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AnnyKZ [126]
2 years ago
13

Please i really need to get this done today..

Physics
2 answers:
serg [7]2 years ago
8 0

The ball at the top of the ramp has potential energy but when it rolls down or when it is rolling down it has kinetic. The ball has kinetic energy when it is moving or rolling down the ramp but there is Potential energy when the ball is at the top or is sitting at the bottom still. when the ball is rolling down the ramp it has heat energy because of how it get speed and it starts to heat up.

Vadim26 [7]2 years ago
8 0

Answer:

The energy of the ball is transformed from potential energy to kinetic energy when it is released at the top of the ramp. When the ball rolls down the ramp, kinetic energy increases and potential energy decreases. The law of conservation of energy says that energy can not be created or destroyed, it can be tranformed or transferred. The total energy of the ball stayed the same because potential and kinetic energy are inversely related.

Explanation:

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A resistor is connected in series with an AC source that provides a sinusoidal voltage of v of t is equal to V times cosine of b
nekit [7.7K]
<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:

p(t)=v(t)i(t)

So the average power is:

P=\frac{1}{T}\intop_{0}^{T}p(t)dt

But:

v(t)=v_{m}cos(\omega t) \\ \\ i(t)=i_{m}cos(\omega t)

So:

P=\frac{1}{T}\intop_{0}^{T}v_{m}cos(\omega t)i_{m}cos(\omega t)dt \\ \\ P=\frac{v_{m}i_{m}}{T}\intop_{0}^{T}cos^{2}(\omega t)dt \\ \\ But: cos^{2}(\omega t)=\frac{1+cos(2\omega t)}{2}

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}

In terms of RMS values:

V_{RMS}=V=\frac{v_{m}}{\sqrt{2}} \\ \\ I_{RMS}=I=\frac{i_{m}}{\sqrt{2}} \\ \\ Then: \\ \\ P=VI

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