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Naddika [18.5K]
3 years ago
9

A source vibrating at constant frequency generates a sinusoidal wave on a string under constant tension. If the power delivered

to the string is quadrupled, by what factor does the amplitude change? 16
Physics
1 answer:
just olya [345]3 years ago
5 0

Answer:

Amplitude changes by factor of 2 means double.

Explanation:

Given:

If the power delivered to the string is quadrupled.

From the formula of power transmitted by a sinusoidal wave on a stretched string is,

  P = \frac{1}{2} \mu  \omega ^{2} A^{2}   v

Where \mu = mass per unit length, \omega  = angular speed, A = amplitude, v = wave speed, P = power delivered.

Here we need only two terms power and amplitude.

       P∝ A^{2}

All other quantities are constant for our problem,

     A = k \sqrt{P}

Where k = constant

Here Power become quadruple means four times,

     A = \sqrt{4P}

     A = 2\sqrt{P}

So amplitude changes by factor of 2 means double

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Answer:

Therefore the surface area of the balloon is increased at 4 cm³/s.

Explanation:

The balloon is being filled with air at a rate of 10 cm³/s

It means the volume of the balloon is increased at a rate 10 cm³/s.

i.e \frac{dv}{dt} =10 cm^3/s

Consider r be the radius of the balloon.

The volume of of a sphere is

v=\frac{4}{3} \pi r^3

Differentiate with respect to t

\frac{dv}{dt} =\frac{4}{3} \pi \times 3r^2\frac{dr}{dt}

\Rightarrow 10 =4\pi r^2\frac{dr}{dt}

\Rightarrow \frac{dr}{dt}=\frac{10}{4\pi r^2}

The surface of area of the balloon is(S) = 4\pi r^2

S=4\pi r^2

Differentiate with respect to t

\frac{dS}{dt} =4\pi\times2r\frac{dr}{dt}

\Rightarrow \frac{dS}{dt} =8\pi r\frac{dr}{dt}

Putting the value of \frac{dr}{dt}

\Rightarrow \frac{dS}{dt} =8\pi r\times\frac{10}{4\pi r^2}

\Rightarrow \frac{dS}{dt} =\frac{20}{ r}

Given that r = 5 cm

[\frac{dS}{dt}]_{r=5} =\frac{20}{ 5}  =4 cm³/s

Therefore the surface area of the balloon is increased at 4 cm³/s.

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Answer:

Astronomer

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A scientist who studies the objects in the sky, including planets, galaxies, black holes, and stars, is called an astronomer. These days, the terms astronomer and astrophysicist are used interchangeably, to talk about any physicist who specializes in celestial bodies and the forces that affect them.

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\\ \sf{:}\!\implies Acceleration=\dfrac{F}{m}

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A car with a velocity of 22 m/s is accelerated at a rate of 1.6 m/s^2 for 6.8s has the final velocity t be 32.88 m/s.

The acceleration means the amount of velocity changing per unit time.

The given data:

initial velocity, u = 22 m/s

time, t = 6.8 s

acceleration, a = 1.6 m/s^2

We will be using the equation of motion:

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The final velocity become 32.88 m/s.

To learn more about Attention here:

https://brainly.in/question/10557838

#SPJ4

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