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iris [78.8K]
3 years ago
5

Suppose you are an astronomer, and a child asks you to explain the light given off by stars. In your own words, describe the rel

ationship between the energy, frequency, and wavelength. Be sure to include information about the electromagnetic spectrum in your discussion.
Physics
2 answers:
lina2011 [118]3 years ago
4 0

The energy, frequency and wavelength of emitted radiations from stars are related as - E = hf =\frac{hc}{\lambda}

EXPLANATION :

Before answering this question, first we have to understand the Planck's quantum theory.

As per Planck's quantum theory, the energy of the emitted radiation is will be equal to the product of Planck's constant with the frequency of the emitted radiation.

Mathematically energy E = hf

Here, h is the Planck's constant and f is the frequency of emitted radiation.

We know that the wavelength is inversely proportional to frequency, and it is calculated as-

                      Wavelength  \lambda=\frac{c}{f}

Putting these values, we get energy E = \frac{hc}{\lambda}

Here, c is the velocity of light.

This formula is used for calculating the energy of radiations coming from stars.

Stars emit different radiations which have different frequencies like gamma ray, U.V ray and visible light.

From the electromagnetic spectrum, we know that more is the frequency of radiation, the more is the energy of the radiation.

Cosmic wave has more energy as it has more frequency. Gamma ray is followed by the cosmic wave which has the second most energy in the electromagnetic spectrum.


                     

Nadya [2.5K]3 years ago
3 0
Well, i would tell the kid that light is electromagnetic radiation which is visible. The frequency and wavelength is just right so the radiation becomes visible, and that visible radiation goes away from the star and reaches earth It also depends on the age of the child and the comprehension level
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Pressure is defined as the force acting perpendicular on an unit area of the surface.


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The thrust per unit area is called Pressure.

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4 0
3 years ago
Suppose a 4.0 kg mass of water gained heat and increased in temperature from 10°C to 15°C. The specific heat capacity of water i
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8 0
3 years ago
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A box is being pulled across a horizontal surface by a 20 N force to the right.
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Answer:

Im pretty sure it's C

Explanation:

The answer A talks about constant speed, which does not correspond to Net Force. The answer B, talks about constant velocity, but talks about how much force apposes the box. The answer C talks about the value of the net force acting on the box, so im pretty sure the answer is C.

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A toy cannon uses a spring to project a 5.35-g soft rubber ball. The spring is originally compressed by 5.08 cm and has a force
Elenna [48]

Answer:

a) the velocity is v=1.385 m/s

b) the ball has its maximum speed at 4.68 cm away from its compressed position

c)  the maximum speed is 1.78 m/s

Explanation:

if we do an energy balance over the ball, the potencial energy given by the compressed spring is converted into kinetic energy and loss of energy due to friction, therefore

we can formulate this considering that the work of the friction force is equal to to the energy loss of the ball

W fr = - ΔE = - ΔU - ΔK = Ui - Uf + Kf - Ki

therefore

Ui + Ki = Uf + Kf + W fr  

where U represents potencial energy of the compressed spring , K is the kinetic energy W fr is the work done by the friction force. i represents inicial state, and f final state.

since

U= 1/2 k x² , K= 1/2 m v²  , W fr = F*L

X= compression length , L= horizontal distance covered

therefore

Ui + Ki = Uf + Kf + W fr

1/2 k xi² + 1/2 m vi² = 1/2 k x² + 1/2 m vf² + F*L

a) choosing our inicial state as the compressed state , the initial kinetic energy is Ki=0 and in the final state the ball is no longer pushed by the spring thus Uf=0

1/2 k X² + 0 = 0 + 1/2 m v² + F*L

1/2 m v² = 1/2 k X² - F*L

v = √[(k/m)x² -(2F/m)*L] = √[(8.07N/m/5.35*10^-3 Kg)*(-0.0508m)² -(2*0.033N/5.35*10^-3 Kg)*(0.16 m)] = 1.385 m/s

b) in any point x , and since L= d-(X-x) , d = distance where is no pushed by the spring.

1/2 k X² + 0 = 1/2 k x² + 1/2 m v² + F*[d-(X+x)]

1/2 m v² =1/2 k X²-1/2 k x² - F*[d-(X-x)] = (1/2 k X²+ F*X) - 1/2k x² - F*x + F*d

taking the derivative

dKf/dx = -kx - F = 0 → x = -F/k = -0.033N/8.07 N/m = -4.089*10^-3 m = -0.4cm

at x m = -0.4 cm the velocity is maximum

therefore is 5.08 cm-0.4 cm=4.68 cm away from the compressed position

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3 years ago
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