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Nearly 400 lbs,by 11.73%.
Answer:
W= -2.5 (p₁*0.0012) joules
Explanation:
Given that p₀= initial pressure, p₁=final pressure, Vi= initial volume=0 and Vf=final volume= 6/5 liters where p₁=p₀ then
In adiabatic compression, work done by mixture during compression is
W= where f= final volume and i =initial volume, p=pressure
p can be written as p=K/V^γ where K=p₀Vi^γ =p₁Vf^γ
W=
W= K/1-γ ( 1/Vf^γ-1 - 1/Vi^γ-1)
W=1/1-γ (p₁Vf-p₀Vi)
W= 1/1-1.40 (p₁*6/5 -p₀*0)
W= -2.5 (p₁*6/5*0.001) changing liters to m³
W= -2.5 (p₁*0.0012) joules
Answer:
Thus induced emf is 0.0531 V
Solution:
As per the question:
Diameter of the loop,
Thus the radius of the loop, R = 0.048 m
Time in which the loop is removed, t = 0.15 s
Magnetic field, B = 1.10 T
Now,
The average induced emf, e is given by Lenz Law:
where
= magnetic flux =
where
A = cross sectional area
Also, we know that:
e = 0.0531 V
The sketch is shown in the figure, where I indicates the direction of the induced current.