Answer:
The temperature T= 648.07k
Explanation:
T1=input temperature of the first heat engine =1400k
T=output temperature of the first heat engine and input temperature of the second heat engine= unknown
T3=output temperature of the second heat engine=300k
but carnot efficiency of heat engine =
where Th =temperature at which the heat enters the engine
Tl is the temperature of the environment
since both engines have the same thermal capacities <em>
</em> therefore 
We have now that
multiplying through by T
multiplying through by 300
-
The temperature T= 648.07k
Answer:
S = 5.7209 M
Explanation:
Given data:
B = 20.1 m
conductivity ( K ) = 14.9 m/day
Storativity ( s ) = 0.0051
1 gpm = 5.451 m^3/day
calculate the Transmissibility ( T ) = K * B
= 14.9 * 20.1 = 299.5 m^2/day
Note :
t = 1
U = ( r^2* S ) / (4*T*<em> t </em>)
= ( 7^2 * 0.0051 ) / ( 4 * 299.5 * 1 ) = 2.0859 * 10^-4
Applying the thesis method
W(u) = -0.5772 - In(U)
= 7.9
next we calculate the pumping rate from well ( Q ) in m^3/day
= 500 * 5.451 m^3 /day
= 2725.5 m^3 /day
Finally calculate the drawdown at a distance of 7.0 m form the well after 1 day of pumping
S = 
where : Q = 2725.5
T = 299.5
W(u) = 7.9
substitute the given values into equation above
S = 5.7209 M
Answer:
24.72 kwh
Explanation:
Electric energy=potential energy=mgz where m is mass, g is acceleration due to gravity and z is the elevation.
Substituting the given values while taking g as 9.81 and dividing by 3600 to convert to per hour we obtain
PE=(108*9.81*84)/3600=24.72 kWh
Answer:
small guitar with no strings?
Explanation:
it would be fun to make i think
Answer:
Reduce manufacturing costs.
Explanation:
Hope This Helps
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