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Answer:</h2>
In circuits, the average power is defined as the average of the instantaneous power over one period. The instantaneous power can be found as:

So the average power is:

But:

So:

![P=\frac{v_{m}i_{m}}{T}\intop_{0}^{T}(\frac{1+cos(2\omega t)}{2} )dt \\\\P=\frac{v_{m}i_{m}}{T}\intop_{0}^{T}[\frac{1}{2}+\frac{cos(2\omega t)}{2}]dt \\\\P=\frac{v_{m}i_{m}}{T}[\frac{1}{2}(t)\right|_0^T +\frac{sin(2\omega t)}{4\omega} \right|_0^T] \\ \\ P=\frac{v_{m}i_{m}}{2T}[(t)\right|_0^T +\frac{sin(2\omega t)}{2\omega} \right|_0^T] \\ \\ P=\frac{v_{m}i_{m}}{2}](https://tex.z-dn.net/?f=P%3D%5Cfrac%7Bv_%7Bm%7Di_%7Bm%7D%7D%7BT%7D%5Cintop_%7B0%7D%5E%7BT%7D%28%5Cfrac%7B1%2Bcos%282%5Comega%20t%29%7D%7B2%7D%20%29dt%20%5C%5C%5C%5CP%3D%5Cfrac%7Bv_%7Bm%7Di_%7Bm%7D%7D%7BT%7D%5Cintop_%7B0%7D%5E%7BT%7D%5B%5Cfrac%7B1%7D%7B2%7D%2B%5Cfrac%7Bcos%282%5Comega%20t%29%7D%7B2%7D%5Ddt%20%5C%5C%5C%5CP%3D%5Cfrac%7Bv_%7Bm%7Di_%7Bm%7D%7D%7BT%7D%5B%5Cfrac%7B1%7D%7B2%7D%28t%29%5Cright%7C_0%5ET%20%2B%5Cfrac%7Bsin%282%5Comega%20t%29%7D%7B4%5Comega%7D%20%5Cright%7C_0%5ET%5D%20%5C%5C%20%5C%5C%20P%3D%5Cfrac%7Bv_%7Bm%7Di_%7Bm%7D%7D%7B2T%7D%5B%28t%29%5Cright%7C_0%5ET%20%2B%5Cfrac%7Bsin%282%5Comega%20t%29%7D%7B2%5Comega%7D%20%5Cright%7C_0%5ET%5D%20%5C%5C%20%5C%5C%20P%3D%5Cfrac%7Bv_%7Bm%7Di_%7Bm%7D%7D%7B2%7D)
In terms of RMS values:

Speed
= (distance covered) / (time to cover the distance)
= (25 m) / (5.0 sec) = 5.0 m/s .
Answer:
The ball has an acceleration of -380 m/s², this means the ball slows down
An acceleration of -380 m/s² is the equivalent of 38.736 g's
Explanation:
Step 1: Data given
Velocity of the baseball at time t=0 = 38 m/s
At time t, the ball stops. This means v = 0
time before stops = 0.1s
Step 2: Calculate the acceleration
v= v0+at
with v= the velocity of the ball at time t = 0. v= 0
with v0 = the velocity of the ball at time t=0. v0 = 38 m/s
with a= the acceleration in m/s²
with t = time in seconds
0 = 38 + a*0.1
a = -380 m/s²
The ball has an acceleration of -380 m/s², this means the ball slows down
An acceleration of -380 m/s² is the equivalent of 38.736 g's