Answer:
1.5 m
Explanation:
Let the distance from the box to the pivot be c.
Let the distance from the pivot to the effort be y.
From the question given above, the following data were obtained:
Effort force (Fₑ) = 7 N
Force of resistance (Fᵣ) = 14 N
Distance from the box to the pivot (c) = 0.75 m
Distance from the pivot to the effort (y) =?
Clockwise moment = Fₑ × y
Anticlock wise moment = Fᵣ × c
Clockwise moment = Anticlock wise moment
Fₑ × y = Fᵣ × c
7 × y = 14 × 0.75
7 × y = 10.5
Divide both side by 7
y = 10.5 / 7
y = 1.5 m
Therefore, the distance from the pivot to the effort is 1.5 m
Complete question:
The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.
Answer:
The exit velocity is 629.41 m/s
Explanation:
Given;
initial temperature, T₁ = 1200K
initial pressure, P₁ = 150 kPa
final pressure, P₂ = 80 kPa
specific heat at 300 K, Cp = 1004 J/kgK
k = 1.4
Calculate final temperature;
k = 1.4
Work done is given as;
inlet velocity is negligible;
Therefore, the exit velocity is 629.41 m/s
It IS <span>PE = (1200 kg)(9.8 m/s²)(42 m) = 493,920 J </span>
Answer: $4,500
Explanation: 1% of $30,000 would be $300. If you multiply that by 15, you get $4,500, so 15% of 30,000 is 4,500.
The answer to the question would be jet streams.
