A bullet is fired from a gun held parallel to the ground. At the exact same moment, a second bullet is dropped from the same height as the gun. Which bullet hits the ground first? The answer to this classic conundrum is not at all intuitive. It depends on a fundamental fact of physics: vertical and horizontal motion are completely independent of one another. While we won’t be firing any guns in this investigation, we will use some everyday objects to demonstrate a basic law of motion that applies to everything from a home run baseball to a rocket launch.
How do the vertical and horizontal motions of a projectile affect one another? Which hits the ground first: a bullet fired horizontally from a gun or a bullet dropped from the same height as the gun?
Skateboard
Coins
Skateboard You probably noticed that the ball went up and came right back down to the person on the skateboard, despite the fact that she was moving the whole time. Coins By listening to the coins, you will hear that they hit the floor at exactly the same time, even though one was shot out while the other dropped straight down. Horizontal and vertical motion are completely independent of one another. If you stand in one spot and toss a ball straight up, it will come back to you. That’s probably not a surprise. But the ball also falls right back into your hands even if you are moving sideways the whole time. The vertical (up and down) motion of the ball knows absolutely nothing about the horizontal (sideways) motion. When you throw the ball up while riding the skateboard, the ball keeps going sideways. Because the ball retains its original horizontal motion after it leaves your hands, it ends up in your hands again as if neither you nor the ball had ever moved sideways in the first place. The coin demonstration shows you the same thing, only a bit more dramatically. Both coins fall from the same height and begin falling at the same time. However, one is fired horizontally off the table while the other drops straight down. Intuitively, you might think that the launched coin should take longer to hit the ground because it has a greater distance to travel, but that’s not the case: they both hit the ground at the same time. Remember: the coin’s vertical motion knows nothing about its horizontal motion, and gravity doesn’t care either way. When figuring out how long it takes for either coin to hit the ground, all that matters is the height from which you’re dropping them. Now you know enough to answer the original question: which bullet hits the ground first? The answer is they both hit the ground at the same time! Disclaimer and Safety Precautions Education.com provides the Science Fair Project Ideas for informational purposes only. Education.com does not make any guarantee or representation regarding the Science Fair Project Ideas and is not responsible or liable for any loss or damage, directly or indirectly, caused by your use of such information. By accessing the Science Fair Project Ideas, you waive and renounce any claims against Education.com that arise thereof. In addition, your access to Education.com's website and Science Fair Project Ideas is covered by Education.com's Privacy Policy and site Terms of Use, which include limitations on Education.com's liability. Warning is hereby given that not all Project Ideas are appropriate for all individuals or in all circumstances. Implementation of any Science Project Idea should be undertaken only in appropriate settings and with appropriate parental or other supervision. Reading and following the safety precautions of all materials used in a project is the sole responsibility of each individual. For further information, consult your state's handbook of Science Safety. QUESTION: ANSWER: QUESTION: ANSWER:
I am not sure I would say that the earth orbits the sun is because it has angular momentum. True, it has angular momentum, but the reason is that the sun exerts a force on it. Technically the straight line path is an orbit, but it has no angular momentum. QUESTION: ANSWER:
Remember, in science you can only argue intelligently if the subject of the argument is well-defined! QUESTION: ANSWER:
I teach Jr. High science. I have a question that I need help with when explaining it to some of my SPED students. If you were playing football on the moon, would it be as hard to stop a 150 kg lineman as it would be on the Earth? ANSWER: (The following is probably too abstract for your students, but I include it for completeness. The force which stops the advancing runner is actually the friction between your feet and the ground. For example, if you were standing on very slippery ice it would be much harder to stop him. The friction between you and the ground is proportional to your weight, so you have less friction on the moon resulting in less ability to stop the runner.)
Is there a general formula for calculating distances and height for motorcycle jumps considering speed, weights etc? Can you please point me in the right direction . I would like to build a motorcross track on my property but looking for a general rule of thumb before I start building piles of sand to find I haven't spaced them correctly. ANSWER: QUESTION: ANSWER:
When you put a candle on a spinning turntable, why does the flame point inwards? ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER:
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A motorcycle hit a car (the car was not moving) and the bike rider (who weighed about 170#) flew about 30' beyond the impact before landing on the ground. About how fast would the bike have to be going for him to fly that distance? ANSWER: QUESTION: ANSWER:
Please settle a debate in my family: would it be possible for a human being to execute a one-armed iron cross on the still rings (gymnastics)? ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: SIMILAR QUESTION: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: Also, can you ask my grandma for the apple dumpling recipe? Her name's Alma, and been dead 15 years now, so I really need that recipe. ANSWER: Your grandmother told me that all recipes are forgotten in the afterlife and she regrets that she cannot help you. In fact, she doesn't really remember you either. Let that be a lesson that you should keep good records of your recipes before you pass to the other side. QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: NOTE ADDED:
ANSWER (CORRECTED, 11/11/12):
The results are shown in the following table:
My earlier answer had an error, namely the time it takes to reach v=1 mm/s is not t≈m but rather t≈1000m. Now the answer to the question seems to clearly be that the more massive object goes farthest. Plotting the three cases for different masses below (choosing a t1 large enough that all masses have the terminal velocity), it is clear that at long enough times the largest mass will always be the farthest ahead at any given time. This conclusion is independent of the values of c or vt.
