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If an engine must do 500 joules of work to pull a 10 kg load up4 meters what is the engine's efficiency?
Wiki User
∙ 15y ago
energy required to lift up 10kg=10x9.81x4=392.4 joule. enegy produced by the engine=500 joule efficiency=392.4/500=78.48%
Wiki User
∙ 15y ago
Wiki User
∙ 13y ago0.80
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400
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joule/sec
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Joules or newtons*meters
A student does 300 joules of work pushing a cart 3 meters due east then does 400 joules of work pushing the cart 4 meters due north the total amount of work done by the student is what?
700 joules
If a engine must do 500 joules of work to pull a 10 kg load up to a height of 4 meters what is the efficiency of the engine?
The efficiency of the engine in this case is the ratio of the required work to create the change in Gravitational Potential energy to the Work done.In this case the work done is 500Jthe Change in potential energy is g*m*hh = change in height = 4mg = acceleration due to earth's gravitational field = 9.81 m/(s*s) (approx.)m = mass of load = 10kgDimensional analysis for g*m*h=PE(final) - PE(Initial)kg*(m/(s*s))*m = JComputation4m*9.81 m/(s*s)*10kg = 392.4J : This is our change in potential energyThe ratio of this to the work done by the engine is the efficiency. The efficiency is a unitless ratio since both the top and the bottom of the equation are in units of Joules.Change in PE / Work done by engine = efficiency ratio392.3J/500J = .7848The efficiency of the engine is .7848
How many joules of work would be accomplished if you move a 200 kg slug 10 meters?
2000 joules
What is the formula to find joules?
To find joules, you have to multiply the force or newtons by distance in meters.
Newtons times meters this is know as what?
joules
