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Varvara68 [4.7K]
3 years ago
11

A trombone has a variable length. When a musician blows into the mouthpiece and causes air in the tube of the horn to vibrate, t

he waves set up by the vibrations reflect back and forth in the horn to create standing waves. As the length of horn is made shorter, what happens to the frequency?
Physics
1 answer:
Lilit [14]3 years ago
6 0

Answer:

The frequency increases.

Explanation:

When the Musician draws the slide in the length of the horn gets shorter, which causes a decrease in the wavelength. A decrease in the wave length results in an increase in frequency.

Note:

The diameter of the horn has an effect on frequency, so a wider horn is effectively a long horn - open end correction ( distance between the the antinode and the open end of a pipe).

Frequency also depends on how hard the musician blows the trombone. The musician can change the frequency with the lip pressure being applied.

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What does the rotational speed of the ring module have to be so that an astronaut standing on the outer rim of the ring module f
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Complete question:

In the movie The Martian, astronauts travel to Mars in a spaceship called Hermes. This ship has a ring module that rotates around the ship to create “artificial gravity” within the module. Astronauts standing inside the ring module on the outer rim feel like they are standing on the surface of the Earth. (The trailer for this movie shows Hermes at t=2:19 and demonstrates the “artificial gravity” concept between t= 2:19 and t=2:24.)

Analyzing a still frame from the trailer and using the height of the actress to set the scale, you determine that the distance from the center of the ship to the outer rim of the ring module is 11.60 m

What does the rotational speed of the ring module have to be so that an astronaut standing on the outer rim of the ring module feels like they are standing on the surface of the Earth?

Answer:

The rotational speed of the ring module have to be 0.92 rad/s

Explanation:

Given;

the distance from the center of the ship to the outer rim of the ring module r, = 11.60 m

When the astronaut standing on the outer rim of the ring module feels like they are standing on the surface of the Earth, then their centripetal acceleration will be equal to acceleration due to gravity of Earth.

Centripetal acceleration, a = g = 9.8 m/s²

Centripetal acceleration, a = v²/r

But v = ωr

a = g = ω²r

\omega = \sqrt{\frac{g}{r}} = \sqrt{\frac{9.8}{11.6} } = 0.92 \ rad/s

Therefore, the rotational speed of the ring module have to be 0.92 rad/s

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3 years ago
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