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I am afraid that this question doesn't make much sense to me. QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: FOLLOWUP QUESTION: ANSWER:
If light in earths atmosphere goes a little slower than light in a vacum, then how come when you turn on a light or laser pointer you don't hear a sonic boom? ANSWER: QUESTION: ANSWER: NOTE ADDED: QUESTION: ANSWER: QUESTION: ANSWER: NOTE ADDED:
Why the energy corresponding to the most probable speed is not the most probable energy? ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER:
This lady lost her legal battle to have the Large Hadron Collider shut down. She believed that the atom smasher could create a black hole and suck up the earth. The courts sided with the scientists. The scientists said it's not possible, and even if it did create a black hole, it would be a micro black hole and collapse in on itself. That seems to go against the laws of physics and quantum physics. We now know that matter sucked up by a black hole is permenantly imprinted on the black holes surface and the size of the black hole increases. Did I miss something? ANSWER: QUESTION: ANSWER: QUESTION: ANSWER:
So, putting this in, we get F(t)=v0cosh(t). Now, the hyperbolic cosine function (cosh) at t=0 is 1, at t=0.1 is 1.005, and at t=0.01 is 1.0001. This is what I meant when I said things would get simpler. For all intents and purposes the force is a constant because the cosh does not vary significantly for the time the ball is in the tube. So, finally, F≈v0. The force you have to exert while a single fast ball is in the tube is 200 N; the force you have to exert while a single slow ball is in the tube is 20 N. But, that is not the whole story because there may be more than one ball in the tube. The distance between fast balls is (200 m/s)/(60 s-1)=3.3 m, so there is only one ball at a time in the tube. But, the distance between slow balls is (20 m/s)/(60 s-1)=1/3 m so there are 6 balls in the tube at a time. The final answer: you need a force of 200 N for the fast ball gun, 120 N for the slow ball gun, the fast ball gun is harder to rotate. NOTE ADDED: QUESTION: ANSWER: QUESTION: ANSWER:
I am a writer putting together a science fiction screenplay. Those who know me say I have an attention to detail--to a fault. There is one particular element I would like to be as accurate as possible. I'm hoping you might be able to help me. Here is the scenario: A spacecraft leaves earth on course to the moon. In order to create an Earth-like gravity inside the ship; the ship accelerates at a constant rate exerting a force on the occupants equal to one G. Half way through the trip the craft will flip, then decelerate for the remainder of the journey. This would give the same sensation of false gravity to the occupants of the craft. So here is the question: If this were possible; how long would it take to actually reach the moon?
Since you are such a stickler for detail, I will give you detail, probably far more than you want! Your scheme of having an acceleration with "constant rate" would work in empty space but not between the earth and the moon because the force causing the acceleration is not the only force on you, the earth's and moon's gravity are also acting. As you go away from the earth the earths gravity gets smaller like 1/r2 where r is the distance from the earth's center, and the moon's gets bigger as you get closer. So, it becomes a complicated problem as to how much force must be applied to keep the acceleration just right for where you are. The picture to the right shows you in your rocket ship. Let's call your mass M. Then there are two forces on you, your weight W down and the force the scale you are standing on exerts on you, F. W gets smaller as you get farther and farther away and you want F to always be what your weight would be on the earth's surface, Mg. So, Newton's second law says that F-W=Ma=Mg-W where a is the acceleration you must have. Note that, for the time being, I am ignoring the moon; that would just complicate things and its force is much smaller than the earth's, at least for the first half of the trip. I want you to understand the complication caused by the fact that W changes as you go farther away. Now, how does W change? W=MMEG/r2=Mg(R/r)2 Where R is the radius of the earth and ME is the mass of the earth. We can now solve for the acceleration the spacecraft would have to have: a=g(1-(R/r)2). I have plotted this in red on the graph above. (The distance to the moon is about 60 earth radii.) Note that for most of the trip the acceleration is just about g. I also calculated the effect the moon would have, blue dashed line, and, except for the very end of the trip, it is pretty negligible. Now that we have taken care of the always-important details, we can try to answer your question. To calculate the time exactly would be very complicated, but, since the required acceleration is g for almost the whole trip, it looks like we can get a real good approximation by just assuming a=g for the first half and a=-g for the second half; your perceived weight (F) will just decrease from twice its usual value when you take off to about normal when you get to about 5 earth radii in altitude. The symmetry of the situation is such that I need only calculate the time for the first half of the trip and double it. The appropriate equation to use is r=r0+v0t+�at2 where r0=R is where you start and v0=0 is the speed you start with. Halfway to the moon is about r=30R=R+�at2 and so, putting in the numbers, I find t≈1.69 hours and so the time to the moon would be about 3.4 hours. You can also calculate the maximum speed you would have to be about 140,000 mph halfway. QUESTION: ANSWER: QUESTION: ANSWER:
My co-workers and I have been in a very heated discussion. They tell me I am wrong and my theory is stupid. Will a bullet fired (perfectly horizontal) and a bullet dropped hit the ground at the same time? My answer is yes, because gravity is constant. ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER:
Next, do the same problem but now let the mass of the rope be 1 kg also, clearly not even approximately massless.
