Kinetic and static friction are both resistive forces
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
If an object that is 86kg on the moon then that is the answer, 86kg.
Answer:
563.86 N
Explanation:
We know the buoyant force F = weight of air displaced by the balloon.
F = ρgV where ρ = density of air = 1.29 kg/m³, g = acceleration due to gravity = 9.8 m/s² and V = volume of balloon = 4πr/3 (since it is a sphere) where r = radius of balloon = 2.20 m
So, F = ρgV = ρg4πr³/3
substituting the values of the variables into the equation, we have
F = 1.29 kg/m³ × 9.8 m/s² × 4π × (2.20 m)³/3
= 1691.58 N/3
= 563.86 N