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maxonik [38]
3 years ago
5

A sound wave travels through a column of hydrogen at STP. Assuming a density of rho = 0.0900 kg/m3 and a bulk modulus of β = 1.4

2 ✕ 105 Pa, what is the approximate speed (in m/s) of the sound wave?
Physics
1 answer:
STALIN [3.7K]3 years ago
4 0

Answer:

speed of sound will be 1256 m/sec

Explanation:

We have given density \rho =0.09kg/m^3

And bulk modulus \beta =1.42\times 10^5Pa

We know that speed of sound in a medium is given by

v=\sqrt{\frac{\beta }{\rho }}, here \beta is bulk modulus and \rho is density

So speed will be v=\sqrt{\frac{1.42\times 10^5}{0.09 }}=12.56\times 10^2=1256m/sec

So speed of sound will be 1256 m/sec

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Answer:

The correct option is;

C. 1,715 m

Explanation:

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Therefore, distance covered by sound in 5 seconds is given by the following equation;

Speed = \frac{Distance}{Time}

\therefore 343 \ m/s= \frac{Distance}{5 \, s}

Hence Distance = 343 m/s × 5 s = 1715 m

To check, we compare the time it would take for the light to cover 1715 m

That is Time = \frac{Distance}{Speed} =  \frac{1715}{299,792,000} = 0.00000572 \, s which is instantaneous hence the distance can be approximated by the time duration for the speed of sound.

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8 0
4 years ago
Outside a spherically symmetric charge distribution of net charge Q, Gauss's law can be used to show that the electric field at
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Answer:

Q at the center of the distribution.

Explanation:

  • The Gauss's law is the law that relates to the distribution of electrical charges to the resulting electrical field. It states that a flux of electricity outside the arabatory closed surface is proportional to the electricitical harg enclosed by the surface.
3 0
3 years ago
The force that keeps objects moving in a circle is called centrifugal force.<br> True<br> False
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Explanation:

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3 years ago
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A roadrunner is running along a straight desert road at a constant velocity of 25 m/s. If a certain coyote wants to capture the
andreyandreev [35.5K]

Answer:

t = 1.42 s and d = 35.5 m

Explanation:

Given that,

Velocity of a roadrunner is 25 m/s

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d=ut+\dfrac{1}{2}at^2\\\\\text{Here, u = 0 and a = g}\\\\d=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2d}{g}} \\\\t=\sqrt{\dfrac{2\times 10}{9.8}} \\\\t=1.42\ s

Let d is the distance traveled. So,

d = vt

d = 25 m/s × 1.42 s

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5 0
3 years ago
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Anon25 [30]

In several of the questions you've posted during the past day, we've already said that a wave with larger amplitude carries more energy.  That idea is easy to apply to this question.

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