Finally, let's do your tug-o-war problem. Part of your confusion is saying that each man exerts a force of 300 N. It is much better to simply say that it is a tie right now so the men must be exerting equal forces in magnitude.
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So, the tension in the rope is still the same everywhere (you could do the above exercise of choosing one man plus half the rope to show that it is also 350 N in the middle) and, for this problem, it turns out to be just the average of the pulls at each end. QUESTION: ANSWER:
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Is it the orbital velocity of an object that enters the earth's atmosphere that causes that object to typically burn up and disentigrate? Could you, in theory, freefall ANSWER: RELATED QUESTION: ANSWER: won't be able to breathe" means. If he did not have his "space suit" he would certainly not be able to breathe until he got down to an altitude around 5 miles up (just not enough air). QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER:
Why when a cyclist is turning round a bend, he tends to lean inwards with the bike? ANSWER: FOLLOWUP QUESTION: Can it be explained without invoking the fictitious force because it may be too technical? Can I say the bike has a tendency to move in a tangential direction to the bend, this causes the bike to topple outward, so leaning inward will prevent the bike from toppling? ANSWER: Στ=0=NLsin θ-FfLcosθ=mgLsin θ-(mv2L/r)cosθ. And so, solving for θ, we get the same answer as above, θ=tan-1(v2/rg). QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER:
What about the structure of alpha particles makes them such a common form of radiation? Why aren't, for instance, lithium nuclei the common form of ionizing radiation? ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER:
My friend and I had a drunken argument. I would like independent council to weigh in (there's $300 on the line) I was given a unique bottle opener by a friend who is a brewer for craft brewery in the northeast. It is a flat piece of wood with a smooth screw embedded in one end ( think ______T_ ) Its measurements in 1/16ths of inches are as follows: 88 long. screw centered 11 from one end. side of screw to end is 9 short side. 75 long side screw lip is 4 from bar bottle cap is 4 high 17 across. The argument is as follows. Person A: There is less force required to open the bottle pressing down with the cap positioned _________XXXT___ (between the screw and the user) Person B: There is less force required to open the bottle pulling up with the cap postioned __________TXXX . With my horrible description of the problem. Can you prove either arguement successfully? ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: ΣMiRi/ΣMi where the symbolΣ means to sum over all the objects. For example, the term ΣMi is the mass of the whole solar system which I will call M. Notice, for example, that if the mass of the sun is much greater than all the other masses, the position of the center of mass is approximately the position of the sun. Now I am going to find the velocity and acceleration of the center of mass by differentiating with respect to time once and twice respectively: Mvcom=M(dRcom/dt)= ΣMi (dRi/dt)= ΣMivi and Macom=M(dvcom/dt)= ΣMi (dvi/dt)= ΣMiai where vi is the velocity of the ith object and ai is its acceleration. Now, we can apply Newton's second law to the acceleration equation: Macom= ΣMiai=ΣFi where Fi is the net force felt by the ith object which is simply Σj≠iFij where Fij is the force on the ith particle due to the presence of the jth particle and this sum runs over j but not the term j=i because an object does not exert a force on itself. So, Macom= ΣΣ j≠iFij. But now look at the double sum: for every term Fij there is also a term in the sum Fji and Fij=-F ji because of Newton's third law (the force of the ith on the jth is equal and opposite of the force of the jth on the ith). So, finally we have acom=0, the center of mass moves with constant velocity and if we happen to have chosen a coordinate system at rest with the center of mass, the center of mass never moves. In other words, the whole solar system, no matter how complicated its motion, orbits around the center of mass. If the total solar system experiences a force F from the rest of the universe, the only difference is that the center of mass now has an acceleration acom=F/M; but you can still say that the solar system orbits around its center of mass, it is just that the center of mass is now "orbiting" around some other point. Maybe that is more detail than you really wanted, but it is a standard derivation in any introductory physics course. The bottom line: any system of interacting objects orbits around its collective center of mass. QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: FOLLOWUP QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: Ψ". The wave function of the system may or may not be an eigenfunction of the operator associated with the observable you are interested in. For some observable z whose associated operator is Z, then Zφn=znφn where zn is the nth eigenvalue and φn is the eigenfunction. Only unless Ψ=φn would you say it is a wave function of the observable. The expectation value of the observable z is defined as <z>≡<Ψ|Z|Ψ>=∫Ψ*ZΨdτ where Ψ* is the complex conjugate of Ψ and the integral is over all space. QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER:
Overall, you win the debate with your brother because, although only under ideal conditions does the man land where he launched, gravity has nothing to do with it. Your bullet on the train idea is an entirely different thing since the bullet which just drops was not fired vertically. QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: �proton-antiproton, electron-electron, electron-positron. What drives the decision of what to make collide depends on lots of considerations. Protons are good because it is pretty easy to get a very intense beam going both directions greatly enhancing your chances of finding what you are looking for. QUESTION: ANSWER: QUESTION: ANSWER:
