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Is that what you’re asking?
F - False.
Its greatest kinetic energy is at the point of release.
It has the least kinetic energy, zero, at its highest point in its path.
B) 7.87 m/s
The gravitational pull is the rate of change of velocity which is the acceleration. Formula for acceleration is;

Given:
• Initial velocity = 0m/s; I dropped the ball, and didn't throw it, so it was at rest firstly
• Time taken = 2.40s
• Acceleration = 3.28m/s^2
We're require to find the final velocity, at which the ball hit the ground with. Ignoring air resistance, keep in mind that the velocity of an object increases as it comes closer to the ground.


Answer: 29.50 m
Explanation: In order to calculate the higher accelation to stop a train without moving the crates inside the wagon which is traveling at constat speed we have to use the second Newton law so that:
f=μ*N the friction force is equal to coefficient of static friction multiply the normal force (m*g).
f=m.a=μ*N= m*a= μ*m*g= m*a
then
a=μ*g=0.32*9.8m/s^2= 3.14 m/s^2
With this value we can determine the short distance to stop the train
as follows:
x= vo*t- (a/2)* t^2
Vf=0= vo-a*t then t=vo/a
Finally; x=vo*vo/a-a/2*(vo/a)^2=vo^2/2a= (49*1000/3600)^2/(2*3.14)=29.50 m
Answer:
W = 1418.9 J = 1.418 KJ
Explanation:
In order to find the work done by the pull force applied by Karla, we need to can use the formula of work done. This formula tells us that work done on a body is the product of the distance covered by the object with the component of force applied in the direction of that displacement:
W = F.d
W = Fd Cosθ
where,
W = Work Done = ?
F = Force = 151 N
d = distance covered = 10 m
θ = Angle with horizontal = 20°
Therefore,
W = (151 N)(10 m) Cos 20°
<u>W = 1418.9 J = 1.418 KJ</u>