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Nana76 [90]
3 years ago
8

Consider a gas in a sealed, rigid container. Which of the following will increase if more gas is pumped into the container at co

nstant temperature?
Check all that apply.

a. the collision frequency of the gas particles
b. the kinetic energy of the gas particles
c. the speed of the gas particles
d. the volumes of the individual gas particles
e. the pressure of the gas
f. the number of gas particles
Physics
1 answer:
leonid [27]3 years ago
3 0
It might be E hope it helps
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kenny6666 [7]

Answer:

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Explanation:

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3 years ago
What length of tube would be required to produce a second tone under the same experimental conditions? Explain your answer.
Alisiya [41]

There is an indirect relationship between length and frequency. The longer the length the pipe has, the higher frequency it is. The shorter the length the pipe has, the lower frequency it is.

<u>Explanation:</u>

The four properties of the string that affect its frequency are length, diameter, tension, and density. These properties are described below: When the length of a string is changed, it will vibrate with a different frequency. Shorter strings have higher frequency and therefore higher pitch.

The longer the tube is the lower the pitch of the note that it can emit. When a tube is heated it expands and so is longer! As the gas in the tube gets warmer the molecules move faster, that means they can carry the vibrations of the sound wave more rapidly and so the pitch goes up.

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3 years ago
A disk of a radius 50 cm rotates at a constant rate of 100 rpm. What distance in meters will a point on the outside rim travel d
Dmitry [639]

Answer:

the distance in meters traveled by a point outside the rim is 157.1 m

Explanation:

Given;

radius of the disk, r = 50 cm = 0.5 m

angular speed of the disk, ω = 100 rpm

time of motion, t = 30 s

The distance in meters traveled by a point outside the rim is calculated as follows;

\theta = \omega t\\\\\theta = (100 \frac{rev}{\min}  \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1\min}{60 s} ) \times (30 s)\\\\\theta = 100 \pi \ rad\\\\d = \theta r\\\\d = 100\pi  \ \times \ 0.5m\\\\d = 50 \pi \ m = 157.1 \ m

Therefore, the distance in meters traveled by a point outside the rim is 157.1 m

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GUESS WHAT? I'LL MAKE YOU BRAINLIEST! EASY QUESTION!!
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Here are the 2 reasons:

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- Heat Exhaustion

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Water Could be Found in the frozen Ice on the moon. it would most likely be underground however
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