when a pitcher throws a baseball, it reaches a top speed of 45 m/s. if the
baseball takes 0.9 seconds to travel from the pitcher to the catcher, what is
its acceleration? (assume the ball is moving at 0 m/s right before it leaves
the pitcher's hand.)

a. 45 m/s2

b. 50 m/s2

c. 55m/s 2

d. 40 m/s2

B. 50 m/s²

Explanation:

Given:

Initial velocity of the ball is,

Final velocity of the ball is,

Time taken by the  ball is,

Acceleration is defined as the rate of change of velocity. So, the velocity of the ball is changing with time. So, acceleration is given as:

Plug in 0 for 'u', 45 for 'v', 0.9 for 't' and solve for acceleration, 'a'. This gives,

Therefore, the acceleration of the ball is 50 m/s².

96.77 amp

explanation:

rl = 250/100 = 2.5 ohms.

egl = 250 + (100 × 0.1) + 2 = 262 v.

at 700 r.p.m., eg2 = 262 × 700/1000 = 183.4 v.

if ia is the new current, then eg2 - 2 - (ia × 0.1) = 2.5 ia

which means this will give ia = 96.77 amp.