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.
Learn more here: brainly.com/question/4280097
Force
Newton, abbreviated as N
Answer:
ω₂ = 93.6 rev / min
Explanation:
ω₀ = 260 rev / min
ω₁ = 0.68*ω₀ = 0.68*(260 rev / min) = 176.8 rev / min
ω₂ = ?
t₁ = 1 min
t₂ = 2 min
We can apply the equation:
ω₁ = ω₀ + α*t₁ ⇒ α = (ω₁ - ω₀) / t₁
⇒ α = (176.8 rev / min - 260 rev / min) / 1 min = - 83.2 rev / min²
then we can use the same formula, knowing the angular acceleration:
ω₂ = ω₀ + α*t₂ ⇒ ω₂ = (260 rev / min) + (- 83.2 rev / min²)*(2 min)
⇒ ω₂ = 93.6 rev / min
The watt<span> (symbol: W) is a unit of power i hope this helps you</span>
