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guapka [62]
3 years ago
9

A source of sound is kept in a jar in a vacuum. Air is slowly introduced in to the jar. What happens to the sound coming out of

the jar? A The loudness of the sound increases gradually. B The loudness of the source decreases gradually. C The sound remains unchanged. D The pitch of the sound increases gradually. E
The pitch of the sound decreases gradually.
Physics
1 answer:
timama [110]3 years ago
3 0
<span>The loudness of the sound increases gradually as the air is slowly introduced in to the jar. This is because sound needs a physical medium and in a vacuum there is none. The air provides that medium and as it is introduced, the transfer of sound energy increases</span>
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Putting this value in equation (i),

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4 x 10^-25 = (9.1x10^-31)^2 x v^2/ 1.6 x 10^-19 x 1 x 10^-3

v comes out to be 8.88 x 10^7 m/s.

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brainly.com/question/15104254?referrer=searchResults

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An airliner arrives at the terminal, and its engines are shut off. The rotor of one of the engines has an initial clockwise angu
Ilia_Sergeevich [38]

(a) 1200 rad/s

The angular acceleration of the rotor is given by:

\alpha = \frac{\omega_f - \omega_i}{t}

where we have

\alpha = -80.0 rad/s^2 is the angular acceleration (negative since the rotor is slowing down)

\omega_f is the final angular speed

\omega_i = 2000 rad/s is the initial angular speed

t = 10.0 s is the time interval

Solving for \omega_f, we find the final angular speed after 10.0 s:

\omega_f = \omega_i + \alpha t = 2000 rad/s + (-80.0 rad/s^2)(10.0 s)=1200 rad/s

(b) 25 s

We can calculate the time needed for the rotor to come to rest, by using again the same formula:

\alpha = \frac{\omega_f - \omega_i}{t}

If we re-arrange it for t, we get:

t = \frac{\omega_f - \omega_i}{\alpha}

where here we have

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Solving the equation,

t=\frac{0-2000 rad/s}{-80.0 rad/s^2}=25 s

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