Answer:
Connect the test light in series with the negative post, and start pulling feed wires. The first to check is the heavy charging wire from the alternator. A bad or leaky diode in an alternator is a very common source of overnight battery drain. Connect wires one at a time to see what lead is drawing current.
Answer:
G = $37,805.65
Explanation:
I found this on another site:
475,000 = 25,000(P/A,10%,6) + G(P/G,10%,6)
475,000 = 25,000(4.3553) + G(9.6842)
9.6842G = 366,117.50
G = $37,805.65
Answer:
F = 0.0022N
Explanation:
Given:
Surface area (A) = 4,000mm² = 0.004m²
Viscosity = µ = 0.55 N.s/m²
u = (5y-0.5y²) mm/s
Assume y = 4
Computation:
F/A = µ(du/dy)
F = µA(du/dy)
F = µA[(d/dy)(5y-0.5y²)]
F = (0.55)(0.004)[(5-1(4))]
F = 0.0022N
Answer:
b) false
Explanation:
We know that Otto cycle is the ideal cycle for all petrol working engine.In Otto cycle all process are consider is ideal ,means there is no any ir-reversibility in the processes.
It consist four processes
1-2:Reversible adiabatic compression
2-3:Constant volume heat addition
3-4:Reversible adiabatic expansion
3-4:Constant volume heat rejection
Along with above 4 processes intake and exhaust processes are parallel to each other.From the P-v diagram we can see that all processes.
But actually in general we are not showing intake and exhaust line then it did not mean that in Otto cycle did not have intake and exhaust processes.