A baseball pitcher loosens up his pitching arm. he tosses a 0.140 kg ball using only the roation of his forearm, 0.270m in length, to accerlate the ball. if the ball starts at rest and is releasedwith a speed of 24 in a time of 0.425, what torque is applied to the ball

2.136 N-m

Explanation:

From the question,

Applying,

Toque = FL Equation 1

Where F = Force applied to the the ball, L = Length of the forearm

From newton's second law of motion,

F = m(v-u)/t Equation 2

Where m = mass of the ball, v = final velocity of the ball, u = initial velocity of the ball, t = time.

Given: m = 0.140 kg, u = 0 m/s (from rest), v = 24 m/s, t = 0.425 s

Substitute into equation 2

F = 0.140(24-0)/0.425

F = 7.91 N.

Also Given: L = 0.270 m

Substitute into equation 1

Torque = 7.91×0.270

Torque = 2.136 N-m

pecialized keyboards

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j