I've decided to run through multiple choice. I am a little uncertain on 12 and 19 since I’ve never actually covered these in my studies. Feel free to pick up on any mistakes if you see them in here. It has been a while for me.
Spoiler
1) D - At the peak of the height, the ball has no vertical velocity, so it can only have horizontal velocity.
2) B - The spectra are slightly shifted, indicating a difference in relative speed. The lines are consistent for both spectra, so the composition is the same.
3) C - They shot alpha particles in their experiment and were expecting them to pass right through with minimal disturbance
4) C - Main sequence is the “stable phase” so P is earlier on, R is a red star, and S has the lowest luminosity.
5) D - we need a constant change to have a non-zero reading, but we only have a brief one.
6) A - λ = b/T can be turned into y = b/x which is a hyperbola with asymptote as T/x = 0, only A shows this.
7) D - Lenz’s law wants to keep the magnet, so a North pole needs to be made, using right-hand grip rule to find what why the current needs to flow shows D as the correct answer.
8.) A - Hubble’s equation used parsec, and he found a linear relationship between them.
9) D - This is just using the formula (or intuition). The orbital period is T^2 = r^3, so bigger radius is a longer period. Potential energy is U = -1/r, so bigger the radius, the closer the answer is to 0. We want the greater total energy, so we want the one closer to 0. The minus in this equation DOES count.
10) A - I = I_max*cos^2(30), since we don’t care what I_0 is, set it to 1. cos^2(30) = 0.75, 0.75/1 = 0.75. It showed light was a wave.
11) C - r^3/t^2 = r^3/t^2 let one of them be earth, the other be the dwarf planet.
12) A - This is a neutron to a proton transformation which is a Beta- decay, which emits an electron
13) B - E = hf -> E = h c/λ = 3.06*10^-19. # of photons = power/E = 0.03/(3.06*10^-19) = B
14) B - v = sqrt(GM/r) double r, we get 1/sqrt(2) = 0.707
15) C - d is the distance between slits, theta angle from slits, so just use trig.
16) D - F = Eq -> mg = Eq -> E = mg/q. However, this is to just make the particle move horizontally (since net force is now 0) so times everything by 2, to mimic mg in the opposite direction. E = 2mg/q
17) D - This one is pretty difficult, and this is the best answer I could come up with. Resistance SLOWS the electrons. Since nothing is changing apart from the speed, it’ll take longer for the electrons to hit the screen, meaning more influenced by gravity, so it should fall down very slightly.
18) B - When the wire enters the field, Lenz’s law states it will try to oppose that change, so the loop with try to make a magnetic field out of the page, this translates to current flowing anti-clockwise - opposite to the battery’s flow, meaning the lightbulb should get dimmer, eliminating A and C. When the loop leaves the field, it’ll do the same thing, but in the opposite direction, so the lightbulb will get brighter than originally. D never gets brighter than that, so B is left.
19) A - From what I’ve gathered from the internet, (Y+Z) - (W+X) will always be positive (unless Z is negative, which can’t happen) meaning that no matter what Z is, this reaction will give off energy. It has nothing to do with binding energy. Again, this one probably is wrong, because I actually have no clue
20) B - A is wrong since the cube is still in a circular motion, so it is accelerating. C is wrong because if this were true, the cube would not be stationary on the blackboard. D is wrong because of the formula F = mu*N, yes they are perpendicular, but it doesn’t make them independent. A more in-depth answer is that the formula for frictional force is a cross product, not a dot product.