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How much power is required to raise a 200-kg crate a vertical distance of 2 m in a time of 4 s?
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A 75kg person and 200kg crate are each parachuted tto earth from a plane which statement is true?
what is it you want to know, please be more specific
If you apply a force of 220 N to the lever how much force is applied to lift the crate?
If the perpendicular distance from the point of application of the force to the fulcrum is x metres and the perpendicular distance from the crate to the fulcrum is y metres, then the force applied on the crate is 220*x/y N.
If it took 5 seconds to lift the crate using a machine what was the power rating of the machine?
We have no way of knowing what power the machine was rated for, but with the information given in the question, we can calculate the power it delivered during the crate-lift: It was (1.96) x (mass of the crate in kilograms) x (distance the crate was lifted in meters) watts.
How much work is done in lifting a 50kg crate a vertical distance of 10 meters?
Potential Energy is given by the fourmulai PE=MGH where M=mass in kilos, G=the force of gravity in Netwons (9.8N) and H=height in meters. So 50*9.8*10=4900joules. A Watt is a unit of power that =1 joule per second. So 4900 joules divided by 5 seconds = 980 Watts, not allowing for losses due to friction etc.
If a worker pushes horizontally on a large crate with a force of 298 newtons and the crate moved 6.5 meters how much work do he do?
Work = force x distance = Newtons x meters = 1937 Joules.
What is the potential energy of a crate lifted with a force of 100 newtons a vertical distance of 2 meters?
Force x distance = 100 x 2 = 200 newton-meters = 200 joules.
If a 75kg person and a 200kg crate are each parachuted to earth from a plane which will hit the the ground first?
Provided that the parachute has the same surface area for both of the parachutist's, the 200kg man will hit the ground first due to the extra weight from the heavier man.
What is the net force of a 200-kg crate sitting at rest on a factory floor?
F=ma F=200kg*9.80665 m/s² = 1980 N
A 75kg person and 200kg crate are each parachuted tto earth from a plane which statement is true?
what is it you want to know, please be more specific
If you apply a force of 220 N to the lever how much force is applied to lift the crate?
If the perpendicular distance from the point of application of the force to the fulcrum is x metres and the perpendicular distance from the crate to the fulcrum is y metres, then the force applied on the crate is 220*x/y N.
If it took 5 seconds to lift the crate using a machine what was the power rating of the machine?
We have no way of knowing what power the machine was rated for, but with the information given in the question, we can calculate the power it delivered during the crate-lift: It was (1.96) x (mass of the crate in kilograms) x (distance the crate was lifted in meters) watts.
What is the power rating of a machine to lift a crate in 5 seconds?
That really depends on the weight of the crate. Also, on how high you want to lift it. Calculate the energy required to lift the crate with the formula for gravitational potential energy: PE = mgh (mass x gravity x height) Then divide this by the 5 seconds to get the minimum power required. (The actual power is somewhat larger, for various reasons - the initial acceleration required, and losses due to friction.)
How much work is done in lifting a 50kg crate a vertical distance of 10 meters?
Potential Energy is given by the fourmulai PE=MGH where M=mass in kilos, G=the force of gravity in Netwons (9.8N) and H=height in meters. So 50*9.8*10=4900joules. A Watt is a unit of power that =1 joule per second. So 4900 joules divided by 5 seconds = 980 Watts, not allowing for losses due to friction etc.
If a worker pushes horizontally on a large crate with a force of 298 newtons and the crate moved 6.5 meters how much work do he do?
Work = force x distance = Newtons x meters = 1937 Joules.
How much work is performed when a 25-kilogram crate is pushed 5 meters with a force of 15 newtons?
Work = (force) x (distance) = (15) x (5) = 75 joules.The mass of the crate is irrelevant.
A crate is at rest on the ground what force or forces are acting on the crate?
The weight of the crate is acting downward on the ground and the ground is exerting a force equal to the weight of the crate upward on the crate.
What is a collapsible crate used for?
A collapsible crate is used for transporting things which would be hard to get out of a regular crate. A collapsible crate can be made flat by folding out all of the sides of the crate.
