An electric motor with a 1200 hp output drives a machine with an efficiency of 60%. What is the energy output of the machine per
1 answer:
Answer:
536,904 J/s
Explanation:
The energy output from motor is the input energy in the machine.
We know that efficiency is percentage energy ouput to energy input, and expressed as
Where n and E represent efficiency and energy respectively, subscripts o and i represent output and input respectively. Since for the machine we have the input energy then the output will be the product of efficiency and input energy
Energy output=0.6*1200 hp=720 hp
Converting hp to J/s we multiply by 745.7
Energy is 720*745.7=536,904 J/s
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Answer:
169/4 ft/min.
Explanation:
See the attached picture.
The current drawn by the series circuit is(R₁ + R₂)/R₁R₂ times the current drawn by the parallel circuit.
Let the resistance of the two lamps are R₁ and R₂.
Then the equivalent resistance in series combination is: R = R₁ + R₂.
And, the equivalent resistance in parallel combination is:
r = R₁R₂/(R₁ + R₂).
So, if the supply voltage is V,
Then, current drown in series combination; = V/R = V/(R₁ + R₂)
And, current drown in parallel combination; = V/r = V(R₁ + R₂)/R₁R₂
So , = [ V/(R₁ + R₂)] /[V(R₁ + R₂)/R₁R₂]
= (R₁ + R₂)/R₁R₂
Hence, the ratio of current drawn in series and current drown in parallel is (R₁ + R₂)/R₁R₂. So, he current drawn by the series circuit is(R₁ + R₂)/R₁R₂ times the current drawn by the parallel circuit.
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