why doesn't your head hurt when raindrops fall on your head despite having an accelation of 9.8 m/s? ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER:
consider a water wheel with mass of 600kg, rotating @ 18 RPM, wheel diameter is 3 meters and the diameter of its axle is 0.1 meter. the wheel is fed with 10 liters of water per second from the top, causing it to rotate and eventually releasing the water at the bottom (3 meters lower). Neglecting efficiency losses due to friction, air drag etc..., the available energy from the water should be MGH: 10 x 9.8 x 3 = 294 joules. However, considering the wheel's diameter, its mass and its RPM, the wheel's kinetic energy (1/2 x moment of inertia x angular velocity squared) is 1200 joules. Where did the extra joules of energy come from? Secondly, the diameter of the axle is only 0.1 meter so this hypothetical wheel has a mechanical advantage of 30 (MA= radius wheel / radius axle). Does this mean the initial energy of water is magnified 30 times at the axle? If not so, then what has been magnified 30 times? ANSWER: FOLLOWUP QUESTION: ANSWER: QUESTION TO QUESTIONER: FOLLOWUP QUESTION: ANSWER:
What is the reason behind right hand rule? I mean the definite current and magnetic field direction relation? Like if current is going the thumb direction magnetic field is directed by curved fingers..why not the opposite direction? ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER:
are the EMF waves mere Oscillations in field values with no movement at all , or , real undulations in the field proper as a ripple moving in space ? ANSWER:
I want to get a tatoo (don't laugh) that expresses Quantum Theory/Quantum Mechanics. Is there a formula for this? Originally I wanted to get the collapsible wave theory, but i've read a whole lot more about this recently and it sounds like a lot of scientists are questioning it and there may be other explanations for this phenomenon. I don't want to get a tatoo of something that may be proven wrong in 10 years if you follow me. So I was thinking I would get a tatoo of simply a formula that expresses Quantum Mechanics. Is there such a thing? ANSWER: Funny, I recently had another question from somebody wanting a physics tattoo! In your case, I can make two suggestions. I would go with the Heisenberg Uncertainty principle which states that you can not know exactly both the position and the momentum (essentially velocity) of a particle; this is a real conceptual keystone of quantum mechanics. In this equation (actually an inequality), Δx is the uncertainty in position, Δp is the uncertainty in momentum (mass times velocity) and ħ is the rationalized Planck's constant (a really tiny number). A second choice, not quite so compact or easy to understand, is Schr�dinger's equation. Without going into too much detail, this is the equation which may be used to determine the wave function Ψ. QUESTION:: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: y1=v01t-�gt2 and v1=v01-gt and y2=v02t-�gt2 and v2=v02-gt. Suppose #1 goes to a height h1 before coming back. The time t to reach h1 is found by knowing that v1=0 at that time, so t=v01/g. Put that time into the equation for y1 and find h1=�v012/g. Now, you tell me that you want #2 to make the full trip up and down in the same time; that means that it will make the trip up in half that time, t=�v01/g. If you now put this time into the equation for y2 (which we would call h2 at this time), you will find that h2=⅛v012/g=�h1. Of practical interest to the juggler is that you must give #1 twice the initial velocity of #2 to achieve the desired heights. QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER:
QUESTION: ANSWER: Parallax.
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Suppose we have Bob the astronaut sitting without a space suit in a spaceship full of air on a mission to Mars. Bob is very fond of balloons and is holding on to a nice, big, red helium balloon via a piece of string. Bob is sitting facing the front of the spaceship. Mission Control decides to slowly accelerate the spaceship. In which direction will the balloon move relative to Bob? Why? ANSWER: "ENHANCED" ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: I'm an eighth grade science teacher who needs a simple explanation and I am pooped trying to locate it on the Internet. I understand the basic idea of an electromagnet. I also understand the basic methods to change the strength of the electromagnet: change the amount of current, number of wire loops, etc. The problem is that one source states that the diameter of the iron core also has an impact on strength. A second source states that the DISTANCE from the core to the wire has an impact. Question 1: Is it only the diameter of the core? Not simply length? If I use a LONGER iron nail with the same diameter as a shorter nail, will that impact the strength of the electromagnet? Why? Question 2: We just take our students to the lab and wrap wire around a nail and hook it to a battery to make an electromagnet. What does distance from the core to the wire do? ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER:
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the weight of your opponent W; the normal force N of the mat up on your opponent; and the frictional force f, the maximum value of which is fmax=μN. For any given situation, Fsinθ-f=0 and N-W-Fcosθ=0. Therefore, the maximum frictional force for this θ, μ, and F is fmax=μ(W+Fcosθ) and so Fmax=μW/(sinθ-μcosθ). That is the answer to your question, it depends on θ, μ, W, and F but not on the radius of the cylinder.
