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goblinko [34]
3 years ago
8

The intensity at a certain distance from a bright light source is 7.20 W/m2 .

Physics
1 answer:
Gemiola [76]3 years ago
8 0

Answer:

A) P_rad.abs = 2.4 × 10^(-8) Pa and P_rad.ref = 4.8 × 10^(-8) Pa

B) P_rad.abs = 2.369 × 10^(-13) atm and P_rad.ref = 4.738 × 10^(-13) atm

Explanation:

A) The formula for radiation pressure for absorbed light is given as;

P_rad = I/c

Where I is the intensity = 7.20 W/m² and c is the speed of light = 3 × 10^(8) m/s

Thus;

P_rad = 7.2/(3 × 10^(8))

P_rad.abs = 2.4 × 10^(-8) Pa

Now formula for radiation pressure for reflected light is given as;

P_rad = 2I/c

Thus;

P_rad = (2 × 7.2)/(3 × 10^(8))

P_rad.ref = 4.8 × 10^(-8) Pa

B) Now, 1.013 × 10^(5) Pa = 1 atm

Thus, for the absorbed surface, we have;

P_rad.abs = (2.4 × 10^(-8))/(1.013 × 10^(5))

P_rad.abs = 2.369 × 10^(-13) atm

For the reflecting surface, we have;

P_rad_ref = (4.8 × 10^(-8))/(1.013 × 10^(5))

P_rad.ref = 4.738 × 10^(-13) atm

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The electric field applies a force on the charged particle: the direction of the force is the same as that of the electric field (for a positive charge).

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W=Fd cos \theta

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In this problem, we have:

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This means that force and displacement are perpendicular to each other, so

\theta=90^{\circ}

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B)

In this case, the charge move upward (same direction as the electric field), so

\theta=0^{\circ}

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cos 0^{\circ}=1

Therefore, the work done by the electric force is

W=Fd

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The electric force is

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The displacement of the particle is

d = 0.660 m

Therefore, the work done is

W=Fd=(1.38\cdot 10^{-3})(0.660)=9.1\cdot 10^{-4} J

C)

In this case, the angle between the direction of the field (upward) and the displacement (45.0° downward from the horizontal) is

\theta=90^{\circ}+45^{\circ}=135^{\circ}

Moreover, we have:

F=1.38\cdot 10^{-3} N (electric force calculated in part b)

While the displacement of the charge is

d = 2.50 m

Therefore, we can now calculate the work done by the electric force:

W=Fdcos \theta = (1.38\cdot 10^{-3})(2.50)(cos 135.0^{\circ})=-2.4\cdot 10^{-3} J

And the work is negative because the electric force is opposite direction to the displacement of the charge.

Learn more about work and electric force:

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