Answer:
E = 2940 J
Explanation:
It is given that,
Mass, m = 12 kg
Position at which the object is placed, h = 25 m
We need to find the potential energy of the mass. It is given by the formula as follows :
E = mgh
g is acceleration due to gravity

So, the potential energy of the mass is 2940 J.
Solution :
Given :
Water have quality x = 0.7 (dryness fraction) at around pressure of 200 kPa
The phase diagram is provided below.
a). The phase is a standard mixture.
b). At pressure, p = 200 kPa, T = 
Temperature = 120.21°C
c). Specific volume




d). Specific energy (
)



e). Specific enthalpy 
At 


f). Enthalpy at m = 0.5 kg


= 1022.91 kJ
Answer and Explanation:
Commutator are used in DC machine commutator is mainly used for the reversing the direction of the current .It is connected to the armature of the DC generator or motor
ADVANTAGE OF COMMUTATOR The main advantage of the commutator in DC motor is to keep keep the direction of the toque always in the same direction by changing the current direction
DISADVANTAGE OF COMMUTATOR : The main disadvantage is due to the friction between the commutator and brushes there is a friction loss.
Answer:
Q=67.95 W
T=119.83°C
Explanation:
Given that
For air
Cp = 1.005 kJ/kg·°C
T= 20°C
V=0.6 m³/s
P= 95 KPa
We know that for air
P V = m' R T
95 x 0.6 = m x 0.287 x 293
m=0.677 kg/s
For gas
Cp = 1.10 kJ/kg·°C
m'=0.95 kg/s
Ti=160°C ,To= 95°C
Heat loose by gas = Heat gain by air
[m Cp ΔT] for air =[m Cp ΔT] for gas
by putting the values
0.677 x 1.005 ( T - 20)= 0.95 x 1.1 x ( 160 -95 )
T=119.83°C
T is the exit temperature of the air.
Heat transfer
Q=[m Cp ΔT] for gas
Q=0.95 x 1.1 x ( 160 -95 )
Q=67.95 W
Answer:
The answer is "
and 157.5 MPa".
Explanation:
In point A:
The strength of its products with both the grain dimension is linked to this problem. This formula also for grain diameter of 310 MPA is represented as its low yield point

Here y is MPa is low yield point, x is mm grain size, and k becomes proportionality constant.
Replacing the equation for each condition:

People can get yo = 275 MPa with both equations and k= 15.5 Mpa
.
To substitute the answer,

In point b:
The equation is 
equation is:

by putting the above value in the formula we get the
value that is= 157.5 MPa