I guess next you have to worry about the mechanics of how you are going to exert this force. Your feet cannot slide on the mat and your weight will always be vertical. QUESTION: ANSWER: FOLLOWUP QUESTION: ANSWER: QUESTION: ANSWER: FOLLOWUP QUESTION: So this what I had; does this make sense from physics standpoint? I read it as "in the absence of any external force, the acceleration is zero" which to means covers the meaning I want, which is "keep moving and you WILL keep moving, but sit still and you WILL keep sitting still". It is a reminder for me to stay active, adventurous and to keep moving.. ANSWER: QUESTION: ANSWER: QUESTION: I am trying to teach quantum physics for the first time. I am not sure what the link is between photoelectric effect and plancks constant or how to link it meaningfully for the students. ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION I am a photographer and general tinker-er. I have started building pinhole cameras and during the course of building them I came across some equations on obtaining the optimal pinhole size. I find I understand the equations well, but I can not find a solid explanation for the 2.44 constant used in the calculation for the Airy Disk (2.44 x light wave length x focal length). Everything I have read is very vague and I want an explanation. Why do we use this number? I read that it is like pie, it just is, but is to what, waves? Is it a speed? What does it represent? I want to understand. ANSWER: This verges on being too technical for this web site, but I always want to encourage folks to "have an enquiring mind", so I will take a stab at it. This question requires that you understand a little about diffraction, interference of light waves. If we shine light through a slit and the slit is very large compared to the wavelength of the light we are looking at, we get an "image" of the slit on a screen. If the rays of light hitting the screen come from very far away compared to the size of the slit, the image of the slit will be the same size as the slit itself. The sun coming in through a window creates an image of the window, the same size, on the wall. So, I now want to make images of smaller and smaller slits; what happens is that we come to a point where as we make the slit smaller, the image starts getting bigger. The figure above shows what the "image" of a very narrow slit looks like (below) and a graph of the intensity (above). It is fairly straightforward to show (see any elementary physics textbook) that asinθ=λ gives the angle of the first dark spot in the pattern for wavelength λ; the geometry showing a and θ are shown to the right. Now, a pinhole is not a slit, but it is very similar and you would expect its pattern to be very similar. Indeed, instead of getting stripes you get a bullseye as shown to the left. Doing the analysis similar to that which leads to the angle of the first minimum gives a slightly different result, asinθ=1.22λ where a is the diameter of the hole. In essence, this is where your factor of 2.44 comes from; where does it come from? It gets pretty technical here! When you solve the pinhole diffraction problem, you work in cylindrical coordinates because the problem has cylindrical symmetry. As is often the case with problems with this symmetry, the solution (for the intensity in this case) involves a Bessel function, a special mathematical function which whole books have been written about. The intensity is given by I=I0[J1(½kasinθ)/(½kasinθ)]2, where J1 is the first order Bessel function and k=2π/λ. Now, we are interested in when the intensity is first zero; the first zero of J1(x) is for x=3.832, so ½kasinθ=(πa/λ)sinθ=3.832, or asinθ=1.22λ. Next, we convert the angle to lengths by approximating (see diagram above) sinθ≈x/R so x=1.22Rλ/a or 2x=2.44Rλ/a; 2x is the diameter of the smallest spot to which a collimated beam of light can be focused. Since you say that you "understand the equations well" except where 2.44 comes from, I guess my task is complete. Apparently (from the little research I did) the optimal size is if the minimum spot size equals the hole size, i.e. 2x=aopt, so aopt=√(2.44Rλ). There is an alternative form of optimum size which is based on the Rayleigh criterion for resolving two spots (which stipulates that the central maximum of one image is on the first minimum of the second) which has aopt=√(3.66Rλ). (A similar problem is the structure of the compound eye of insects, elegantly discussed in Feynman's Lectures on Physics, Vol. 1.) QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: ANSWER: QUESTION: ANSWER: QUESTION: We say that the moon has the gravitational force one sixth in comparison to that of the earth. If the earth has held an atmosphere with its gravitational force which is 480 kilometers high, why can�t the moon even hold an atmosphere one sixth of the height of the atmosphere which the earth has, that is, 80 kilometers? ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: �because my calculations proved it". Calculations do not prove something, measurements do. QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: FOLLOWUP QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER:
The object starts at rest and so the distance it travels in time t is x1(t)=�a1t2 and its velocity is v1(t)=a1t. Now the object starts decelerating (physicists do not like that word!) beginning with a speed of v0=v1(t1)=a1t1 so the distance it travels in time t is x2(t)=a1t1t-�a2t2 and its velocity is v2(t)=a1t1-a2t (note the minus sign because it is slowing down). Now, v2(t2)=0=a1t1-a2t2 (it comes to rest at the end), so t2=a1t1/a2. The total distance traveled is s=x1(t1)+x2(t2)=�a1t12+a1t1t2-�a2t22=�a1t12+a12t12/a2-�a2(a1t1/a2)2=�a1t12(1+(a1/a2)). Final results, using your numbers:
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Because of the way quarks bind, there is no such thing as a free quark and therefore it is difficult to talk about its rest mass. To zeroth approximation, it takes infinite energy to unbind a quark so it would have infinite mass. QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER:
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Hi, I had a question about airships (zeppelins, blimps, etc.). We all know that they work by filling the balloon with a gas that is lighter than the surrounding air (hydrogen, helium, hot air, etc.). My question is, what would happen if you filled the balloon with nothing (i.e. a vacuum)? I understand that in blimps and balloons you need the air to give the balloon structure, but what about in the case of a zeppelin or other rigid-structure airship that has a solid frame that provides structure? Nothing, having no mass, is certainly lighter than air, so would that cause lift, just like with helium? ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: � 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the 133Cs atom �" (ref.: Wikepedia). The meter is defined in terms of the second and the speed of light. A" � metre is the length of the path travelled by light in vacuum during a time interval of 1 ⁄ 299,792,458 of a second �" (ref.: Wikepedia). You should not be bothered by this because the speed of light is a universal constant, no matter who measures it, the speed is always the same. QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER:
