to correct. Quick research showed that the Chicxulub impact which nearly wiped out all life on earth had waves up to about 1500 meters which is like 15 football fields. These waves spread out and quickly shrank in size so that by the time they hit land they were only tens of meters high. But that is pretty big still and very destructive. The water from that impact went inland hundreds of miles, so you would doubtless be affected quite seriously, depending on your altitude and the terrain between you and the coast. But, that would only be the start of your trouble. Because of debris thrown into the atmosphere most sunlight would be blocked from the surface for years; most vegetation would die disrupting all food chains. Recent evidence has been found that such a large impact would cause tsunamis all over the world, not just near the point of impact.

QUESTION:
I am a retiree worked in the Railways as a locomotive electrician I am always interested to know why what when and how of technical issues and, problems. One problem always intrigue me why does space shuttle have to make such dangerous high speed reentry into earth atmosphere is always fraught with danger, why can’t reentry be done at below subsonic speed.

ANSWER:
The speed of the shuttle in orbit is about 17,000 mph. To reduce the speed to, say, 1000 mph would require an enormous amount of fuel. Furthermore, as you reduced your speed the shuttle would drop down into the atmosphere long before you were entirely slowed down anyway. The presence of an atmosphere is a cheap way to slow down without expenditure of much energy.

QUESTION:
When a hollow cylinder filled with a liquid is rolled down an inclined slope, does the liquid rotate in the opposite direction to the direction of rotation of the cylinder? Also what is the relationship between viscosity of the liquid and the final velocity of the cylinder at the end of the slope?

ANSWER:
In the diagram I have shown only three of the forces in the problem: (Note that there is a drum-fluid force pair at every point of contact between the drum and fluid.)

Now, to understand how this system behaves for various situations, I will look at a couple of extreme situations quantitatively. I will assume that the mass of the fluid is M and the mass of the drum is m and that the radius of each is R. Suppose that the fluid has an extremely high viscosity, i.e. it is essentially a solid.

If the viscosity is 0, there is no shear force which can get the any of the fluid inside its outer surface rotating, so the result will be the same as for a solid with no friction with the drum.

Finally let's allow values intermediate between 0 and ∞ for the viscosity. In these cases the fluid will have different angular velocities depending how far from the axis it is. If you wait long enough, have a long enough incline, eventually the whole fluid will be rotating with the same angular velocity as the drum, just like the solid case. But, for a given length incline, the speed at the bottom will decrease from v₀ to v_∞ as the viscosity increases from zero.

QUESTION:
I know you say "no homework questions," but this is a case where I believe a textbook answer is wrong. Your answer will not affect my grade. The question: A heavy frog and a light frog jump straight up into the air. They push off in such a way that they both have the same kinetic energy just as they leave the ground. Air resistance is negligible. Which frog will go higher? I said that the lighter frog would go higher as it is moving at a greater speed. The textbook said that both frogs would reach the same height as they had the same kinetic energy. Could you please help me?

ANSWER:
An object which has a kinetic energy K with a vertical velocity at y=0 in a uniform gravitaitional field will rise (neglecting air drag) to a height h=K/(mg). You are right and your text is wrong.

QUESTION:
I'm writing a work of fiction and was wondering the physics regarding explosives like dynamite on mars for the purpose of excavation. Would explosives like that even work without oxygen?

ANSWER:
Most explosives, dynamite, nitroglycerin, plastic explosives, ammonium nitrate fertilizer, etc., contain the oxygen they need to explode.

QUESTION:
Would the laws of physics allow people to build a hose or tubular structure large enough to get all the way out into the vacuum of space to then use the same vacuum energy to remove water from areas with rising sea levels?

ANSWER:
The highest that a vacuum can lift water at atmospheric pressure is about three quarters of a meter (760 mm).

QUESTION:
Some co-workers and I were having a discussion about special relativity and we got caught up on a question we were hoping you could help answer. If you (a traveler) were 10 light years away from a destination and you could spontaneously begin moving at the speed of light towards that destination, from the traveler's frame of reference how much time would pass according to their clock before they reached that destination? Also, how much time would a stationary perspective at your initial position experience passing before they observed you reach the destination? For simplicity we are assuming there is no relative velocity between the initial position and the destination.

