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olga nikolaevna [1]
3 years ago
11

What is the wavelength of a light of frequency 7.21?

Physics
1 answer:
erik [133]3 years ago
4 0
MEMORIZED E=h*v h=6.626x10-34J*s INFORMED v=7.21x1014S-1CALCULATE E=h*v  E=(6.626x10-34J*s)*(7.21x1014s-1) The "s" cancels out. s-1=1/s so you get s/s  so you are left with Solution 4.78 10-19 J OR .478 aJ  <span>Apex - 467 nm ^.^  hopefully thats the correct thing</span>
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dlinn [17]

Answer:

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Explanation:

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Why is radiation fog more likely near the center of an anticyclone than near the center of a cyclone?
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Radiation fog is the fog that is formed when the heat absorbed the Earth's surface is released into the atmosphere producing fog. This only occurs when the air is clear and calm. In the center of an anticyclone, the conditions of the air are clear and calm which is favorable for the formation of radiation fog. The center of cyclones, on the other hand, is turbulent and cloudy which prevents the formation of radiation fogs.
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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:

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

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

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

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 %.

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

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

4 0
2 years ago
The two blocks in oscillate on a frictionless surface with a period of 1.5 s. The upper block just begins to slip when the ampli
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Answer:

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\mu = Coefficient of static friction = ?

a = acceleration of the block

m = mass of the block

Maximum acceleration of the block is given as

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frictional force is given as

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As per newton's second law

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