2.10-kg box is moving to the right with speed 8.50 m>s on a horizontal, frictionless surface. At t = 0 a horizontal force is
applied to the box. The force is directed to the left and has magnitude F1t2 = 16.00 N>s22t2. (a) What distance does the box move from its position at t = 0 before its speed is reduced to zero? (b) If the force continues to be applied, what is the speed of the box at t = 3.00 s?
2 answers:
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Answer:
= 3,126 m / s
Explanation:
In a crash exercise the moment is conserved, for this a system formed by all the bodies before and after the crash is defined, so that the forces involved have been internalized.
the car has a mass of m = 1.50 kg a speed of v1 = 4.758 m / s and the mass of the train is M = 3.60 kg and its speed v2 = 2.45 m / s
Before the crash
p₀ = m v₁₀ + M v₂₀
After the inelastic shock
= m
+ M
p₀ =
m v₀ + M v₂₀ = m + M
We cleared the end of the train
M = m (v₁₀ - v1f) + M v₂₀
Let's calculate
3.60 v2f = 1.50 (2.15-4.75) + 3.60 2.45
= (-3.9 + 8.82) /3.60
= 1.36 m / s
As we can see, this speed is lower than the speed of the car, so the two bodies are joined
set speed must be
m v₁₀ + M v₂₀ = (m + M)
= (m v₁₀ + M v₂₀) / (m + M)
= (1.50 4.75 + 3.60 2.45) /(1.50 + 3.60)
= 3,126 m / s
We know
How he find:-
The owner will measure speed of light through the index(Diamond or rutile).Then using calculations he fill find which is the velocity of diamond or rutile
<h3> For diamond</h3>