The tension in the string with friction would be the biggest because of the involvement of the force of gravity. This would result in that the friction force that is acting on the system. There is no friction in the frictionless system, and only the force of gravity is relevant.
Answer:
=3.5 m/s
Explanation:
y = x tanθ - 1/2 g x² / (u²cos²θ )
y = 0.25 , x = 0.5, θ = 40°
.25 = .50 tan40 - .5 x 9.8x x²/ u²cos²40
.25 = .42 - 2.0875/u²
u = 3.5 m / s.
It would still be eastward because The direction of the velocity vector is always in the same direction as the direction which the object moves.
"the field of force surrounding a body of finite mass in which anotherbody would experience an attractive force that is proportional to theproduct of the masses and inversely proportional to the square of thedistance between <span>them."
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Answer:
a) 0.05s
b) 4000N
Explanation:
a)When car is stopped its final velocity become zero
U- 10 m/s
V- 0 m/s
S - 0.25 m
t -?
S = (v+u)*t/2
0.25 =(10+0)*t/2
t = 0.05s
b) If we happened to calculate the avarage force we have to consider about acceleration
V= 0
U = 10
t = 0.05 s
a =?
V = U + at
0 = 10 -a * 0.05
a = 200 m/s2
F = m *a
= 20 * 200
= 4000N
