Both A and B technicians are correct because both might be used to test fuses, according to technician B.
<h3>What is continuity?</h3>
The behavior of a function at a certain point or section is described by continuity. The limit can be used to determine continuity.
From the question:
We can conclude:
The technician claims that you may check for continuity using both an ohmmeter and a self-powered test light. Both might be used to test fuses, according to technician B.
Thus, both A and B technicians are correct because both might be used to test fuses, according to technician B.
Technician A says both an ohmmeter and a self-powered test light may be used to test for continuity. Technician B says both may be used to test fuses. Who is correct?
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Answer:
a)-True
Hope this helps even tho theres no school right now
Answer:
Explanation:
Work, U, is equal to the force times the distance:
U = F · r
Force needed to lift the weight, is equal to the weight: F = W = m · g
so:
U = m · g · r
= 20.4kg · 9.81 
· 1.50m
= 35.316 
= 35.316 W
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.
Answer:
The work of the cycle.
Explanation:
The area enclosed by the cycle of the Pressure-Volume diagram of a Carnot engine represents the net work performed by the cycle.
The expansions yield work, and this is represented by the area under the curve all the way to the p=0 line. But the compressions consume work (or add negative work) and this is substracted fro the total work. Therefore the areas under the compressions are eliminated and you are left with only the enclosed area.
