Answer:
a. 25000 J
b. 2500 J/s
Explanation:
Given,
Distance ( s ) = 50 m
Force ( f ) = 500 N
a.
To find : -
Work done ( W ) = ?
Formula : -
W = fs
W
= 500 x 50
= 25000 J
Therefore,
the work done by the force the horse exerts is
25000 J.
b.
To find : -
Power ( P ) = ?
Formula : -
W = Pt
P = W / t
P
= 25000 / 10
= 2500 J/s
Therefore,
the power produced if the movement took 10 s
is 2500 J/s.
PWC operators must keep in mind that a jet drive needs moving water through the drive nozzle for maneuverability. In other term, you must have power applied in order to uphold steering control. If the engine shuts off during operation or If you release the throttle to idle, you will lose all steering control. In either situation, the PWC will remain in the direction it was headed before the throttle was released or the engine was shut off. Action of the steering control will have no outcome. If you are approaching a dock, shore, or other vessel at a speed bigger than you can control and you release the throttle to idle or shut off the engine, you won’t have maneuvering capability and the PWC will stay its forward movement.
It should be at the very top since it has more space to fall which gives it more potential energy
Answer:
Explanation:
If friction is neglected, the wheel cannot roll and can only slide frictionlessly and will have the same velocity at the bottom of the ramp as if it had been in free fall as it has converted the same amount of potential energy.
mgh = ½mv²
v = √(2gh) = √(2(9.81)(2.00)) = 6.26418... = 6.26 m/s
However if we do not ignore all friction and the wheel rolls without slipping down the slope, the potential energy becomes linear and rotational kinetic energy
mgh = ½mv² + ½Iω²
mgh = ½mv² + ½(½mR²)(v/R)²
2gh = v² + ½v²
2gh = 3v²/2
v = √(4gh/3) =√(4(9.81)(2.00)/3) = 5.11468... = 5.11 m/s
Answer:
Though there is no chart on my screen, I can give you the correct information to solve the question without knowing what it looks like. Assuming that this track has slopes and or ramps of any kind, the place where it would have the most potential energy would be at the highest point of this ramp, or and the highest point on the track. This would also mean that it would fall the farthest. The highest kinetic energy would be at the lowest part of the ramp, after its used the slope to its full ability. But this would also be before it hits flat ground, or starts going up another ramp. This would be the point where it would be going its fastest.
Explanation:
