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A 140 W lightbulb emits 4% of its energy as electromagnetic radiation. What is the radiation pressure (in N/m2) on a perfectly absorbing sphere of radius 14 m that surrounds the bulb

Answers

Answer 1
Answer:

Answer: 7.578x10^-12

Explanation:

First, we find the power:

Power P = 140x4/100 =5,6W

Distance r = 14m

Then,

Intensity I = P/4πr2

= 5.6/(4π x 14 x 14)

=. 2.27 x10^3 W/m2

Radiation pressure:

P(rad) = I/c =0.00227÷{3 x 10^8)

=7.578x10^-12 N/m2
Answer 2
Answer:

Answer:

Pr=7.57*10^{11}Pa

Explanation:

We can solve this problem by taking into account the expression

P_r=(IA)/(c)

where I is the irradiance, c is the speed of light and A is the area.

We have that the power is 140W, but only 4% is electromagnetic energy, that is

P=140W=140(J)/(s)\n0.4*140J=56J

56J is the electromagnetic energy.

The area of the bulb is

A_b=4\pi r^2=4\pi (14m)^2=2463m^2

The radiation pressure is

P_r=(56)/(2463m^2*3*10^8m/s^2)=7.57*10^(-11)Pa

hope this helps!!

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Answers

Answer:

42.67N

Explanation:

Step one:

Given

mass m= 0.32kg

intital velocity, u= 14m/s

final velocity v= 22m/s

time= 0.06s

Step two:

Required

Force F

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Answers

W=f*d
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Answers

Answer:

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Explanation:

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Answers

Answer:

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Explanation:

Final answer:

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Explanation:

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