Answer:
A. We have that radius r = 4.00m intensity I = 8.00 W/m^
total power = power/ Area ( 4πr2)= 8.00 w/m^2( 4π ( 4.00 m)2=1607.68 W
b) I = total power/ 4πr2= 8.00 W/m2 ( 4.00 m/ 9.5 m)2= 1.418 W/m2
c) E = total power x time= 1607 . 68 W x 1s= 1607.68 J
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
Answer: Potassium iodide
Explanation: their you go
Explanation:
Gravitational potential energy
= mgh
= (2kg)(10N/kg)(5m)
= 100J.