Five lamp, each labbled "6V,3W" are operated at normal brightness. What is the total energy supplied to the lamps in five second
1 answer:
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Answer:
W = 19.845 J
Explanation:
Work is defined as W = Fdcos, where F is the force exerted and d is the distance. Because the direction the ball is falling is the same direction as the force itself,
= 0 deg, and since cos(0) = 1, this equation is equivalent to W = Fd. In this case, the force exerted is the weight force, which is equivalent to m * g. Substituting you get:
W = mgd = 0.810 kg * 9.8 m/s^2 * 2.5m
W = 19.845 J
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.