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Ivahew [28]
3 years ago
5

When a container decreases in size, what will happen to an amount of gas in the container??

Physics
1 answer:
alexgriva [62]3 years ago
7 0
The gas would also decrease in size since the container lost gas to decrease the size of the container.
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Suppose a small planet is discovered that is 16 times as far from the Sun as the Earth's distance is from the Sun. Use Kepler's
mamaluj [8]

Answer:

23376 days

Explanation:

The problem can be solved using Kepler's third law of planetary motion which states that the square of the period T of a planet round the sun is directly proportional to the cube of its mean distance R from the sun.

T^2\alpha R^3\\T^2=kR^3.......................(1)

where k is a constant.

From equation (1) we can deduce that the ratio of the square of the period of a planet to the cube of its mean distance from the sun is a constant.

\frac{T^2}{R^3}=k.......................(2)

Let the orbital period of the earth be T_e and its mean distance of from the sun be R_e.

Also let the orbital period of the planet be T_p and its mean distance from the sun be R_p.

Equation (2) therefore implies the following;

\frac{T_e^2}{R_e^3}=\frac{T_p^2}{R_p^3}....................(3)

We make the period of the planet T_p the subject of formula as follows;

T_p^2=\frac{T_e^2R_p^3}{R_e^3}\\T_p=\sqrt{\frac{T_e^2R_p^3}{R_e^3}\\}................(4)

But recall that from the problem stated, the mean distance of the planet from the sun is 16 times that of the earth, so therefore

R_p=16R_e...............(5)

Substituting equation (5) into (4), we obtain the following;

T_p=\sqrt{\frac{T_e^2(16R_e)^3}{(R_e^3}\\}\\T_p=\sqrt{\frac{T_e^24096R_e^3}{R_e^3}\\}

R_e^3 cancels out and we are left with the following;

T_p=\sqrt{4096T_e^2}\\T_p=64T_e..............(6)

Recall that the orbital period of the earth is about 365.25 days, hence;

T_p=64*365.25\\T_p=23376days

4 0
3 years ago
1. A cyclist accelerates from 0 m/s to 9 m/s in 3 seconds. What is his<br>acceleration?​
Lapatulllka [165]

The cyclist accelerates from 0 m/s to 9 m/s in 3 seconds with an acceleration of 3 m/s².

Answer:

Explanation:

Acceleration exerted by an object is the measure of change in speed or velocity of that object with respect to time. So the initial and final velocities play a major role in determining the acceleration of the cyclist. As here the initial velocity of the cyclist is the speed at rest and that is given as 0 m/s. Then after 3 seconds, the velocity of the cyclist changes to 9 m/s.

Then acceleration = change in velocity/Time.

Acceleration = \frac{Change in velocity}{Time taken}

Acceleration = (9-0)/3=9/3=3 m/s².

So the cyclist accelerates from 0 m/s to 9 m/s in 3 seconds with an acceleration of 3 m/s².

3 0
3 years ago
The human eye is sensitive to yellow-green light having a frequency of about 5.5x 10^14 Hz (a wavelength of about 550 nm) what i
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Answer:The human eye is sensitive to yellow-green light having a frequency of about 5.5*10^{14} ... What is the energy in joules of the photons associated with this light? ... As the wavelength and frequency of a wave are related, we can find the energy ... In order to find this value, we need Planck's Constant, h=6.626×10−34 J⋅s h ...

Explanation:

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3 years ago
When we look at a star
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I'm not really sure what specific answer they're looking for, but if it's an open-ended question, then let's think about it this way...

A light year, is the distance it takes for light to travel in a year. If an object is 50,000 light years away, then by the time the light travels to us, 50,000 years has passed. We are looking at a 50,000 year old image of that object. (ignoring gravity and spatial expansion fun stuffs)

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Sebutkan aplikasi gerak osilasi fisis
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