Answer:
Magnitude of the average force exerted on the wall by the ball is 800N
Explanation:
Given
Contact Time = t = 0.05 seconds
Mass (of ball) = 0.80kg
Initial Velocity = u = 25m/s
Final Velocity = 25m/s
Magnitude of the average force exerted on the wall by the ball is given by;
F = ma
Where m = 0.8kg
a = Average Acceleration
a = (u + v)/t
a = (25 + 25)/0.05
a = 50/0.05
a = 1000m/s²
Average Force = Mass * Average Acceleration
Average Force = 0.8kg * 1000m/s²
Average Force = 800kgm/s²
Average Force = 800N
Hence, the magnitude of the average force exerted on the wall by the ball is 800N
Answer:
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Answer:



Explanation:
g = Acceleration due to gravity = 
= Angle of slope = 
v = Velocity of child at the bottom of the slide
= Coefficient of kinetic friction
= Coefficient of static friction
h = Height of slope = 1.8 m
The energy balance of the system is given by

The speed of the child at the bottom of the slide is 
Length of the slide is given by


The force energy balance of the system is given by

The coefficient of kinetic friction is
.
For static friction

So, the minimum possible value for the coefficient of static friction is
.
Answer:
The difference between the velocity graph made walking at a steady rate means that its the same value in time, that means there's no slope on the graph, so its acceleration is 0
On the other hand, if the velocity is increasing with time, the slope of the graph becomes positive, which means that the acceleration of the particle is positive.
Answer:
Derivation of Conservation of Momentum
Applying Newton's third law, these two impulsive forces are equal and opposite i.e. is equal to the change in momentum of the first object. is equal to the change in momentum of the second object. This relation suggests that momentum is conserved during the collision.
Explanation:
Hope it helps!!!