ANSWER:
Check out my site ground rules. "I no longer answer questions which are based on premises like: "...if we could travel faster than the speed of light..." or "...ignoring the fact that nothing can go faster than the speed of light...", "...if I were traveling at the speed of light..., etc." The laws of physics forbid anything with mass from traveling with the speed of light. However, I think I can address your question if 99.99% of c is the speed of your traveler. The traveler sees the distance L=10 ly length-contracted to the distance L'=10√(1-.9999²)=0.0141 ly so the time of travel would be 0.00141 yr=12.4 hr. The stationary observer would see the time elapsed to be 10x0.9999=9.999 yr.

QUESTION:
Einstein's General Theory of Relativity explains gravity as a warping of spacetime by matter or energy. All the attempts to picture this warping show the "rubber sheet" distorted by the earth or an apple on your book. It appears as though the distortion occurs on one plane (membrane), but does it?

QUESTION:
Does light have or hold a shape? For example if a laser is emitted from a square shaped opening does the light beam have a square shape or revert to cylindrical or

other three dimensional shape?

ANSWER:
No beam of light holds its shape because of Huygen's principle—every point on a wave front acts like a point source of the waves, so those points on the edges of the beam emit light with components transverse to the direction of the beam. Laser light through a square aperature is shown in the figure. The central spot is saturated, much brighter than the fringes.

QUESTION:
Hey, I was doing some thinking and wondered;why does a plastic bag loaded with groceries burst when given a sudden jerk upwards?

ANSWER:
Suppose that the bag can hold a total load of 20 lb. Then you hold the bag while a friend loads it up with 20 lb of groceries, and the bag will not break. How much upward force are you exerting to hold up the bag full of groceries? Obviously 20 lb. But suppose that you now want to lift the bag to put it up on a table. To do that you have to exert an upward force greater than 20 lb and so the bag will experience a force greater than its design load and break. Even if you had only 10 lb in the bag, if you pulled with a force greater than 20 lb the bag would break. A "sudden jerk" implies that you are pulling upward with a large force.

QUESTION:
How do I calculate velocity of an object when I know its mass and the height from which it fell

ANSWER:
If air drag is negligible, an example of which would be a bowling ball dropped from up to several tens of meters high, it is easy to calculate the velocity v: v=√(2gh) where g=9.8 m/s² is the acceleration due to gravity and h is the height. Note that the speed does not depend on the mass. For example, drop a bowling ball from h=10 m; v=√(2x9.8x10)=14 m/s=46 ft/s=31 mph. However, if air drag is important the object will speed up at first, but the faster it goes the greater the drag so eventually it falls with constant speed called the terminal velocity, v_t=√(mg/D) where m is the mass and D is a constant which depends on the geometry of the object, the density of the air, etc. The drag force due to air drag is proportional to the square of the velocity of the object and D is the proportionality constant, F_drag=Dv². For example, for a crumpled up piece of paper I estimate that the terminal velocity would be about v_t=1.6 m/s=3.6 mph and, if dropped from 10 m, it would quickly accelerate to this speed and fall most of the rest of the way down with constant velocity.

QUESTION:
if we found the atomic structure/components of a single-cell organism, or bacteria/virus, would placing the components in an atom smasher create life in the form of those organisms? Would it create life at all?

ANSWER:
Certainly not. An accelerator ("atom smasher") generates very energetic particles, (electrons or protons or many other possible ions) which, when they encounter your components (atoms and molecules) would either scatter off without changing anything or would actually break electron bonds and destroy the molecules; so you would not cause the components to assemble themselves into cells but would rather cause a chaotic mess.

QUESTION:
As we know, E=mc^2 and E=hc/lambda So hc/lambda=mc^2 or. h/lambda × c = m If mass of photon is 0, then after calculating, what should we get as the value of 'lambda'? Don't you think value of 'lambda' should be undefined as lambda =h/mc (according to the above equation) and 'm' is 0. Please explain?

ANSWER:
E is not mc² for a massless particle. The energy of a photon for light with frequency f is E=hf=hc/λ as you correctly state in your question.

QUESTION:
To apply some force we need to spend some energy, like to move a object distance s with a force f we must spend f*s energy. Now, every object has gravitation force. Like our earth, It constantly pulling object towards itself with it's gravitational force, does this mean earth is constantly loosing energy, and if yes where this energy is coming from ?

ANSWER:
It is incorrect to say that exerting a force takes energy. Only if the force does work on an object does the energy of the object change. For example, when you drop a ball its weight (the force which the earth exerts on it) does work on it and is converted into kinetic energy and the ball is moving when it hits the ground. A book sits on a table and the earth is exerting a force on it; but no energy is being used even though the force is acting.