What makes particles, atoms, planets, galaxies and everything else spin? ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: ANSWER: QUESTION: ANSWER: QUESTION: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: FOLLOWUP QUESTION: ANSWER: QUESTION:
I have received many answers saying that P = 0 as a consequence of the definition of mechanical work. But this cannot be true since no gravity affects the satellite. Energy is needed to change a linear path in space and a circular orbit is one such motion. In absence of gravity no particle with mass will loop in a circle without any energy. ANSWER: QUESTION: ANSWER: ANSWER: QUESTION: Kind of a simple question, I think, but just curious. If I check in at an airport with my luggage, someone said if I lay my bag down it will weigh less than verticle. Obviously the thought may not get charged for over 50lbs or whatever if I lay it down. If you're married you probably understand where I'm coming from:) ANSWER: QUESTION: ANSWER: FOLLOWUP: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: FOLLOWUP QUESTION: ANSWER:
The other is the horizontal position which must end up at y=h where h is how far vertically the basket is above the launch point; h=v0tsinθ-�gt2 where g is the acceleration due to gravity (9.8 m/s2). So, you have two equations with three unknowns, v0, θ, and t. The first thing to do is reduce it to one equation with two unknowns by solving the R equation for t and substituting that into the h equation. The result is h=Rtanθ-�gR2/(v0cosθ)2. Now, think about it�you cannot mathematically know both θ and v0 because you have only one equation; but you shouldn't expect to from a physical point of view either. If you change v0 you will have to change θ too to hit the target. So, there are an infinite number of right answers here, not one unique answer. Nevertheless, if you program your robot to always shoot with the same velocity, you can solve for the angle. So, you need to solve the above equation for θ and that will tell you how to aim. It is not trivial algebra, but not real hard either and I do not intend to present the general solution because it is complicated. Be aware that there will be two solutions for every situation which you can easily understand, I think. Think of a rifle shooting at a target where you have to aim a little above the target to hit it; if you also shot way up in the air, there would be some large angle where it would come back down through the target. I will leave the general solution to you; I will do one simple special case to illustrate, R=4 m, h=2 m, v0=10 m/s, and I will approximate g as 10 m/s2: 2=4sinθ/cosθ-0.8/cos2θ. Rearranging and simplifying (and using sinθ=√(1-cos2θ)), 0.4+cos2θ=2cosθ√(1-cos2θ). Now square this equation and rearrange: 5cos4θ-3.2cos2θ+0.16=0. This is a quadratic equation in cos2θ. Thus, cos2θ=(0.585 or 0.055) and cosθ=(0.765 or 0.234) and θ=(40.10 or 76.50). Of course, you could also fix θ and solve for v0. This is easier. For example, if you fixed θ at 40.10, 2=4tan40.10-80/(v0cos40.10)2 which leads to, guess what, v0=10 m/s. However, you will run into trouble here because since you are playing basketball, the ball must be coming down when it gets to the basket, and that will not always be the case for a randomly chosen set of R, h, and θ. For example, in the example I did, 40.10 would not work because the ball would be going up when it got to the basket (vy=1.2 m/s); for the other solution, 76.50, the ball would be going down (vy=-7.4 m/s) like you need to make the basket. I got these by solving for the times from the x equation and then using the velocity equation (vy=v0sinθ-gt). You have your work cut out for you! QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: Again, this is a rough approximation because neither is really a simple pendulum as I have used to try to make it 11-year old friendly (you can easily make simple pendula and show him that the shorter gets to the bottom quicker). In real life, these are called physical pendula and you have to worry about how the mass is distributed in detail (moment of inertia if you have ever studied physics), not just where the center of mass is, but I think my explanation should give a good qualitative overview. So, the answer to your final question is that one bat does fall faster but not because it is lighter but because its mass is differently distributed. *You can easily find where the center of mass of each bat is by finding where each balances. QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: FOLLOWUP QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION:N: ANSWER: QUESTION: What would happen if ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: *It can actually be a little broader than that since, to change the momentum of a system, an external force must deliver an impulse which is, essentially, some external force times the time over which it acts [I do not know your level, but impulse is ∫F(t)dt]. QUESTION: ANSWER:
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Therefore, if it is not originally moving upwards, it can never reach a point either level with or higher than where it was launched from. QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION:
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My question is regarding an elevator in free fall. I had a debate with my uncle about what would happen to a person in a free falling elevator. I asked him if the person would float. My uncle said no. He pointed out that a ball would not float if put on the floor of the elevator and the cable was cut, so why would a person? To see what would happen to a ball in a free falling elevator, I put a ball on a folder (the folder representing the floor of an elevator), and I dropped the folder with the ball. The ball stayed on top of the folder as they both fell to the ground. This indicated that what my uncle said about a ball staying on the floor of a free falling elevator was likely correct. However, when I researched the subject on the internet, I came across many articles that said that a person would float in a free falling elevator. So I'm a bit confused. Why would a person float but a ball wouldn't? So my question is: 1) Would a person stand on the floor of a free falling elevator or would he float above the floor in a free falling elevator (i.e. weightlessness)? 2) If the person would stand on the floor of a free falling elevator, please explain why. If the person would float above the floor, please explain why. ANSWER: Now let's talk about the more realistic case of including air friction. The elevator would feel a drag force (like a parachute) which got bigger the faster it went and eventually would fall with a constant velocity called the terminal velocity. You, however, are at rest relative to the air you are in, so you would at first feel weightless but feel weightier and weightier and then, when terminal velocity was reached, feel normal, all the while standing on the floor. But, I don't think the second scenario is what you were thinking about because it is not what would be called free fall. QUESTION: ANSWER:
