How does uniform motion relate to velocity and acceleration?
1 answer:
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The work done is 400 J
Explanation:
The work done by you in pushing the box along the slope is given by
where
F is the magnitude of the force applied
d is the distance covered by the box along the slope
Here we have the following:
F = 200 N is the magnitude of the force applied
is the distance covered, where
h = 1 m is the vertical rise
is the slope of the plane
Substituting and solving, we find
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Answer:
a)P₂ =4 bar
b)W= - 1482.48 KJ
It means that work done on the system.
c)S₂ - S₁ = 3.42 KJ/K
Explanation:
Given that
T₁ = 300 K ,V₁ = 3 m³ ,P₁=2 bar
T₂ = 600 K ,V₂=V₁ 3 m³
Given that tank is rigid and insulated.It means that volume of the gas will remain constant.
Lets take the final pressure = P₂
For ideal gas P V = m R T
P₂ =4 bar
Internal energy
ΔU = m Cv ΔT
Cv=0.71 KJ/kg.k for air
m= 6.96 kg
ΔU= 6.96 x 0.71 x (600 - 300)
ΔU=1482.48 KJ
From first law
Q= ΔU + W
Q= 0 Insulated
W = - ΔU
W= - 1482.48 KJ
It means that work done on the system.
Change in the entropy
S₂ - S₁ = 3.42 KJ/K