# Tag Info

45

I am sorry to say, but your colleague is right. Of course, air friction acts in the same way. However, the friction is, in good approximation, proportional to the square of the velocity, $F=kv^2$. At terminal velocity, this force balances gravity, $$m g = k v^2$$ And thus $$v=\sqrt{\frac{mg}{k}}$$ So, the terminal velocity of a ball 10 times as ...

44

Start by considering what is seen by the people watching you from the Earth. Nothing can travel faster than the speed of light, $c$, so the quickest you could get to Kepler 186f would be if you were travelling at $c$ in which case it would take 490 years. In practice it would take longer than this because you have to accelerate from rest when you leave the ...

31

like even when light gets on the moon why does the space appears dark from the moon? For the same reason it appears dark from the Earth (when flying at an altitude of 80,000 feet or so): Image credit: View from the SR-71 Blackbird. The fact is, we can't 'see space' from the Earth's surface during the day because the atmosphere is 'in the way'- the ...

28

This diagram shows the Earth rotating round the Sun at it's orbital velocity $v$. That is the centre of the Earth is orbiting around the Sun at velocity $v$. NB the scale is rather fanciful - don't take it literally! I'll also assume the orbit is circular, and for convenience I'll ignore the Earth's rotation i.e. assume it's tidally locked. To calculate ...

23

Olin Lanthrop suggested a plausible approach but there was a lot of (inaccurate) guessing in his answer. I was going to write this as a comment to his answer but it got too long. Note - in the below I round to no more than 2 significant figures - the nature of the problem doesn't support more. Let's take the famous Sherman tank as our example. A brief ...

22

The reason is because the time taken for the two trips are different, so the average speed is not simply $\frac{v_1 + v_2}{2}$ We should go back to the definition. The average speed is always (total length) ÷ (total time). In your case, the total time can be calculated as \begin{align} \text{time}_1 &= \frac{120 \mathrm{miles}}{40 \mathrm{mph}} ...

22

The best way to solve it would be experimentally, by doing the run several times, with calibrated instrumentation by the roadside to measure your speed. The acceleration won't have been constant, so that's not an assumption we can use. Knowing the 0-60 time capability won't really help; it could be different when accelerating up hill, compared to on the ...

20

Ball 1 will drop faster in air, but both balls will drop at the same speed in vacuum. In vacuum, there is only the gravitational force on each ball. That force is proportional to mass. The accelleration of a object due to a force is inversely proportional to its mass, so the mass cancels out. Each ball will accellerate the same, which is the ...

19

As you have probably noticed, the moon is tidally locked with the earth so that we always see the same side. You can look up in the sky and watch sunlight move across the moon's face. From the surface of the moon this change in illumination would look just like the day/night cycle on Earth ... except that it's roughly a month long. Until the advent of radar ...

16

I took a look at the clip and my take on it that the object of the exercise was not to slow down the tank, but to move it sideways and land in the lake a half a mile away. Hannibal says, "rotate the main gun to 82º" which I take to be sticking out sideways. Background Information The facts I could find were: M1A2 Abrams ...

16

This scientific problem – well, a more general one – has been solved in the following paper: http://arxiv.org/abs/1204.0162 Because it's legal in my country to move backwards in time, I remember the future event – one minute from now – in which Andrew Gibson will mention that he has this paper hanging in his physics lounge. He will curse me. 11 minutes ...

16

Your calculation is incorrect. $\text{Work} = \text{Force} \cdot \text{displacement} = F \cdot s$ The above product is a "dot" or "scalar" product, which means we only consider the displacement that occurs in the direction of the Force, which in the case of gravity is downwards. Can we set this vertical displacement to 0? No we cannot, and here is why: ...

15

Electric forces are created by and act on, both moving and stationary charges; while magnetic forces are created by and act on only moving charges.

15

If you write down the formulas for kinetic energy, $$E_k = \frac{1}{2} m v^2$$ and momentum $$p = mv$$ you see that you can write the energy in terms of momentum via $$E_k = \frac{p^2}{2m}$$ So, if two objects have the same energy $E_k$, they only have the same momentum if they also have the same mass. Since the bull has a much larger mass than the bullet, ...

15

The factor of two is correct as far as the integral goes; it comes from the unphysical situation of having your test mass exactly on the thin shell. Intuitively, you get the average of the "just outside" result (as if mass is concentrated at the centre) and the "just inside" result of zero. A more physical thing to do would be to `regulate' the calculation ...

