<h2>
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:

I would say that the answer is C
Answer:
Explanation:
Given ,
dv / dt = k ( 160 - v )
dv / ( 160 - v ) = kdt
ln ( 160 - v ) = kt + c , where c is a constant
when t = 0 , v = 0
Putting the values , we have
c = ln 160
ln ( 160 - v ) = kt + ln 160
ln ( 160 - v / 160 ) = kt
(160 - v ) / 160 = 
1 - v / 160 = 
v / 160 = 1 - 
v = 160 ( 1 -
)
differentiating ,
dv / dt = - 160k 
acceleration a = - 160k 
given when t = 0 , a = 280
280 = - 160 k
k = - 175
a = - 160 x - 175 
a = 28000 
when a = 128 t = ?
128 = 28000 
= .00457
Answer:
Explanation:
Theoretical efficiency = T₁ - T₂ / T₁ where T₁ and T₂ is absolute temperature of hot and cold end of the heat engine.
= 600 / (273 + 700 )
= 600 / 973
= .6166
operating efficiency = 40% of .6166
= .4 x .6166
= .2466 = 24.66 %
efficiency = work output / heat input
= 5000 / heat input = .2466
heat input = 5000 / .2466
= 20275.75 J .
HEAT EXTRACED = 20275.75 J.