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zzz [600]
3 years ago
12

Light requires 4.5 years to travel from the nearest star to earth. If we could travel there in a spaceship going 90% of the spee

d of light, the trip would require 5.0 years according to clocks on earth. How much time would pass for the passengers in the ship
Physics
1 answer:
weeeeeb [17]3 years ago
7 0

Answer:

Time according to earth clock (T0) = 0.22 years (Approx)

Explanation:

Given:

Time taken by light = 4.5 years

Time taken by ship = 5 years

Speed of light = c

Speed of ship (v) = 90% of c = 0.9c

Find:

Time according to earth clock (T0) = ?

Computation:

Time dilation is ,

T(Difference) = \frac{T0}{\sqrt{1-\frac{v^2}{c^2} } }\\\\ (5-4.5)= \frac{T0}{\sqrt{1-\frac{(0.9c)^2}{c^2} } }\\\\ 0.5=\frac{T0}{\sqrt{1-0.81} }\\\\T0 =0.2179

Time according to earth clock (T0) = 0.22 years (Approx)

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A factory worker pushes a crate of mass 31.0 kg a distance of 4.35 m along a level floor at constant velocity by pushing horizon
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Answer:

a. 79.1 N

b. 344 J

c. 344 J

d. 0 J

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Explanation:

a. Since the crate has a constant velocity, its net force must be 0 according to Newton's 1st law. The push force F_p by the worker must be equal to the friction force F_f on the crate, which is the product of friction coefficient μ and normal force N:

Let g = 9.81 m/s2

F_p = F_f = \mu N = \mu mg = 0.26 * 31 * 9.81 = 79.1 N

b. The work is done on the crate by this force is the product of its force F_p and the distance traveled s = 4.35

W_p = F_ps = 79.1*4.35 = 344 J

c. The work is done on the crate by friction force is also the product of friction force and the distance traveled s = 4.35

W_f = F_fs = -79.1*4.35 = -344 J

This work is negative because the friction vector is in the opposite direction with the distance vector

d. As both the normal force and gravity are perpendicular to the distance vector, the work done by those forces is 0. In other words, these forces do not make any work.

e. The total work done on the crate would be sum of the work done by the pushing force and the work done by friction

W_p + W_f = 344 - 344 = 0 J

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