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Sidana [21]
4 years ago
5

If two planets orbit a star, but planet B is twice as far from the star as planet A, planet A will receive ____ times the flux t

hat planet B receives.
Physics
1 answer:
Tresset [83]4 years ago
7 0

Answer:

The nearest plant (A) receives 4 times more radiation from the farthest plant

Explanation:

The energy emitted by the star is distributed on the surface of a sphere, whereby intensity received is the power emitted between the area of ​​the sphere

                I = P / A

               P = I A

The area of ​​the sphere is

               A = 4π r²

Since the amount of radiation emitted by the star is constant, we can write this expression for the position of the two planets

               P = I₁ A₁ = I₂ A₂

               I₁ / I₂ = A₂ / A₁

 Suppose index 1 corresponds to the nearest planet,

            r2 = 2 r₁

            I₁ / I₂ = r₁² / r₂²

            I₁ / I₂ = r₁² / (2r₁)²

            I₁ / I₂ = ¼

           4 I₁ = I₂

The nearest plant (A) receives 4 times more radiation from the farthest plant

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I believe its C

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A traveling electromagnetic wave in a vacuum has an electric field amplitude of 50.9 V/m. Calculate the intensity ???? of this w
Sliva [168]

Answer:

3.44 W/m²

1.134 J

Explanation:

E₀ = Intensity of electric field = 50.9 V/m

I = Intensity of electromagnetic wave

Intensity of electromagnetic wave is given as

I = (0.5) ε₀ E₀² c

I = (0.5) (8.85 x 10⁻¹²) (50.9)² (3 x 10⁸)

I = 3.44 W/m²

A = Area = 0.0277 m²

t = time interval = 11.9 s

Amount of energy is given as

U = I A t

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5 0
3 years ago
Will mark the brainliest
AysviL [449]

Answer:

The impulse transferred to the nail is 0.01 kg*m/s.

Explanation:

The impulse (J) transferred to the nail can be found using the following equation:

J = \Delta p = p_{f} - p_{i}

Where:                                                                

p_{f}: is the final momentum

p_{i}: is the initial momentum

The initial momentum is given by:

p_{i} = m_{1}v_{1_{i}} + m_{2}v_{2_{i}}

Where 1 is for the hammer and 2 is for the nail.

Since the hammer is moving down (in the negative direction):

v_{1_{i}} = -10 m/s

And because the nail is not moving:

v_{2_{i}}= 0                      

p_{i} = m_{1}v_{1_{i}} = 0.25 kg*(-10 m/s) = -2.5 kg*m/s

Now, the final momentum can be found taking into account that the hammer remains in contact with the nail during and after the blow:

p_{f} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}}

Since the hammer and the nail are moving in the negative direction:

v_{1_{f}} = v_{2_{f}} = -9.7 m/s

p_{f} = -9.7 m/s(7 \cdot 10^{-3} kg + 0.25 kg) = -2.49 kg*m/s

Finally, the impulse is:

J = p_{f} - p_{i} = - 2.49 kg*m/s + 2.50 kg*m/s = 0.01 kg*m/s

Therefore, the impulse transferred to the nail is 0.01 kg*m/s.

I hope it helps you!                                                                                                                                                                                                                                                                                                                                                                                                                    

7 0
3 years ago
The output voltage of a voltage amplifier has been found to decrease by 20% when a load resistance of 1 k is connected. What is
yKpoI14uk [10]

We use the voltage division problem  between the load resistance, amplifier output resistance as

V_{out} = V_{amlifire} \times \frac{R_{load} }{R_{load} + R_{out} }.

Here, V_{out} is the output voltage, V_{amlifire} is the amplifier voltage, R_{load} is the load resistance and R_{out} is the amplifier output resistance.

Therefore,

1-\frac{20}{100} = \frac{1 \ k\Omega }{1 \ k\Omega +R_{out} } \\\\ R_{out} = \frac{1 \ k\Omega }{0.8} -1 \ k\Omega =1250 \Omega -1000 \Omega =250 \Omega.

Thus, the amplifier output resistance is 250 \ \Omega.

4 0
3 years ago
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