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1

You are differentiating with respect to $t$, and by the chain rule $${d \over dt}(f(x)) = {df \over dx}{dx \over dt}={df \over dx} \dot{x}$$ You have just missed the $\dot x$ part resulting from proper application of the chain rule!


0

There are a variety of ways to do this problem: a. Use the law of cosines to find the length of the third line. You will need to figure out what the interior angle between the two known lines is based on the angle each one makes with the ${0}^o$ direction. Then use the law of sines to find the angle. b. Find the $x$ components (parallel to ${0}^o$) and $y$ ...


1

Check the bounds in the second integral, and plug in values at the extremes. When $x=a$, the far corner of the triangle, you should be integrating $y$ from $0$ to $0$, but the integral runs from $0$ to $a \sqrt{3}$. This is because you used the equation $y = \sqrt 3 x$ for the upper bound of the second half of the triangle, when it only applies to the first ...


2

I'm assuming you mean the region $\{(\alpha,\delta) : \alpha_1\leq \alpha\leq \alpha_2, \delta_1\leq \delta\leq \delta_2\}$. Measuring $\alpha$ and $\delta$ in radians, we can find the desired formula by integrating in spherical coordinates: \begin{align*} \int_{\alpha_1}^{\alpha_2} \int_{\delta_1}^{\delta_2} \cos \delta\ d\delta\ d\alpha ...


0

Rotation about the center of the sphere and about the center of mass of a hemisphere is different. You have to move the rotation point to the center of the sphere using the parallel axis theorem to arrive at the equivalency equation.


1

I would use my knuckle, tap each sphere, and listen for the tone. The solid sphere would have a higher pitch tone.


2

Probably only once or twice, as a very small gravitational disturbance (a nearby electron) would send the photon into an orbit with a necessary velocity not = to C.


10

There are three kinds of orbits that encompass all cases: bounded $E < E_{crit}$, extreme $E = E_{crit}$ and unbounded orbits $E > E_{crit}$. As Akano has said, the extreme case can circulate around the `photon sphere' for ever, because the energy is tuned to exactly repel the inward pull. You can still have an unbounded orbit circulate around as ...


1

The photon sphere lies outside the event horizon ($r_p = 3GM/c^2 > r_s = 2GM/c^2$), so a noncaptured photon can orbit a black hole as many times as it wants. Since at the photon sphere the gravitational potential of the black hole is at a peak, it is unstable, so a photon coming in from infinity and having just the right trajectory can spiral inward ...


6

Andynitrox gives a great answer, but it sounds like you are considering a ramp to be equipment. However, because the moment of inertia describes the rotational motion of each object, you actually don't need the spheres to move translationally at all. If you just take each sphere and place it on a flat surface and try to spin it about the vertical axis ...


15

If you are not allowed "any equipment" I suppose that eliminates using a ramp and letting them roll down it (@andynitrox's otherwise good answer). However, assuming your hands are not "equipment", if you took one ball in each hand and tried rotating your hands back and forth as fast as you could, you would find that the frequency you could achieve with the ...


30

Let both roll down an inclined plane. The hollow sphere will accelerate slower than the solid one (due to their different moments of inertia). For a solid sphere, the moment of inertia is $$I = \frac{2}{5}mr^2$$ with mass $m$ and radius $r$. For a hollow sphere it is $$I = \frac{2}{3}mr^2$$ The hollow sphere therefore has a greater moment of inertia and ...


1

A common misunderstanding of quantum mechanics is the belief that EVERYTHING in the world is quantized, but this is simply not true. For example the position of a free particle is not quantized but may take on any value.



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