QUESTION:
If there are two nuclear objects with the same mass but they both emit nuclear energy at different times... Why?? Like one after 60 min and the other after 120 min.

ANSWER:
I presume you are comparing two identical (not just mass, but also atomic number) nuclei which are radioactive. The reason is that the "lifetime" of a radioactive nucleus is not how you should look at it because decay time is meaningful only in a statistical sense, not applicable to single nuclei. That is why we talk about "half life" which quantifies the time which it takes for half of a large population of emitters to decay. For example, if some radioactive sample has a half life of 60 minutes, half will have decayed in 60 minutes and 3/4 of the entire population will be gone in 120 minutes. But you cannot predict the time it takes for a single nucleus to decay.

QUESTION:
Can you tell me which interatomic force keeps the magnetic moments of many atoms parallel to each other in a feromagnet?

ANSWER:
It is the exchange interaction. The summary explanation from Wikepedia is "In simple terms, the electrons, which repel one another, can move 'further apart' by aligning their spins, so the spins of these electrons tend to line up. This difference in energy is called the exchange energy." Note that, if the outer electrons in adjacent atoms can get farther apart, they reduce their energy and will therefore tend to align their spins. To get the details of the exchange interaction in ferromagnetism, see the Wikepedia article.

QUESTION:
When you throw a baseball with a right-handed spin, the ball breaks to the left. I know it has resistance from the threads on the ball. So if you shoot a bullet out of a rifle at a long distance, even though the bullet is smooth, and the bullet is spinning to the right, shouldn't the bullet move somewhat to the left?

ANSWER:
The ball in the figure is spinning as shown. The ball will drag a layer of air around in the direction it is spinning. If the ball is at rest, anywhere you look on the ball the air will be moving with the same speed. Now, suppose the ball is now moving to the left as I have drawn it; that is the same as if there were a wind blowing to the right. So the air passing over the top of the ball will be going more slowly than that passing over the bottom; so by Bernouli's principle the pressure is lower at the bottom and the ball will experience a downward force. Now suppose the picture represents instead a cylinder which you are observing, looking at one end of the cylinder. This would be representative of your spinning bullet. But now the bullet is moving directly away from you and from the bullet's point of view there is a wind blowing toward you. Now every point on the surface of the cylinder sees exactly the same wind, one component tangent to the surface and perpendicular to the velocity and the other moving opposite the direction of the velocity; therefore there is no force due to the air (except air drag which just slows it down).

Now, you might ask, "what about the baseball which can be made to break right or left?" The baseball is not thrown so that its spin axis is parallel to the velocity direction like the bullet spins, so it can have lateral forces on it.

QUESTION:
How do you calculate the torque on a 40mm gear that is being turned by a 200mm length of rack? The rack is moving at 200mm per minute and it is being pushed by a hydraulic piston which exerts 4 tom force. I could submit a sketch if I am not clear enough.

ANSWER:
The speed is irrelevant in calculating the torque, except that it is constant. The torque is just Fr=(4000 kg)(9.8 m/s²)(0.04 m)=1568 N·m=1157 ft·lb. (Note that I have converted the mass 4000 kg to force in newtons using F=mg.) I have assumed that the force is in metric tons.

QUESTION:
Hi, My wife and I have a sheet of LED Christmas lights up across the window. We both see them as advertised, bluish white lights. However we see transient blue and yellow ones that are distinctly so, almost like illuminated antibiotic capsules. However we cannot make them appear, and as soon as we notice them they go white. It seems like the bicolour images appear fleetingly when not thinking or concentrating on the lights and disappear as soon as we do perceive them as bicoloured. I tried photographing them but cant reproduce the effect there either. I wondered if it was a quantum phenomenon that collapses on cognitive or photographic measuring but I'm sure this must be too bog a set up for that unless maybe there is a quantum effect in the brain. As doctor I cannot explain it with the knowledge I have.

ANSWER:
Here is the best I can do: A simple light emiting diode comes with a particular wavelength (color) but you usually want white light. The most common way this is achieved is to use the output from a blue or near-ultraviolet LED to excite a phosphor of Ce:YAG (Y₃Al₅O₁₂:Ce) which absorbs some of the light from the LED and then emits yellow light. I would guess that, since both yellow and blue are present, the design of the device could be such that at certain orientations one would see only blue or yellow. As the bulbs in your sheet move slightly, you might be looking at just the right orientation that you separately see the blue and yellow. You might try taking a video to see if you can catch the phenomenon. I am pretty sure that your "quantum effect in the brain" is not what is happening!

