<h2>
Person must have 8.18 m/s to catch the ball</h2>
Explanation:
Consider the vertical motion of ball
We have equation of motion s = ut + 0.5at²
Initial velocity, u = 12 m/s
Acceleration, a = -9.81 m/s²
Displacement, s = -25 m
Substituting
-25 = 12 x t + 0.5 x -9.81 x t²
4.905 t² -12t - 25 = 0
t = 3.79 sec
Ball hits ground after 3.79 seconds.
So person need to cover 31 m in 3.79 seconds
Consider the horizontal motion of person
We have equation of motion s = ut + 0.5at²
Initial velocity, u = ?
Acceleration, a = 0 m/s²
Displacement, s = 31 m
Time, t = 3.79 seconds
Substituting
31 = u x 3.79 + 0.5 x 0 x 3.71²
u = 8.18 m/s
Person must have 8.18 m/s to catch the ball
Answer: 0 km/h
Explanation:
As a vector, the plane's velocity is 100 km/h (west) - 100 km/h (east) = 0 km/m.
To an observer on the ground, the plane will be standing still.
Answer:
fdghgjkk , mmn xgwvdkqgeb e keqwkclhehjq qbqlq.bq;q
Explanation:
Answer:
c. V = k Q1 * Q2 / R1 potential energy of Q1 and Q2 separated by R
V2 / V1 = (R1 / R2) = 1/4
V2 = V1 / 4
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

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

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=> 