My musical instrument has been broken during an air travel, but its container, a hard case with thick foam padding inside and outside is absolutely intact! how is this possible ? ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER:
One final proviso: remember where I said "traveling with constant speed in a straight line"? Well, this is never rigorously true because the earth is rotating and therefore Newton's laws are only approximately true. The effects of this (centrifugal and coriolis forces) are so tiny that any ball throwing will be unmeasurably small. However, for very long-range problems they can be important. These "forces" are what are result in the circulation of weather systems. QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: Theoretically, if you were traveling through space in an object e.g., a spaceship and going at an incredible speed in one direction what would happen to you, inside, if it made an immediate 90 degree change in direction? wouldyou smash against the inside of the craft? ANSWER: ≈22,000 mph, just a little faster than the shuttle goes. Now, at that speed let's make a 900 turn around a curve of radius 1 km=103 m. Then the spaceship and all its contents experience a centripetal acceleration of ac=v2/R=(104)2/103=105 m/s2. This is ten thousand times the acceleration due to gravity which means that it would take a force of 10,000 times your weight to move you in this circle. I would hate to see you after this maneuver, certainly you would not be recognizable. QUESTION: ANSWER: QUESTION: Here is my question: Is "weight shift" a misnomer, or an ill concept of physics (vehicle dynamics to be specify)? The weight of an object is determined by the gravitational pull, regardless of what motion the object is doing, so how can the weight of different sections of the car change during maneuver? I think if one was to say force acting on different sections of the car during maneuver is different, then it is correct; but to say the weight shifted during maneuver, I can't get it. ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER:
I am using "Shockwatch" indicators to measure the impact of baseballs on a helmet. I am comparing the impact of a baseball with and without a helmet. These are impact indcators from 10G to 100G. How do I make the data meaningful converting the G's to _________. Would it be newtons of force? I don't know. I am a 7th grade life science teacher helping one of my students. ANSWER: Important conversions: 1 N (Newton)=0.225 lb (pound), 1 kg (kilogram) weighs 2.2 lb QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: RELATED QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: FOLLOWUP QUESTION: ANSWER: 6.6�10-34 m2kg/s.
Galaxies recede from us at a velocity proportional to their distance. Why then is Andromeda on a collision course with our beloved Milky Way? ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: Finally, a couple of real-world provisos. Of course, N is really distributed over the whole area, but the dynamics can be done by assuming it effectively acts all at one point just like we assume the weight acts all at the center of mass. And, the rolling friction might not really act at the surface of contact since it arises from the deformation of the wheel and it might not be purely horizontal since it is not directly a force due to the contact with the floor. So, there are still some idealizations in my analysis, but there are always idealizations when dealing with friction. And, the problem could have been equally well done assuming the ground, not the wheel was being deformed. One could also have done the analysis by summing torques about the center of mass and using the parallel axis theorem Icm=I-mL2. I would like to acknowledge a very useful discussion over pizza with friends and colleagues Edwards, Love, Meltzer, and Anderson. QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: SIMILAR QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: FOLLOWUP QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER: QUESTION: ANSWER:
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I have dealt with this kind of question before, but maybe it is time to revisit the whole thing with one answer. First, consider the ideal situation where there is no friction of any kind. Physicists do not like English units, so I am going to convert everything to SI units: 1000 ft=305 m, 425 lb=193 kg. I assume that you do not want all the details of my calculations, just the pertinent results. The time does not depend at all on what the mass is (if friction plays no role). There is an acceleration down the incline which is a=�g. I find that the time to the bottom is about 11.2 s and the speed at the bottom is about 54.9 m/s=123 mph. I assume you are not crazy enough to be in a soap box car going that speed, so friction must play a role. There are two kinds of friction you have to consider:
So, the bottom line is that if you go fast enough for air drag to be important (and I suspect you do), the heavier of two otherwise identical cars should win. QUESTION:If there is a yacht on a lake and it drops its anchor overboad. What happens to the water level in the lake. Apparently it falls. I have no idea why. Shouldn't it stay the same using common sense. ANSWER:This is a little complicated, so bear with me. First, you have to understand Archimedes' principle which states that if you have an object in a fluid, the fluid exerts an upward force on it and the magnitude of that force (called the buoyant force) is equal to the weight of the fluid which the object displaced. So, the boat plus anchor floating on the water have displaced a volume of water whose weight is the same as the weight of the boat plus anchor. Now, remove the anchor. Now the water is holding up only the weight of the boat, so the water will have fallen equivalent to removing water whose weight equals that of the anchor; but, the anchor is