QUESTION:
when ch4 is combusted it reacts with oxygen to form 1x co2 plus 2xh2o have you now created these molecules ? So do we create water from the combustion process that did not exist prior ?

ANSWER:
Your question is impossible to answer. Burning methane certainly does have water as a byproduct, but you cannot say that it "did not exist prior". The whole earth is a huge chemical factory where plants and animals are continually turning water into other molecules, methane among them. So who can say where the carbon and hydrogen in a particular methane molecule came from? Most likely from water molecules.

QUESTION:
If we have equal amounts of hot and cold tap water, which will freeze faster? And why? Please give a valid reason for the underlying effect and it's implications.

QUESTION:
I am a semi retired engineer who has begun his quest to peruse one of my favorite subjects from decades ago. As I understand it, electric and magnetic fields are "energetic". That implies that a particle (presumably a photon) carries this energy throughout the field. Q: what is the frequency of this photon (or am I completely missing the boat).

ANSWER:
The fact that electric and magnetic fields contain energy does not imply "…that a particle … carries this energy…" Classical electromagnetism predicts an energy density in electromagnetic fields. This is most easily deduced in electromagnetic (EM) waves. If sunlight shines on your face, your face warms. If a radio waves fall on an antenna, electrons are set in motion by the waves. Now, quantum electrodynamics (QED) quantizes the EM field and the quanta are photons. If you have an EM wave of frequency f, the energies of the photons which comprise it are E=hf where h is Planck's conatant.

QUESTION:
What elements existed in the universe when the first stars ignited after the big bang?

ANSWER:
Big bang theory tells us that the elements present right after the big bang were hydrogen, helium, and small amounts of lithium and berylium.

QUESTION:
A spaceship is 240,000 miles from the earth. At t = 0, the ship has an initial velocity of 0 and starts to accelerate toward the earth due to earths gravity. The ship is in free fall with respect to the earth. How long will it take for the ship to reach earth? What is the velocity of the ship as a function of time as the ship approaches the earth? What is velocity of the ship as a function of distance from the earth? The problem I am having is that the gravitational acceleration is not 9.8 meters per second squared. The grav_acc is constantly changing as the ship approaches the earth. All the equations I have found assume that the grav_acc is constant. My interest in this problem stems from the apollo command module returning to the earth in free fall.

ANSWER#1:
This is a pretty mathematical question, so I will answer it in pieces as I get the time to work it out. To find the time it takes to fall, you start with Newton's second law, ma=m[d²r/dt²]=-GMm/r²=dv/dt and integrate it once to get v=dr/dt=√[2GM(R-r)/(Rr)] where, for your question, R is the distance from the center of the earth to the point from which it was dropped, r is where it is at the time t(r), G is the universal gravitation constant, and M is the mass of the earth. (The mass m of the ship does not matter.) (You can more easily get v(r) from energy conservation.) Next integrate v(r)=dr/dt to get t(r)=-2.69x10⁵{0.5tan^-1((2x-1)/(2√[x(1-x)]-√[x(1-x)]-π/4} where x=r/R; this is not a simple integral to perform, but you can find good integrators on the web, for example Wolfram Alpha. Once you have gotten the indefinite integral you are not finished, you need to find the integration constant which will put the ship at rest at x=r/R=1; that is where the π/4 comes from. Putting in your numbers, I find t=4.22x10⁵ s=117 hr=4.89 days. The figure shows the graph of t(x).

QUESTION:
Ok so let's just go with a big bird (stay with me).. imagine it was inside an airtight container and because it’s big it's powerful enough to lift this container. Would it be able to fly through air normally? as the containers not open to outside air surely flight isn't possible.. like an aircraft in a similar situation not being able to draw in air to generate thrust... Providing the bird has enough space to flap its wings too Thanks for reading my dumb thoughts

ANSWER:
This is mostly Newton's third law (N3) stuff. The bird has a weight W_bird, the box W_box, and the air W_air; the total weight I will denote as W.

The bottom line here is that any forces which happen inside a system, forces between individual pieces of a system, all cancel out because of N3 and can be ignored. This is a case where the general case is easier to understand than a specific case. If a system consists of a large number of objects, the force between any pair of objects will add to zero because of N3, so only external forces need be considered; in the case being discussed above, the only forces external to the system are its weight and the force the ground exerts.

