A crane lifts an I-beam up the side of a building. The crane's power output is 1750W for 20 seconds. After 20 seconds the I-beam
1 answer:
Explanation:
This is one of those great problems that is harder to read than to solve!
Remember the following...
Power is the amount of work done over time.
and that work is Force times distance.
and that Force is just mass times acceleration.
So... nesting all the equations together we might get something that looks like.
We know the power, the time, the acceleration (at this given time) and the mass. We can then solve for d.
Plug and chug!
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Answer:
h = 50.49 m
Explanation:
Data provided:
Speed of skier, u = 2.0 m/s
Maximum safe speed of the skier, v = 30.0 m/s
Mass of the skier, m = 85.0
Total work = 4000 J
Height from the starting gate = h
Now, from the law of conservation of energy
Total energy at the gate = total energy at the time maximum speed is reached
where, g is the acceleration due to the gravity
on substituting the values, we get
or
170 + 833.85 × h = 4000 + 38250
or
h = 50.49 m
Answer: Add an incline or grade to the road track.
Explanation:
Refer to the figure shown below.
When a vehicle travels on a level road in a circular path of radius r, a centrifugal force, F, tends to make the vehicle skid away from the center of the circular path.
The magnitude of the force is
F = mv²/r
where
m = mass of the vehicle
v = linear (tangential) velocity to the circular path.
The force that resists the skidding of the vehicle is provided by tractional frictional force at the tires, of magnitude
μN = μW = μmg
where
μ = dynamic coefficient of friction.
At high speeds, the frictional force will not overcome the centrifugal force, and the vehicle will skid.
When an incline of θ degrees is added to the road track, the frictional force is augmented by the component of the weight of the vehicle along the incline.
Therefore the force that opposes the centrifugal force becomes
μN + Wsinθ = W(sinθ + μ cosθ).
Explanation:
Refer to the figure shown below.
When a vehicle travels on a level road in a circular path of radius r, a centrifugal force, F, tends to make the vehicle skid away from the center of the circular path.
The magnitude of the force is
F = mv²/r
where
m = mass of the vehicle
v = linear (tangential) velocity to the circular path.
The force that resists the skidding of the vehicle is provided by tractional frictional force at the tires, of magnitude
μN = μW = μmg
where
μ = dynamic coefficient of friction.
At high speeds, the frictional force will not overcome the centrifugal force, and the vehicle will skid.
When an incline of θ degrees is added to the road track, the frictional force is augmented by the component of the weight of the vehicle along the incline.
Therefore the force that opposes the centrifugal force becomes
μN + Wsinθ = W(sinθ + μ cosθ).