You do a certain amount of work on an object initially at rest, and all the work goes into increasing the object’s speed. If you
1 answer:
Answer:
v_{f2} = √ 2 v_{f}
Explanation:
Work and kinetic energy are related by
W = ΔK
The formula for work is
W = F . d
The formula for kinetic energy is
K = ½ m v²
W = - K₀
Since the body starts from rest, the initial velocity is ro and therefore the kinetic energy is zero.
W = K_{f}
W = ½ m v_{f}²
v_{f} = √ 2W / m
If W = 2 W₀
v_{f2} = √ 2 (2W₀) / m
v_{f2} = √ 2 √2W₀ / m
v_{f2} = √ 2 v_{f}
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Answer:
390 J
Explanation:
m = 3 kg
u = 16 i + 2 j
(a) Magnitude of velocity = = 16.1245 m/s
KEi = 1/2 m v^2 = 0.5 x 3 x 16.1245 = 390 J
(b) v = 18 i + 14 j
Magnitude of velocity = = 22.804 m/s
KEf = 1/2 m v^2 = 0.5 x 3 x 22.804 = 780 J
According to the work energy theorem
Work done = change in KE = KEf - KEi = 780 - 390 = 390 J
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