Answer:
W= 8120 KJ
Explanation:
Given that
Process is isothermal ,it means that temperature of the gas will remain constant.
T₁=T₂ = 400 K
The change in the entropy given ΔS = 20.3 KJ/K
Lets take heat transfer is Q ,then entropy change can be written as
Now by putting the values
Q= 20.3 x 400 KJ
Q= 8120 KJ
The heat transfer ,Q= 8120 KJ
From first law of thermodynamics
Q = ΔU + W
ΔU =Change in the internal energy ,W=Work
Q=Heat transfer
For ideal gas ΔU = m Cv ΔT]
At constant temperature process ,ΔT= 0
That is why ΔU = 0
Q = ΔU + W
Q = 0+ W
Q=W= 8120 KJ
Work ,W= 8120 KJ
Answer:
modulus of elasticity for the nonporous material is 340.74 GPa
Explanation:
given data
porosity = 303 GPa
modulus of elasticity = 6.0
solution
we get here modulus of elasticity for the nonporous material Eo that is
E = Eo (1 - 1.9P + 0.9P²) ...............1
put here value and we get Eo
303 = Eo ( 1 - 1.9(0.06) + 0.9(0.06)² )
solve it we get
Eo = 340.74 GPa
Answer:
Electromechanical systems or devices are systems or devices that involves the interaction between electrical systems and mechanical systems in which the motion of mechanical parts is converted to electrical energy or made to interact with energy or in which electrical energy is converted to mechanical energy or interacts with a moving mechanical system
Therefore;
Electromechanical systems convert <u>electrical energy</u> input into a <u>mechanical energy</u> output
Explanation:
Answer:
Backdoor
Explanation:
The back door fragment in a program allows user to access backdoor information without necessarily following the common security procedures needed. In this case, once the programmer keys in the username he or she logs in without putting password. Therefore, this is a backdoor fragment.
