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The ramp does 480J of useful work with an efficiency of 80% .
<h3>What is efficiency of work done ?</h3>
- Efficiency is the ratio of the useful energy released by a system to the input energy .
- Mathematically, efficiency of energy = out put energy/ input energy
<h3>
What is the useful work done by the ramp having efficiency 80% and an input work done 600J?</h3>
- The efficiency =output work done/ input work done
- 80% =output work done/ 600J
- output work done =( 80×600)/100
=480J
Thus, we can conclude that the useful work done by the ramp is 480J.
Learn more about efficiency of energy here:
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Answer:
I = 9.82 10⁻⁷ W / m²
Explanation:
The intensity of the sound wave is the energy of the wave between the order per unit area of the same
I = P / A = E / T A
the energy is calculated by integrating the mechanical energy in a period, where the mass is changed by the density and ‘s’ is the amplitude of the sound wave
I = ½ ρ v (w s)²
I = ½ 1.35 328 (2π 530 2.00 10⁻⁸)²
I = 221.4 (4.435 10⁻⁹)
I = 9.82 10⁻⁷ W / m²
Answer:
Efficiency = 77%
Explanation:
Input energy = 570 J
Output energy = 440 J
To find the efficiency;
Substituting into the equation, we have;
Efficiency = 77.19 ≈ 77%
Therefore, the efficiency of the engine is 77 percent.
