I am very sorry I don’t know
        
             
        
        
        
Answer:
Total momentum = 16 Kgm/s
Explanation:
Let the momentum of the two balls be A and B respectively. 
Momentum A = 16 kgm/s
Momentum B = 0 kgm/s (since the ball is at rest).
Total momentum = A + B
Total momentum = 16 + 0
Total momentum = 16 Kgm/s
Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.
Mathematically, momentum is given by the formula;
 
 
 
        
                    
             
        
        
        
Answer:
P = 2439.5 W = 2.439 KW
Explanation:
First, we will find the mass of the water:
Mass = (Density)(Volume)
Mass = m = (1 kg/L)(10 L)
m = 10 kg
Now, we will find the energy required to heat the water between given temperature limits:
E = mCΔT
where,
E = energy = ?
C = specific heat capacity of water = 4182 J/kg.°C
ΔT = change in temperature = 95°C - 25°C = 70°C
Therefore,
E = (10 kg)(4182 J/kg.°C)(70°C)
E = 2.927 x 10⁶ J
Now, the power required will be:

where,
t = time = (20 min)(60 s/1 min) = 1200 s
Therefore,

<u>P = 2439.5 W = 2.439 KW</u>
 
        
             
        
        
        
Answer:
magnitude of the frictional torque is 0.11 Nm
Explanation:
Moment of inertia I = 0.33 kg⋅m2
Initial angular velocity w° = 0.69 rev/s = 2 x 3.142 x 0.69 = 4.34 rad/s
Final angular velocity w = 0 (since it stops)
Time t = 13 secs
Using w = w° + §t
Where § is angular acceleration
O = 4.34 + 13§
§ = -4.34/13 = -0.33 rad/s2
The negative sign implies it's a negative acceleration. 
Frictional torque that brought it to rest must be equal to the original torque. 
Torqu = I x §
T = 0.33 x 0.33 = 0.11 Nm
 
        
             
        
        
        
Answer:
155.38424 K
2.2721 kg/m³
Explanation:
 = Pressure at reservoir = 10 atm
 = Pressure at reservoir = 10 atm
 = Temperature at reservoir = 300 K
 = Temperature at reservoir = 300 K
 = Pressure at exit = 1 atm
 = Pressure at exit = 1 atm
 = Temperature at exit
 = Temperature at exit
 = Mass-specific gas constant = 287 J/kgK
 = Mass-specific gas constant = 287 J/kgK
 = Specific heat ratio = 1.4 for air
 = Specific heat ratio = 1.4 for air
For isentropic flow

The temperature of the flow at the exit is 155.38424 K
From the ideal equation density is given by

The density of the flow at the exit is 2.2721 kg/m³