Answer:
Option d
Explanation:
When we throw an object in the upward direction, we provide it with certain initial velocity due to which it covers a certain distance up to the maximum height.
While the object is moving in the upward direction, its velocity keeps on reducing due to the acceleration due to gravity which acts vertically downwards in the opposite direction thus reducing its velocity.
So, the maximum height attained by the object is the point where this upward velocity of the body becomes zero and after that the object starts to fall down.
Their linear inertia is equivalent to their masses. Let the inertia of the first moose be m₁ and the second be m₂.
m₁u + m₂u = (m₁ + m₂) x 1/3 u
3m₁ + 3m₂ = m₁ + m₂
3 m₁/m₂ + 3 = m₁/m₂ + 1
m₁/m₂ = 2
The ratio of their inertias is 2
Answer:
692.31 N
Explanation:
Applying,
F = ma............... Equation 1
Where F = Average force required to stop the player, m = mass of the player, a = acceleration of the player
But,
a = (v-u)/t............ Equation 2
Where v = final velocity, u = initial velocity, t = time.
Substitute equation 2 into equation 1
F = m(v-u)/t............ Equation 3
From the question,
Given: m = 75 kg, u = 6.0 m/s, v = 0 m/s (to stop), t = 0.65 s
Substitute these values into equation 3
F = 75(0-6)/0.65
F = -692.31 N
Hence the average force required to stop the player is 692.31 N
