Tim is pushing a box with a 7 N force. The weight of the box is 5 N. A frictional force
1 answer:
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Answer:
Explanation:
Let i be the angle of incidence and r be the angle of refraction .
From the figure
Tan ( 90 - i ) = 2.5 / 8
cot i = 2.5 / 8
Tan i = 8 / 2.5 = 3.2
i = 72.65°
From snell's law
sini / sin r = refractive index
sin 72.65 / sinr = 1.333
sin r = .9545 / 1.333
= .72
r = 46⁰
From the figure
Tan r = d / 4
Tan 46 = d /4
d = 4 x Tan 46
= 4 x 1.0355
=4.14 m .
Answer:
Restoring force of the spring is 50 N.
Explanation:
Given that,
Spring constant of the spring, k = 100 N/m
Stretching in the spring, x = 0.5 m
We need to find the restoring force of the spring. It can be calculated using Hooke's law as "the force on a spring varies directly with the distance that it is stretched".
F = 50 N
So, the restoring force of the spring is 50 N. Hence, this is the required solution.