C.Earth is the largest and most dense of the terrestrial planets
Answer:
v = 0.92 c
Explanation:
Here, we will use the time dilation formula from Einstein's theory of relativity to find the speed of traveling of the friend:

where,
v = speed of traveling = ?
c = speed of light
t = time of return = 10 years
t₀ = time passed on earth = 4 years
Therefore,

<u>v = 0.92 c</u>
Complete Question
The complete question is shown on the first uploaded image
Answer:
a
The torque produced by the pile of rocks is
b
The distance of the single for equilibrium to occur is
Explanation:
From the question we are told that
The mass of the left rock is 
The mass of the rock on the right 
The distance from fulcrum to the center of the pile of rocks is 
Generally the torque produced by the pile of rock is mathematically represented as

Substituting values
Generally we can mathematically evaluated the distance of the the single rock that would put the system in equilibrium as follows
The torque due to the single rock is

At equilibrium the both torque are equal

Making
the subject of the formula

Substituting values
Complete Question
An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 70 cm long and has a mass of 4.0 kg. Assume, a bit unrealistically, that the athlete's arm is uniform.
What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor? Include the torque due to the steel ball, as well as the torque due to the arm's weight.
Answer:
The torque is 
Explanation:
From the question we are told that
The mass of the steel ball is 
The length of arm is 
The mass of the arm is 
Given that the arm of the athlete is uniform them the distance from the shoulder to the center of gravity of the arm is mathematically represented as

=>
=>
Generally the magnitude of torque about the athlete shoulder is mathematically represented as

=> 
=> 
Since the stone was dropped from height, its initial velocity = 0 m/s
Using v² = u² + 2gs.
Where g ≈ 10 m/s², u = initial velocity = 0 m/s, s = height from drop = 2.5 m
v² = u² + 2gs
v² = 0² + 2*10*2.5
v² = 0 + 50
v² = 50
v = √50
v ≈ 7.07 m/s
Hence velocity just before hitting the ground is ≈ 7.07 m/s