Answer:
ΔδB = 5.7 m
Explanation:
Given
mA = 20 Kg
mB = 30 Kg
vA = 3 m/s
vA0 = vB0 = 0 m/s (the blocks released from rest)
Since the cable length (L) is constant
3*δA + δB = L
3*ΔδA = - ΔδB
If we apply
d(3*δA + δB)/dt = dL/dt
3*vA + vB = 0 ⇒ 3*vA = - vB
vB = -3*(3 m/s) ⇒ vB = - 9 m/s
We apply Conservation of Energy
TA0 + TB0 + vA0 + vB0 = TA + TB + vA + vB
0 + 0 + 0 + 0 = 0.5*mA*vA² + 0.5*mB*vB² + WA*ΔδA + WB*ΔδB
0 = 0.5*mA*vA² + 0.5*mB*(-3*vA)² + WA*(ΔδB/3) - WB*ΔδB
0 = 0.5*20 Kg*(3 m/s)² + 0.5*30 Kg*(-9 m/s)² + (20 Kg*9.81 m/s²)*(ΔδB/3) - (30 Kg*9.81 m/s²)*ΔδB
⇒ ΔδB = 5.7 m
The pic shown can help us to understand the question.
Answer:
576.21kJ
Explanation:
#We know that:
The balance mass
so,
#Also, given the properties of water as;
#We assume constant properties for the steam at average temperatures:
#Replace known values in the equation above;
#Using the mass and energy balance relations;
#We have : we replace the known values in the equation as;
#Hence,the amount of heat transferred when the steam temperature reaches 500°C is 576.21kJ
Answer:
minimum flow rate provided by pump is 0.02513 m^3/s
Explanation:
Given data:
Exit velocity of nozzle = 20m/s
Exit diameter = 40 mm
We know that flow rate Q is given as
where A is Area
minimum flow rate provided by pump is 0.02513 m^3/s
