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4 years ago

11

If pressure is increased from 200 kPa to 300 kPa, and the original volume of gas was 1.5 L, what is the new volume? Assume the t

emperature and number of particles are constant.
a. 0.5 L
b. 1 L
c. 2 L
d. 3 L

1 answer:

Oxana [17]4 years ago

6 0

Answer:

The answer to your question is: V2 = 1 l

Explanation:

Data

P1 = 200 kPa

P2 = 300 kPa

V1 = 1.5 l

V2 = ?

Formula

P1V1 = P2V2

V2 = (P1V1) / P2

V2 = (200 x 1.5) / 300

V2 = 1 l

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The universal gravitation law and Newton's second law allow us to find that the answer for the relation of the rotation periods of the satellites is:

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$F = G \frac{Mm}{r^2}$

Where G is the universal gravitational constant (G = 6.67 10⁻¹¹ $\frac{N m^2 }{kg^2}$ ), F the force, m and m the masses of the bodies and r the distance between them

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Where F is the force, m the mass and the acceleration

In this case the body is the satellites of Jupiter and the planet,

$G \frac{Mm}{r^2} = m a$

Suppose the motion of the satellites is circular, then the acceleration is centripetal

a = $\frac{v^2}{r}$ r

Where v is the speed of the satellite and r the distance to the center of the planet

we substitute

$G \frac{Mm}{r^2} = m \frac{v^2}{r} \\G \frac{M}{r} = v^2$

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In the case of a complete orbit, the time is called the period.

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$T_{Io}^2 = (\frac{4 \pi ^2}{GM} ) \ r_{Io}^3$ TIo² = (4pi² / GM) rIo³

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$r_{Eu} = 2 \ r_{Io}$

We calculate

$T_{Eu} = ( \frac{4\pi ^2 }{GM} (2 \ r_{Io})^3\\T_{Eu} = ( \frac{4 \pi ^2 }{GM}) r_{Io} \ 8$

$T_{Eu}^2 = 8 T_{Io}^2$

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$\frac{T_{Eu}}{T_{Io}} = 2.83$

Learn more about universal gravitation law and Newton's second law here:

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3 years ago

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evablogger [386]

<h2>Hello!</h2>

The answer is: 8.62m

<h2>Why?</h2>

There are involved two types of mechanical energy: kinetic energy and potential energy, in two different moments.

<h2>First moment:</h2>

Before the ball is thrown, where the potential energy is 0.

<h2>Second moment: </h2>

After the ball is thrown, at its maximum height, the Kinetic Energy turns to 0 (since at maximum height,the speed is equal to 0) and the PE turns to its max value.

Therefore,

$E=PE+KE$

Where:

$PE=m.g.h$

$KE=\frac{1*m*v^{2}}{2}$

<em>E</em> is the total energy

<em>PE</em> is the potential energy

<em>KE</em> is the kinetic energy

<em>m</em> is the mass of the object

<em>g</em> is the gravitational acceleration

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<em>v</em> is the velocity of the object

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Hence,

The height at the second moment (maximum height) is 8.62m

Have a nice day!

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4 years ago