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sladkih [1.3K]
3 years ago
11

A piano wire with mass 2.60g and length 84.0 cm is stretched with a tension of 25.0 N. A wave with frequency 120.0 Hz and amplit

ude 1.6 mm travels along the wire. Calculate the average power carried by the wave. What is the average power if the wave amplitude is halved?
Physics
1 answer:
likoan [24]3 years ago
4 0

Answer:

Power will be 0.2023 watt

And when amplitude is halved then power will be 0.0505 watt

Explanation:

We have given mass of the Piano wire m = 2.60 gram = 0.0026 kg

Length of wire l = 84 cm = 0.84 m

So mass density \mu =\frac{m}{l}=\frac{0.0026}{0.84}=0.0031kg/m

Tension in the wire T = 25 N

Frequency f = 120 Hz

So angular frequency \omega =2\pi f=2\times 3.14\times 120=753.6rad/sec

And amplitude A = 1.6 mm = 0.0016 m

We have to find the generated power

Power is given by P=\frac{1}{2}\sqrt{\mu T}\omega ^2A^2=\frac{1}{2}\times \sqrt{0.0031\times 25}\times 753.6^2\times 0.0016^2=0.2023watt

From the relation we can see that power P\ \propto\ A^2

So if amplitude is halved then power will be \frac{1}{4} times

So power will be equal to \frac{0.2023}{2}=0.0505watt

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(a) The equivalent resistance of three parallel resistors is (R₁R₂R₃)/(R₁R₂ + R₁R₃ + R₂R₃).

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<h3>Equivalent resistance of three parallel resistors</h3>

The equivalent resistance of three parallel resistors is calculated as follows;

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\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_2}\\\\R_{eq} = \frac{R_1R_2R_3}{R_1R_2 + R_2R_3 + R_1R_3}

<h3>Total Circuit Current </h3>

The total circuit current is calculated as follows;

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Learn more about parallel circuit here: brainly.com/question/80537

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