<h2>Explanation:</h2><h3>3. </h3>
When light bounces back, it is <em>reflected</em>. (That's why you see your <em>reflection</em> in a mirror.) When light is bent from the path it is taking, it is <em>refracted</em>. The only answer choice that makes correct use of these terms is the third choice:
- Part of the ray is <em>refracted</em> into ray B; part of the ray is <em>reflected</em> as ray R.
_____
<h3>4.</h3>
The index of refraction is the ratio of the sine of the angle of incidence to the sine of the angle of refraction. Both angles are measured from the normal to the surface. The angle of refraction here is 12.5° less than the angle of incidence, 44°, so is 31.5°. Then the index of refraction of the medium is ...
n = sin(44°)/sin(31.5°) = 0.69466/0.52250 = 1.3299 ≈ 1.33
- none of the offered choices is correct. The closest is 1.34.
Answer:
B
Explanation:
Is the equilibrium stable or unstable for the third charge
The best conclusion that can be drawn is that D) A current does not flow in the wire
Answer:

Explanation:
The total energy of the satellite when it is still in orbit is given by the formula

where
G is the gravitational constant
m = 525 kg is the mass of the satellite
is the Earth's mass
r is the distance of the satellite from the Earth's center, so it is the sum of the Earth's radius and the altitude of the satellite:

So the initial total energy is

When the satellite hits the ground, it is now on Earth's surface, so

so its gravitational potential energy is

And since it hits the ground with speed

it also has kinetic energy:

So the total energy when the satellite hits the ground is

So the energy transformed into internal energy due to air friction is the difference between the total initial energy and the total final energy of the satellite:
