Answer:
One
For surface-mounted and pendant-hung luminaires, support rods should be placed so that they extend about _one___
<h3>what is supported mounted?</h3>
- A structure that holds up or serves as a foundation for something else. Support is a synonym for mounting.
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Answer:
The process from a liquid to a vapor.
Explanation:
Water is evaporated by heating up and turning into a vapor.
Answer:
a) m=336.18N
b) Vn=16.67m/kmol
Vm=0.1459m^3/kg
Explanation:
To calculate the mass of the octane(m):
Number of mole of octane (n) =0.3kmol(given)
Molarmass of octane (M) =114.23kg/kmol
m=n*M
m=(0.3kmol)*(114.23kg/kmol)
m=34.269kg
To calculate for the weight of octane(W):
W=g*m
W=(9.81m/s^2)*(34.269kg)
W=336.18N
b) For specific volumes of Vn and Vm:
Given volume of octane (V) =5m^3
Vm=V/m
Vm=5m^3/34.269kg
Vm=0.1459m^3/kg
And Vn will be :
Vn=V/m=5m^3/0.3kmol
Vn=16.67m/Kmol
Therefore, the answers are:
a) m=336.18N
b) Vn=16.67m/kmol
Vm=0.1459m^3/kg
Answer:
h = 10,349.06 W/m^2 K
Explanation:
Given data:
Inner diameter = 3.0 cm
flow rate = 2 L/s
water temperature 30 degree celcius




at 30 degree celcius 

Re = 106390
So ,this is turbulent flow



SOLVING FOR H
WE GET
h = 10,349.06 W/m^2 K
Answer:
The diameter is 50mm
Explanation:
The answer is in two stages. At first the torque (or twisting moment) acting on the shaft and needed to transmit the power needs to be calculated. Then the diameter of the shaft can be obtained using another equation that involves the torque obtained above.
T=(P×60)/(2×pi×N)
T is the Torque
P is the the power to be transmitted by the shaft; 40kW or 40×10³W
pi=3.142
N is the speed of the shaft; 250rpm
T=(40×10³×60)/(2×3.142×250)
T=1527.689Nm
Diameter of a shaft can be obtained from the formula
T=(pi × SS ×d³)/16
Where
SS is the allowable shear stress; 70MPa or 70×10⁶Pa
d is the diameter of the shaft
Making d the subject of the formula
d= cubroot[(T×16)/(pi×SS)]
d=cubroot[(1527.689×16)/(3.142×70×10⁶)]
d=0.04808m or 48.1mm approx 50mm