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3 years ago

Who wants to be freinds

Physics

2 answers:

Oxana [17]3 years ago

7 0

Answer:

why not

Explanation:

brilliants [131]3 years ago

5 0

Hello there, Goodmorning and God Bless...Here is your answer to your following question:

yessss i do!!!

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The dances created and performed collectively by the ordinary people.

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3 years ago

Read 2 more answers

A small steel wire of diameter 1.0mm is connected to an oscillator and is under a tension of 5.7N . The frequency of the oscilla

Fiesta28 [93]

Answer:

a) $P = 2 \pi^2 (57 Hz)^2 (0.0054 m)^2 \sqrt{\pi(0.0005m)^2 (7800 \frac{kg}{m^3}) (5.7N)}= 0.349 W$

b) For this case we se that $P \propto A^2 f^2$

Since the power is constant but the frequency is doubled we will see that $A^2 \propto \frac{P}{f^2}$

So the original amplitude is $A_i \propto \sqrt{\frac{P}{f^2}}$

And if the frequency is doubled we have that:

$A^2_f \propto \frac{P}{(2f)^2}= \frac{P}{4f^2}$

$A_f \propto \sqrt{\frac{P}{4f^2}}= \frac{1}{2} \sqrt{\frac{P}{f^2}}$

So then we will see that the amplitude would be reduced the half and for this case:

$A_f = \frac{0.54cm}{2}= 0.27 cm = 0.0027 m$

Explanation:

For this case we have the following data given:

$D= 1mm = 0.001 m$ represent the diameter

$r = D/2= 0.0005m$ represent the radius

$T= 5.7 N$ represent the tension

$f = 57 Hz$ represent the frequency of the oscillator

$A= 0.54 cm = 0.0054 m$ represent the amplitude of the wave

Part a

For this case we can assume that the power transmitted to the wave is the same power of the oscillator. and we have the following formula for the power:

$P= 2 \pi^2 \rho S v f^3 A^2$

This expression can be written in different ways:

$P= 2 \pi^2 \rho S \sqrt{\frac{T}{\mu}} f^2 A^2$

$P= 2\pi^2 \rho S \sqrt{\frac{T}{\rho S}} f^2 A^2$

$P= 2 \pi^2 f^2 A^2 \sqrt{S \rho T}$

Where f is the frequency , T the tension $rho= 7800 \frac{kg}{m^3}$ the density of the steel, A the amplitude and $S= \pi r^2$ the area, so then we have everuthing in order to replace and we got:

$P = 2 \pi^2 (57 Hz)^2 (0.0054 m)^2 \sqrt{\pi(0.0005m)^2 (7800 \frac{kg}{m^3}) (5.7N)}= 0.349 W$

Part b

For this case we se that $P \propto A^2 f^2$

Since the power is constant but the frequency is doubled we will see that $A^2 \propto \frac{P}{f^2}$

So the original amplitude is $A_i \propto \sqrt{\frac{P}{f^2}}$

And if the frequency is doubled we have that:

$A^2_f \propto \frac{P}{(2f)^2}= \frac{P}{4f^2}$

$A_f \propto \sqrt{\frac{P}{4f^2}}= \frac{1}{2} \sqrt{\frac{P}{f^2}}$

So then we will see that the amplitude would be reduced the half and for this case:

$A_f = \frac{0.54cm}{2}= 0.27 cm = 0.0027 m$

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4 years ago

What do light and energy tell us about the universe?<br><br> need a paragraph

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3 years ago

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Asalam a lekum my brother

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3 years ago