Calculating the RMS Value of Sinewave Voltage

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The discussion revolves around calculating the back emf of a generator with specific parameters, focusing on the correct use of RMS values in the context of sinewave voltage. The user expresses confusion about whether to calculate the RMS speed or frequency, leading to clarification that RMS speed is not applicable. It is emphasized that sinewave voltage can be defined by peak values or RMS values, and the user is encouraged to check if their calculations align with textbook answers. Ultimately, understanding the distinction between these values is crucial for solving the problem correctly.
Vladi

Homework Statement


I keep telling myself that this problem should be easy, but I keep getting the wrong answer. Any help is appreciated.
A generator has an armature with 500 turns, which cut a flux of 8.00 mWb during each rotation. Compute the back emf it develops when it runs as a motor at 1500 rpm.

Homework Equations


emf= 2pi*(N)(A)(B)(f)*cos(2pi)
Flux=B*A

The Attempt at a Solution


This is the best I could come up with.
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You have found the maximum value. Try finding the rms value.
 
kuruman said:
You have found the maximum value. Try finding the rms value.

If the rms speed is calculated with the following equation: Vrms=Vmax/sqrt(2), does that mean I must plug in the rms speed within the equation? I will give it a shot. Also, how do you know I should be calculating for the rms value? Thank your for your time. I meant to say rms frequency.
 
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Vladi said:

If the rms speed is calculated with the following equation: Vrms=Vmax/sqrt(2), does that mean I must plug in the rms speed within the equation? I will give it a shot. Also, how do you know I should be calculating for the rms value? Thank your for your time. I meant to say rms frequency.

No, there is no such thing as RMS speed. A sinewave voltage can be specified by its peak value, its peak-to-peak value, or its RMS value. Does one of these agree with the textbook answer?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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