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2 years ago

6

The gas in a cylinder has a volume of 3 liters at a pressure of 117 kpa. the pressure of the gas is increased to 219 kpa. assumi

ng the temperature remains constant, what would the new volume be?

1 answer:

solniwko [45]2 years ago

8 0

V = (1 / p) × p0×V0 = (1 / 219kPa) × 117kPa × 3L = 1.06L

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a) Total power output: $3.845\cdot 10^{26} W$

b) The relative percentage change of power output is 1.67%

c) The intensity of the radiation on Mars is $540 W/m^2$

Explanation:

a)

The intensity of electromagnetic radiation is given by

$I=\frac{P}{A}$

where

P is the power output

A is the surface area considered

In this problem, we have

$I=1360 W/m^2$ is the intensity of the solar radiation at the Earth

The area to be considered is area of a sphere of radius

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Therefore

$A=4\pi r^2 = 4 \pi (1.5\cdot 10^{11})^2=2.8\cdot 10^{23}m^2$

And now, using the first equation, we can find the total power output of the Sun:

$P=IA=(1360)(2.8\cdot 10^{23})=3.845\cdot 10^{26} W$

b)

The energy of the solar radiation is directly proportional to its frequency, given the relationship

$E=hf$

where E is the energy, h is the Planck's constant, f is the frequency.

Also, the power output of the Sun is directly proportional to the energy,

$P=\frac{E}{t}$

where t is the time.

This means that the power output is proportional to the frequency:

$P\propto f$

Here the frequency increases by 1 MHz: the original frequency was

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so the relative percentage change in frequency is

$\frac{\Delta f}{f_0}\cdot 100 = \frac{1}{60}\cdot 100 =1.67\%$

And therefore, the power also increases by 1.67 %.

c)

In this second case, we have to calculate the new power output of the Sun:

$P' = P + \frac{1.67}{100}P =1.167P=1.0167(3.845\cdot 10^{26})=3.910\cdot 10^{26} W$

Now we want to calculate the intensity of the radiation measured on Mars. Mars is 60% farther from the Sun than the Earth, so its distance from the Sun is

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Now we can find the radiation intensity with the equation

$I=\frac{P}{A}$

Where the area is

$A=4\pi r'^2 = 4\pi(2.4\cdot 10^{11})^2=7.24\cdot 10^{23} m^2$

And substituting,

$I=\frac{3.910\cdot 10^{26}}{7.24\cdot 10^{23}}=540 W/m^2$

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On Earth, plasma exists in the ionosphere, in flames, and in chemical and nuclearexplosions. Matter in a controlled thermonuclea

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The cost of developing thermonuclear power with plasmabe defended because D. It can provide an inexpensive power source.

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It should be noted that this thermonuclear power with plasmabe posses the characteristics which make it to exist in the ionosphere, and it can be felt in the flames as well; as in the chemical and nuclearexplosions.

In conclusion the power can be seen as an inexpensive source power because the p[roduction of this power cn be found in most of the thing that can be found around us as discused above.

Therefore, option D is correct.

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Answer:

Explanation:

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E = 300 / 5 x 10⁻³

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F = 1.6 X 10⁻¹⁹ X 60 X 10³ = 96 X 10⁻¹⁶ N.

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a = 96 x 10⁻¹⁶/ mass = 96 x 10⁻¹⁶ / 9.1 x 10⁻³¹

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s = 2.5 x 10⁻³ m

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t = $\sqrt{\frac{2s}{a\\ } }$

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Answer:how long it took you to get there and the distance between your house and her house

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2 years ago

The minute and hour hands of a clock have a common axis of rotation and equal mass. The minute hand is long, thin, and uniform;

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