Complete question:
Part A:) The fictional rocket ship Adventure is measured to be 50 m long by the ship's captain inside the rocket.When the rocket moves past a space dock at 0.5c , space-dock personnel measure the rocket ship to be 43.3 m long. The rocket ship Adventure travels to a star many light-years away, then turns around and returns at the same speed. When it returns to the space dock, who would have aged less: the space-dock personnel or ship's captain?
Part B: What is the momentum of a proton traveling at 0.62 c ?
Answer
a)Who would have aged less=The Captain would have aged less
b) 
Explanation:
From the question we are told that
Length measured by captain 
Speed of rocket past tje space dock 
Length measured by space-dock personnel 
a)
Generally time moves slower when moving at speed of light, due to time dilation or variation.
Who would have aged less=The Captain would have aged less
b)
Generally the equation for Relativistic Momentum is mathematically given as



Acceleration = Change in Velocity / time
a = (v - u) / t
Where v = final velocity in m/s
u = initial velocity in m/s
t = time in seconds.
a = acceleration in m/s²
A proper record of the changes in velocity with the corresponding time would help find the acceleration.
Answer
given,
mass of copper rod = 1 kg
horizontal rails = 1 m
Current (I) = 50 A
coefficient of static friction = 0.6
magnetic force acting on a current carrying wire is
F = B i L
Rod is not necessarily vertical


the normal reaction N = mg-F y
static friction f = μ_s (mg-F y )
horizontal acceleration is zero


B_w = B sinθ
B_d = B cosθ
iLB cosθ= μ_s (mg- iLB sinθ)





B = 0.1 T
Answer:
It comes out the positive side of the battery and goes in to the negative side of the battery
Explanation:
There are already electrons in wires in a circuit before you add the battery. By adding the battery, you're giving the electrons the energy it needs to move along the circuit.
In a series circuit, the circuit is one continuous loop so there is only one path for the electrons to go - out of the positive side of the battery and around the circuit then goes back into the negative side of the battery.
However, with a parallel circuit, there are two or more ways the electrons can go so they take the path of least resistance. The electrons still go out the positive side of a battery but along the circuit, the electrons will go through the path of least resistance ( I tend to think of it like a net with holes in it - the lower the resistance the bigger the holes for the electrons to go through so more can fit in a set amount of time ) but the electrons still go out of the positive side and in through the negative
If net external force acting on the system is zero, momentum is conserved. That means, initial and final momentum are same → total momentum of the system is zero.