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leonid [27]
3 years ago
11

A cylindrical tank has a tight-fitting piston that allows the volume of the tank to be changed. The tank originally contains 0.1

10 m3 of air at a pressure of 0.355 atm. The piston is slowly pulled out until the volume of the gas is increased to 0.390 m3 . If the temperature remains constant, what is the final value of the pressure?
Physics
1 answer:
Shkiper50 [21]3 years ago
5 0

Answer:

0.10013 atm

Explanation:

Applying Boyle's Law,

P'V' = PV................... Equation 1

Where P' = Initial pressure of air, V' = Initial volume of air, P = Final pressure of air, V = Final volume of air.

make P the subject of the equation

P = P'V'/V..................... Equation 2

Given: P' = 0.355 atm, V'  0.110 m³, V = 0.390 m³

Substitute into equation 2

P = 0.355(0.11)/0.39

P = 0.10013 atm.

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The answer is Period
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3 years ago
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Of the planets with atmospheres, which is the warmest?<br> a. Venusb. Earthc. Marsd. Jupiter
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In the case of Mars, its atmosphere is thin, mostly composed of Carbon Dioxide and Diatomic Nitrogen, which allow a temperature oscillation of 210K.

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7 0
3 years ago
A uniform solid sphere has mass m= 7 kg and radius r= 0. 4 m. What is its moment of inertia about an axis tangent to its surface
lilavasa [31]

The moment of inertia of a uniform solid sphere is equal to 0.448 kgm^2.

<u>Given the following data:</u>

Mass of sphere = 7 kg.

Radius of sphere = 0.4 meter.

<h3>How to calculate moment of inertia.</h3>

Mathematically, the moment of inertia of a solid sphere is given by this formula:

I=\frac{2}{5} mr^2

<u>Where:</u>

  • I is the moment of inertia.
  • m is the mass.
  • r is the radius.

Substituting the given parameters into the formula, we have;

I=\frac{2}{5} \times 7 \times 0.4^2\\\\I=2.8 \times 0.16

I = 0.448 kgm^2.

Read more on inertia here: brainly.com/question/3406242

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2 years ago
Optical tweezers use light from a laser to move single atoms and molecules around. Suppose the intensity of light from the tweez
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Answer:

a= 4.4×10 m/s^2

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P = 1000/ 3×10^8

P = 3.33×10^(-6) Pa

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F = 3.33×10^{-6}×6.65×10(-29)

= 2.22×10^{-6}

acceleration a  = F/m = 2.22×10^{-6}/ 5.10×10^{-27}

a= 4.4×10 m/s^2

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