Im not 100% sure you have to tell me if im wrong or not.
D
B
C
Answer:
Explanation:
Given:
Steam Mass rate, ms = 1.5 kg/min
= 1.5 kg/min × 1 min/60 sec
= 0.025 kg/s
Air Mass rate, ma = 100 kg/min
= 100 kg/min × 1 min/60 sec
= 1.67 kg/s
A.
Extracting the specific enthalpy and temperature values from property table of “Saturated water – Pressure table” which corresponds to temperature at 0.07 MPa.
xf, quality = 0.9.
Tsat = 89.9°C
hf = 376.57 kJ/kg
hfg = 2283.38 kJ/kg
Using the equation for specific enthalpy,
hi = hf + (hfg × xf)
= 376.57 + (2283.38 × 0.9)
= 2431.552 kJ/kg
The specific enthalpy of the outlet, h2 = hf
= 376.57 kJ/kg
B.
Rate of enthalpy (heat exchange), Q = mass rate, ms × change in specific enthalpy
= ms × (hi - h2)
= 0.025 × (2431.552 - 376.57)
= 0.025 × 2055.042
= 51.37455 kW
= 51.38 kW.
Answer:
the answer would be 2
Explanation:
it would be 2 because if u look at the diagram the darkest arrow is pointsin towards earth and the moon and when the moon is infront of the sun it cause's an eclispe
Answer:
8000J
Explanation:
The kinetic energy of the car lost during breaking are converted to thermal energy and are gained by the brakes.
Kinetic energy loss by car = thermal energy gained by brakes.
∆K.E = ∆T.E ....1
The Kinetic energy loss by car can be expressed as;
∆K.E = K.E1 - K.E2
Initial K.E = K.E1 = 10000J
Final K.E = K.E2 = 2000J
∆K.E= 10000J - 2000J = 8000J
From equation 1,
∆K.E = ∆T.E
∆T.E = 8,000J
thermal energy gain by brakes = 8,000J
Answer:
19.6m/s
Explanation:
A Rock falling off a cliff can be modeled as an object starting with zero velocity moves with constant acceleration for certain period of time, for such motion following equation of motion can be used.
here in our case 
because object starts off from rest and 
is acceleration because of gravity ( Motion under gravity).
and of course t = 2 second.
Now by substituting all this information in equation of motion we get.
that would be the velocity of rock as it would hit the ground.
Note! We have assumed that there is no air resistance.
A rock falling off a cliff can be modeled as an object starting with zero velocity moves with constant acceleration for a certain period of time, for such motion following equation of motion can be used.
here in our case because object starts off from rest and is acceleration because of gravity ( Motion under gravity).
and of course t = 2 seconds.
Now by substituting all this information in equation of motion we get.
V = 19.6m/s
that would be the velocity of rock as it would hit the ground.
Note! We have assumed that there is no air resistance.
