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4 years ago

Sound waves tend to move in?

Physics

1 answer:

3241004551 [841]4 years ago

5 0

<span>A sound wave is a means of transporting energy without transporting matter.</span>

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physical feature of a wave is related to the depth of the wave base is The circular orbital motion

B. The wave base is the depth, and the still water level is the horizontal level

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3 years ago

Which of these is equivalent to the law of conservation of energy?

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The Law of the Conservation of Energy is stating that the total mechanical energy is always conserved or in simpler terms, not used or saved.

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According to the picture, how would the gravitational potential energy of the car change if its weight was doubled? A) The gravi

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3 years ago

Radiation from the Sun The intensity of the radiation from the Sun measured on Earth is 1360 W/m2 and frequency is f = 60 MHz. T

Zina [86]

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:

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

$r=1.5\cdot 10^{11} m$ (distance Earth-Sun)

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:

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

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

$f_0 = 60 MHz$

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 %.

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

$r'=(1+0.60)r=1.60r=1.60(1.5\cdot 10^{11})=2.4\cdot 10^{11}m$

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,

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

Learn more about electromagnetic radiation:

brainly.com/question/9184100

brainly.com/question/12450147

#LearnwithBrainly

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3 years ago