The work done by the heat engine is 40 kCal.
The given parameters;
- input heat of the engine, Q₁ = 70 kCal
- output heat of the engine, Q₂ = 30 kCal
To find:
- the work done by the heat engine
The work done by the heat engine is the change in the heat energy of the engine;
W = Q₂ - Q₁
Substitute the given parameters and solve work done (W)
W = 70 kCal - 30 kal
W = 40 kCal
Thus, the work done by the heat engine is 40 kCal.
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Answer:
2a.) Wavelength = 1.8 m
2b.) F = 66.67 Hz
3a.) Find the attached file
3b.) Wavelength = 0.6 m
Explanation:
Given that the
Length L = 0.9m
Wavelength (λ) = 2L/n
Where n = number of harmonic
If n = 1, then
Wavelength (λ) = 2L = 2 × 0.9 = 1.8 m
b.)
If waves travel at a speed of 120m/s on this string, what is the frequency
associated with the longest wave (first harmonic)?
Given that V = 120 m/s
V = Fλ
But λ = 2L, therefore,
F = V/2L
F = 120/1.8
F = 66.67 Hz
3. b.) If there are two node, the position will be in 3rd position which is 3rd harmonic
Using the same formula,
Wavelength (λ) = 2L/n
Where n = 3
Wavelength (λ) = 2 × 0.9/3
Wavelength (λ) = 0.6 m
Answer:

Explanation:
initial frequency of revolution is given as

now initial angular speed is



Similarly final angular speed is given as



Now angular acceleration is given as


