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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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A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0
otez555 [7]

The given question is incomplete. The complete question is as follows.

A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each second. The push force has a horizontal component of 20 N and a vertical component of 25 N downward. Calculate the coefficient of kinetic friction between the box and the floor.

Explanation:

The given data is as follows.

    F_{1} = 20 N, F_{2} = 25 N, a = -0.9 m/s^{2}

             W = 83 N

         m = \frac{83}{9.81}

             = 8.46

Now, we will balance the forces along the y-component as follows.

       N = W + F_{2}

           = 83 + 25 = 108 N

Now, balancing the forces along the x component as follows.

       F_{1} - F_{r} = ma

        20 - F_{r} = 8.46 \times (-0.9)

             F_{r} = 7.614 N

Also, we know that relation between force and coefficient of friction is as follows.

             F_{r} = \mu \times N

          \mu = \frac{F_{r}}{N}

                    = \frac{7.614}{108}

                    = 0.0705

Thus, we can conclude that the coefficient of kinetic friction between the box and the floor is 0.0705.

7 0
3 years ago
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C cost (I need points)

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If a cart is accelerating downhill under a net force of 25 N, what additional force would cause the cart to have a constant velo
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No additional force is required because it's already going downhill
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Imagine that the earth and the-moon have positive charges of the same magnitud. How big isäºthe charge necesary to produce an el
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Answer:

5.7 x 10^12 C

Explanation:

Let the charge on earth and moon is q.

mass of earth, Me = 5.972 x 10^24 kg

mass of moon, Mm = 7.35 x 10^22 kg

Let d be the distance between earth and moon.

the gravitational force between them is

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The electrostatic force between them is

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According to the question

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3.25 \times 10^{25}=q^{2}

q = 5.7 x 10^12 C

Thus, the charge on earth and the moon is 5.7 x 10^12 C.

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