3 years ago

Calculate the RMS value of an AC signal with a peak amplitude of 10V and a frequency of 10kHz

Physics

1 answer:

madreJ [45]3 years ago

8 0

Answer:

63.01jkh

Explanation:

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A transverse wave on a rope is given by y(x,t)= (0.750cm)cos(π[(0.400cm−1)x+(250s−1)t]). part a part complete find the amplitude

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The amplitude of a wave corresponds to its maximum oscillation of the wave itself.
In our problem, the equation of the wave is
$y(x,t)= (0.750cm)cos(\pi [(0.400cm-1)x+(250s-1)t])$
We can see that the maximum value of y(x,t) is reached when the cosine is equal to 1. When this condition occurs,
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and therefore this value corresponds to the amplitude of the wave.

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Answer:

a. $\frac{dx}{dt}=20ft/s$

b. $\frac{d(x+L)}{dt}==25ft/s$

Explanation:

Using the triangle theorem both triangle the woman makes between the light so the rate of change of length can use geometry first

$tan(\beta)=\frac{24ft}{L+x}=\frac{6ft}{x}$

Solve to find the rate relation

$x=\frac{24}{6}*L$

$x=4*L$

Now the rate of the change rate

$\frac{dx}{dt}=4*\frac{dL}{dt}$

$\frac{dx}{dt}=4*5ft/s=20ft/s$

Finally the rate of her shadow moving

$\frac{d(x+L)}{dt}=\frac{dx}{dt}+\frac{dL}{dt}$

$\frac{d(x+L)}{dt}=20ft/s+5ft/s=25ft/s$

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The answers are B, D, E!

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