By doing this he increases the time due to which the rate of change momnetemdecreases. that means the ball comes to rest gently and does not hurt the fielder. but, if does not move his hand backwards, then the ball would come to rest in a fraction of second with a high rate of change of momentum. the ball will exert a lot of force on the hands of the fielder which will hurt him.
Answer
speed of the molecules
s₁ = v t
when velocity is doubled
s₂ = (2 v)t
= 2 s₁
they will hit the wall of container two times as often.
the momentum of molecule
p₁ = mvr
p₂ = m(2v)r = 2(mvr)
= 2 p₁
the momentum change is two times as great.
force is change in momentum
Δp = F(Δt)
mv-(-mv) = 2 mv
F α v
therefore average force impart to the wall on each collision is two times
p α v²
here the velocity is doubled it means pressure becomes four times.
Answer:
11250 N
Explanation:
From the question given above, the following data were obtained:
Normal force (R) = 15000 N
Coefficient of static friction (μ) = 0.75
Frictional force (F) =?
Friction and normal force are related by the following equation:
F = μR
Where:
F is the frictional force.
μ is the coefficient of static friction.
R is the normal force.
With the above formula, we can calculate the frictional force acting on the car as follow:
Normal force (R) = 15000 N
Coefficient of static friction (μ) = 0.75
Frictional force (F) =?
F = μR
F = 0.75 × 15000
F = 11250 N
Therefore, the frictional force acting on the car is 11250 N
Take 110 and divide it by 2 and there's your answer
