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
=> 
=>
consider the motion of the tennis ball in downward direction
Y = vertical displacement = 400 m
a = acceleration = acceleration due to gravity = 9.8 m/s²
v₀ = initial velocity of the ball at the top of building = 10 m/s
v = final velocity of the ball when it hits the ground = ?
using the kinematics equation
v² = v²₀ + 2 a Y
inserting the values
v² = 10² + 2 (9.8) (400)
v = 89.11 m/s
Answer:
the answer is c
Explanation:
it's c because the moon has to be a full moon to be a solar eclipse when the sun moon and earth line up
Instead of asking this question go and study
Answer:
192.08J
19.6m/s
Explanation:
Since there will be no potential energy when the ball is on the ground, the change in potential energy is equal to the potential energy at the start when the ball is 19.6m above the ground.
PE=mgh
=(1)(9.8)(19.6)
=192.08J
v²=u²+2as, where v is the final velocity, u is initial velocity, a is acceleration and s is distance. Initial velocity is 0 since it starts at rest.
v²=u²+2as
v²=0²+2(9.8)(19.6)
v=√384.16
=19.6m/s
