The efficiency of the device is 30 %
Explanation:
The efficiency of a heat engine is given by:
where
W is the work done by the engine
is the heat in input to the engine
For the device in this problem, we have:
W = 120 J is the work done
is the heat in input
Substituting, we find the efficiency:
which corresponds to an efficiency of 30%.
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Answer:
2,100J
Explanation:
Workdone=Force x distance
=700 x 3
=2,100 J
Answer:
a) 1.3 rad/s
b) 0.722 s
Explanation:
Given
Initial velocity, ω = 0 rad/s
Angular acceleration of the wheel, α = 1.8 rad/s²
using equations of angular motion, we have
θ2 - θ1 = ω(0)[t2 - t1] + 1/2α(t2 - t1)²
where
θ2 - θ1 = 53.2 rad
t2 - t1 = 7s
substituting these in the equation, we have
θ2 - θ1 = ω(0)[t2 - t1] + 1/2α(t2 - t1)²
53.2 =ω(0) * 7 + 1/2 * 1.8 * 7²
53.2 = 7.ω(0) + 1/2 * 1.8 * 49
53.2 = 7.ω(0) + 44.1
7.ω(0) = 53.2 - 44.1
ω(0) = 9.1 / 7
ω(0) = 1.3 rad/s
Using another of the equations of angular motion, we have
ω(0) = ω(i) + α*t1
1.3 = 0 + 1.8 * t1
1.3 = 1.8 * t1
t1 = 1.3/1.8
t1 = 0.722 s
Answer: The correct answer is "Instrument A is placed closer to Sam than instrument B".
Explanation:
The sound can be soft or loud. Loudness depends on the amplitude of the sound wave. Higher the amplitude, more will be loudness. Lower the amplitude, lesser will be loudness.
Pitch depends on the frequency.
In the given problem, the instruments A and B generate sound waves of the same amplitude and at the same time.
Loudness depends on the sound energy produced as the energy of the sound is directly proportional to the square of the amplitude. It also depends on the distance between the source and the receiver.
Sam records a louder sound from instrument A than from instrument B. It means that there is mismatch in loudness. It can happen due to the placement of the instrument A closer to Sam than instrument B.
Therefore, the correct option is "Instrument A is placed closer to Sam than instrument B".