14

Here is a mathematical derivation. We use the sign convention $(+,-,-,-)$ for the Minkowski metric $\eta_{\mu\nu}$. I) First recall the fact that $SL(2,\mathbb{C})$ is (the double cover of) the restricted Lorentz group $SO^+(1,3;\mathbb{R})$. This follows partly because: There is a bijective isometry from the Minkowski space ...

13

You basically just need to be careful about the distinction between velocity and speed. In particular, you say that Won't the particles change velocity when exposed to the magnetic field, and therefore change KE? A change in velocity is not necessarily accompanied by a change in speed, and it's the speed that determines the kinetic energy. The ...

13

If the cage is completely closed, it doesn't make a difference if the bird is hovering inside it or if it sits on the ground. When flying, the bird pushes air to the ground which will exert a downward force on the cage exactly equal to the weight of the bird. This is a direct consequence of the conservation of momentum and Newton's second & third law. ...

12

Deriving the relativistic equations for constant acceleration would be a formidable problem for most non-physicists. If you want to see how it's done then look at Gravitation by Misner, Thorne and Wheeler, chapter 6. For most of us the best option is just to look at John Baez's excellent article on the relativistic rocket. The relevant equation is: $$d = ... 12 A common mistake when students begin the study of the quantum harmonic oscillator is to try to convert everything to integrals. The thing is, in most curricula, the QHO is also used as a way to secretly acquaint you with bra-ket notation, and all the conveniences it offers. In reality, you shouldn't need any integrals at all here. \lvert n \rangle is a ... 12 Another way of solving such problems is to go to another reference frame, where you obviously don't have enough energy. For example you've got a 5 MeV photon, so you think that there is plenty of energy to make e^-e^+ pair. Now you make a boost along the direction of the photon momentum with v=0.99\,c and you get a 0.35 MeV photon. That is not ... 11 I agree with @Ron Maimon that these ETS questions are problematic. But this is (i think) the reasoning they go with. Unlike @Mike's assumption you should not take the normal average, but as stated in the question the weighted average. A weighted average assigns to each measurement x_i a weight w_i and the average is then$$\frac{\sum_iw_ix_i}{\sum_i ...

11

There is a nice reason for this, which Witten often explains. Imagine that your three dimensional space is the boundary of a four-dimensional space, for example, you can imagine that space is the surface z=0 of regular four dimensional space x,y,z,t. Further, you can imagine that space is closed into a sphere, which doesn't affect things except for some ...

11

It seems that the question (v1) is caused by the fact that there are two different notions of the commutator: One for group theory: $$\tag{1} [A,B] ~:=~ ABA^{-1}B^{-1}$$ (or sometimes $[A,B] := A^{-1}B^{-1}AB$, depending on convention), which is relatively seldom used in physics. One for rings/associative algebras: $$\tag{2} [A,B]:=AB-BA,$$ which is ...

11

In a nutshell, the problem with OP's choice of operators $\hat{p}_j$ and $\hat{H}$ is that they are not selfadjoint wrt. to the pertinent measure $\mu$. In other words, the usual integration by part method to prove selfadjointness does not work. Here are some more details. Let us put the constants $m=1=R$ for simplicity. Then the Lagrangian reads ...

11

Other answers & comments cover the difference in acceleration due to friction, which will be the largest effect, but don't forget that if you are in an atmosphere there will also be buoyancy to consider. The buoyancy provides an additional upward force on the balls that is equal to the weight of the displaced air. As it is the same force on each ball, ...

11

Do the math. I'm no military expert, so I'll guess at some parameters, but I think it will show the effect is so small that it doesn't matter even if the numbers were considerably more favorable. Let's say the tank weighs 50 tons, which I think is rather light for a tank. That's 100,000 pounds, which puts its mass at 45,000 kg. I don't know what the mass ...

10

I doubt if your textbook makes it explicit, but the only sacred tenet in here is to respect dimensional homogeneity. One can make no sense of the sum of quantities with different dimensions.

10

Let's simplify. Let's eliminate the Moon. Let's get rid of the Sun temporarily. Let's replace the Earth with an equivalent mass-and-density perfect sphere of iron that is neither moving linearly nor spinning or revolving in any way. We place two 1KG iron test masses on opposite sides of the Iron Earth, suspended 1 M above the surface by identical ...

10

tl;dr: Velocity required: 1680 m/s Time to hit you: 6500 seconds Part 1: Velocity required (Using Google search values) Radius of moon = 1737.4 kilometers Mass of moon = 7.34767309E22 kilograms Assuming perfectly circular motion of the bullet, and no air resistance, and ignoring gravitational effects of other planets / objects in space, and using simple ...

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