The component of the crate's weight that is parallel to the ramp is the only force that acts in the direction of the crate's displacement. This component has a magnitude of
<em>F</em> = <em>mg</em> sin(20.0°) = (15.0 kg) (9.81 m/s^2) sin(20.0°) ≈ 50.3 N
Then the work done by this force on the crate as it slides down the ramp is
<em>W</em> = <em>F d</em> = (50.3 N) (2.0 m) ≈ 101 J
The work-energy theorem says that the total work done on the crate is equal to the change in its kinetic energy. Since it starts at rest, its initial kinetic energy is 0, so
<em>W</em> = <em>K</em> = 1/2 <em>mv</em> ^2
Solve for <em>v</em> :
<em>v</em> = √(2<em>W</em>/<em>m</em>) = √(2 (101 J) / (2.0 m)) ≈ 10.0 m/s
Answer:
Battery will run for t = 90 s
Explanation:
As we know that rate of flow of charge is known as electric current
So we will have



now we have



Answer:
20.47 m/s
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
g = Acceleration due to gravity = 9.81 m/s² = a

Total height of the fall is 29.3577 m
Height the ball reached above the building is 

Time taken to reach the point from where the ball was thrown is 2.0866 s
This will also be the time it takes the ball to reach the maximum height

The initial velocity with which the rock was thrown was 20.47 m/s
Answer:
energy would become four times
Explanation:
Kinetic energy K= 0.5mv^2
⇒
now, v'= 2v

v'=
hence, speed can be doubled by four folding the energy or by decreasing mass to one fourth.