Answer:
s = it+1/2 at²
s= 8×3+1/2 (10)(3)²
s = 24+45
s= 69
the object was thrown from a height of 69 meters
 
        
             
        
        
        
Answer:
A) - 1.8 m/s
Explanation:
As we know that whole system is initially at rest and there is no external force on this system
So total momentum of the system must be conserved
so we will have

now plug in all data into above equation



so correct answer is
A) - 1.8 m/s
 
        
             
        
        
        
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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