QUESTION:
What would the atmospheric pressure be if, the oceans levels were to rise say, by 3000 meters? What would the physiological effects be on living humans, animals, plants above that sea level? What would average weather, temperature be? [ You know, Mount Ararat, Noah's Ark. Could they survive at all? Where would all that water come from? ]

ANSWER:
You would be increasing the radius of the earth by 0.05% so the atmospheric pressure at the new sea level would be essentially unchanged. There would be no physiological effects on life except most of the land would no longer be land, only regions currently over 3000 m above sea level would remain. The oceans play a major role in the weather (think of hurricanes, el nino, el nina), but I am not an earth scientist so I wouldn't speculate on weather changes. I do not know where that water would come from but not from current ice because if all ice melted the oceans would rise only about 70 m.

QUESTION:
Hello, I am Kayra, and I am a 9th grade student. I am studying about the speed of light but I cannot find the answers to my questions. I understand snell's law and n=c/v but the problem is, is density or the size of the molecules that determines the speed of light in different mediums. Like for example, the density of ethanol is less than water, but light moves faster in water than in ethanol. I really am wondering what determines the speed of light in these solutions. If you can help I would be really grateful.

ANSWER:
As you probably know if you have been studying light, it is composed of oscillating electric and magnetic fields. As the light passes through a medium these fields exert forces on the atoms in the medium. Mostly it is the negatively charged electrons in the molecules which respond because they have so much less mass than the positively charged nuclei. For example, the electric field alternately pushes and pulls on the electric charge distributions which respond by oscillating with the same frequency as the light itself. Now, we know that if you have an oscillating charge distribution, it will radiate light with the same frequency which will be identical to the incident light so now we have both the light that came in and the light from the oscillating molecules which both travel, in the medium, with the speed of light in a vacuum c. But when you add the two together you are left, astonishingly, with light of the same frequency but which travels at a speed slower than c. So you see, neither the density nor size of the molecules is relevant; the speed you end up with is simply determined by how the particular molecule responds to electric and magnetic fields. I think any more quantitative explanation is probably beyond the mathematical capabilities of a ninth grader, but if you want to explore this further you can look at Wikepedia articles on refractive index and the Ewald-Oseen extinction theorem.

QUESTION:
As I was driving this week I thought of this question: what would happened if I grabbed a person, who is standing still, while I am moving at 50 mph in a car. Would the person instantly move at 50 mph? The person would not accelerate due to the car moving already at 50 mph.

ANSWER:
Most likely you would break your arm and the person would experience a force as well but since your arm would bend away, the force on the person would modest and maybe break a few ribs. Suppose we think about a really strong steel arm sticking out the side of the car. If it hits in a way that he "sticks" to it he would be accelerated up to 50 mph very quickly. During the collision time his chest would be crushed maybe by a few inches, probably doing enough interior damage to kill him.

QUESTION:
Hypothetically, could gamma rays be used as a method of sending and receiving information from one person to another telepathically. I know gamma rays are harmful to people. But can a gamma ray be altered on a subatomic level to prevent the bad stuff from the gamma ray harming the person but still allow the info to be received or sent?

ANSWER:
One gamma ray cannot carry very much information, not even a single word or letter. You would need a whole bunch of them, the idea presumably being to modulate them somehow like radio waves. Then you are faced with having enough to injure. And, no, you cannot "alter" them to remove the "bad stuff" because the bad stuff is the ability of gamma rays to damage cells. You would also be faced with the problem of how they would be generated in my brain and detected in yours.

QUESTION:
A log is floating downstream. How would you calculate the drag force acting on it?

QUESTION:
I've actually had this question since I saw a video telling me that a bowling ball and a feather will fall at the same speed, in a vacuum room. But let's say, I released some pure H2O steam into the chamber. Will it also fall at the same speed as that of a bowling ball and a feather?

ANSWER:
If you haven't already noticed, I have recently answered a question about this video. It is not immediately relevant to your question, but you might be interested. Regarding your question, note that the feather and bowling ball are at rest at the beginning of the experiment. However, if you were able to look microscopically, all the water molecules are in motion when you release moving with a whole range of speeds and directions. However, if you were able to follow the center of mass of all those molecules, it would fall right alongside the other two fallers.