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likoan [24]
3 years ago
11

A powerful searchlight shines on a man. The man's cross-sectional area is 0.500m2 perpendicular to the light beam, and the inten

sity of the light at his location is 35.5kW/m2. He is wearing black clothing, so that the light incident on him is totally absorbed. What is the magnitude of the force the light beam exerts on the man? Do you think he could sense the force found in part (A)? a. The force is large enough to be felt by the man. b. The force the light beam exerts is much too small to be felt by the man.
Physics
1 answer:
babymother [125]3 years ago
8 0

Answer:

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵N

(b) the force the light beam exerts is much too small to be felt by the man.

Explanation:

Given;

cross-sectional area of the man, A = 0.500m²

intensity of light, I = 35.5kW/m²

If all the incident light were absorbed, the pressure of the incident light on the man can be calculated as follows;

P = I/c

where;

P is the pressure of the incident light

I is the intensity of the incident light

c is the speed of light

P = \frac{I}{c} =\frac{35500}{3*10^8} = 1.18*10^{-4} \ N/m^2

F = PA

where;

F is the force of the incident light on the man

P is the pressure of the incident light on the man

A is the cross-sectional area of the man

F = 1.18 x 10⁻⁴ x 0.5 = 5.9 x 10⁻⁵ N

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵ N

Therefore, the force the light beam exerts is much too small to be felt by the man.

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Answer:

21870.3156 N

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F=ma\\\Rightarrow F=8.34\times 10^3\times 1.02234\\\Rightarrow F=8526.3156\ N

Weight of the craft

W=mg\\\Rightarrow W=8.34\times 10^3\times 1.6\\\Rightarrow W=13344\ N

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The thrust needed to reduce the velocity to zero at the instant when the craft touches the lunar surface is 21870.3156 N

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What is the weight of a 8-kg substance in N, kN, kg·m/s2, kgf, lbm·ft/s2, and lbf?
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Answer:

W = 78.48N\\W =0.0784kN\\W = 8kgf\\

W= 3.6lbf\\W= 115.2 lbm*ft/s2

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if

m=8kg=3.6lb\\

and g=9.81 m/s2=32.16 ft/s2

and

W=m*g

we can just replace de mass and gravity and we have

W = 8kg * 9.81 \frac{m}{s^{2} } =78.48N\\W = \frac{78.48N}{1000} =0.0784kN\\W = 8kgf\\

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According to Gay Lussacs law which states that at constant volume, pressure of an ideal gas is directly proportional to it's absolute temperature.

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