It is known that the connecting rod AB exerts on the crank BCa 2.5-kN force directed down andto the left along the centerline of
AB. Determine the moment of this force about C for the two casesshown.
1 answer:
Answer:
M_c = 61.6 Nm
Explanation:
Given:
F_a = 2.5 KN
Find:
Determine the moment of this force about C for the two casesshown.
Solution:
- Draw horizontal and vertical vectors at point A.
- Take moments about point C as follows:
M_c = F_a*( 42 / 150 ) *88
M_c = 2.5*( 42 / 150 ) *88
M_c = 61.6 Nm
- We see that the vertical component of force at point A passes through C.
Hence, its moment about C is zero.
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Answer:
Check the explanation
Explanation:
Energy alance of 2 closed systems: Heat from CO2 equals the heat that is added to air in
1x0.723x=3x0.780x
⇒
= 426.4 °K
The initail volumes of the gases can be determined by the ideal gas equation of state,
=
= 0.201
The equilibrium pressure of the gases can also be obtained by the ideal gas equation
= 1x(8.314 28.97)x426.4+3x(8.314 44)x426.4
(0.201+1.275)
= 246.67 KPa = 2.47 bar
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