denser than water, so the volume of the water of equal weight is bigger than the volume of the anchor. Now, putting the anchor into the water, the water level will go up equivalent to adding a volume of water equal to the volume of the anchor. So the net effect has been a fall followed by a smaller rise, a net fall. QUESTION: I understand that atoms are mostly empty space and that what we understand as matter "touching" is not really touching at all but the magnetic force of electrons repelling each other. My question is: if all this is true, how do we get bitten by mosquitoes or ticks or anything for that matter? How can something penetrate us if this "touching" sensation is actually magnetic repulsion? ANSWER:You have it wrong. It is not a magnetic force but the repulsive electrostatic force between the electrons in "touching" atoms. But, the force is not infinite and if you push hard enough, particularly if what you are pushing has an edge or a point which can "pry apart atoms" at the surface, you can penetrate. QUESTION: What is the magnetic field of a permenant bar magnet made of? I have researched this, but cannot find a clear answer. Leading to the question...How does a bar magnets strength dissapate over time? How long will my fridge magnet stay attached to my fridge? Where does the energy come from and go to? ANSWER: When a ball is thrown up in the air, does it always form a parabola? Is there any possible way that the ball could go to the top and not fall back down in a symmetrical parabola shape? ( Air resistance/ wind or something could affect this I suppose, but what about in a vaccum?) ANSWER: If a light bulb doesn�t operate at a maximum efficiency, then is the law of conservation of energy correct? ANSWER: WHAT IS MATTER? I understand that matter is composed of atoms, and that atoms are made of subatomic particles. This is the answer that most physics textbooks and google searches yield. It seems like physicists are afraid to answer this question. To say that matter is made of wave-particles is no answer at all. If it's a wave then a wave of what? If a particle, a particle of what? PLEASE I MUST KNOW!! I have lost more than a few hours of sleep over this. ANSWER: How is it that my laptop is able charge as soon as I plug it in with out it shutting off? Does it work like a car where first it runs off the battery then a relay switches it to gas? ANSWER: I've been told that when a gas is cooled, it takes up less volume, and therefore its density increases. I've also been told that a gas always fills its volume (i.e. a container). These two ideas, however, seem to contradict one another. On the one hand, cooling a gas causes it to contract and decrease in volume, which would increase its density. However, on the other hand, a gas will always fill its volume, which means its volume doesn't change when cooled. Therefore, if the volume doesn't change when cooled since a gas will always fill its volume, then its density would not increase. In other words, if a gas will always fill its volume, then cooling it would not increase its density since the gas always fills its volume and therefore remains at the same volume. In a closed container, if a gas expands to fill the volume of the container, then how could cooling the gas increase its density? How could a gas fill its container/volume on the one hand, but then on the other hand, when its cooled, decrease in volume and therefore increase in density? If a gas fills its container/volume, it wouldn't increase in density when cooled since it FILLS its volume. Therefore, since it fills its volume, the volume of the gas would remain constant. Therefore, since the volume of the gas would remain constant, its density would not increase when cooled. So my question is, how could a gas increase in density when cooled if that same gas always fills the volume of the container it's in? ANSWER:That's a pretty long question which requires only a short answer. If the amount of gas remains constant, the relation between pressure, volume, and temperature is PV/T=constant. The situation you describe has a constraint that volume remains a constant, so P/T=constant which means that as the temperature goes down the pressure goes down; and, the volume stays the same. So, if you cool a gas in a rigid container, the density remains the same. QUESTION: I would like confirmation about the velocity of objects moving in a circle. If an object moves in a full circle at a constant speed, is the average velocity zero? One of my friends stated that it wouldn't be zero, and that I was confusing velocity with displacement. But since velocity has direction, I thought that if I were to divide the circle into, say, ten pieces, then the velocities of each independant piece of the circle would cancel each other out because the two segments of the circle that are opposite each other would have opposite signs because they are going in opposite directions. Can you help us settle this? ANSWER:The average velocity is a vector and is the displacement vector divided by the elapsed time. Since the displacement is zero, the average velocity is zero. The average speed is not zero but is the circumference of the circle divided by the elapsed time. QUESTION: Everything radiates heat. Nothing "radiates cold". The rate at which energy is radiated is determined by the temperature of the object. Your body radiates heat faster than the freezer does, so the freezer absorbs more of your radiation than you do its radiation and you feel colder because heat has left your body. QUESTION: Skydiver free falls from plane with a loaded gun, and shoots gun while free falling. Does the bullet being fired travel downward any quicker then say a bullet travelling just being dropped? Well, during a heated discussion with co-workers, I was trying to explain how I thought the bullet fired downwards would not go any faster than a bullet just dropped by someone skydiving ANSWER:OK, let's say that two skydivers are side by side each falling with the same speed. Because of air drag, both bullets will eventually both reach "terminal velocity" which will be the same for both (assuming enough time passes before the ground is reached). But the shot bullet will have a big head start over the dropped bullet. The shot bullet will begin with a speed above the terminal velocity and slow down until it reaches it. The dropped bullet will begin with a speed below the terminal velocity and speed up until it reaches it. Certainly, the shot bullet will reach the ground first. QUESTION: if i where riding a bike with a flashlight on the front end and i was going 10mph and knowing that the speed of light is constant wouldnt that mean that the light had to slow down 10 mph ? ANSWER: To see questions and answers from longer ago, link here. |