Answer:
A) s=1/2at^2
t=√(2s/a)=√(2x400)/10.0)=9.0s
B) v=at
v=10.0x9=90m/s
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

=> 
=> 
Answer:
1. 24375 N/C
2. 2925 V
Explanation:
d = 12 cm = 0.12 m
F = 3.9 x 10^-15 N
q = 1.6 x 10^-19 C
1. The relation between the electric field and the charge is given by
F = q E
So, 

E = 24375 N/C
2. The potential difference and the electric field is related by the given relation.
V = E x d
where, V be the potential difference, E be the electric field strength and d be the distance between the electrodes.
By substituting the values, we get
V = 24375 x 0.12 = 2925 Volt
<span>A spatial pattern arranges main points according to physical direction or location.</span>
Answer:
coefficient of static friction of the surface and the normal force
Explanation:
The coefficient of static friction of the surface and the normal force exerted on the surface given by equation F = μR