Question 6 (15 points)
1 answer:
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Answer:
t_{out} = t_{in}, t_{out} =
Explanation:
This in a relative velocity exercise in one dimension,
let's start with the swimmer going downstream
its speed is
The subscripts are s for the swimmer, r for the river and g for the Earth
with the velocity constant we can use the relations of uniform motion
= D /
D = v_{sg1} t_{out}
now let's analyze when the swimmer turns around and returns to the starting point
= D /
D = v_{sg 2} t_{in}
with the distance is the same we can equalize
t_{out} = t_{in}
t_{out} = t_{in}
This must be the answer since the return time is known. If you want to delete this time
t_{in}= D /
we substitute
t_{out} = \frac{v_s - v_r}{v_s+v_r} ()
t_{out} =
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
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.