That is a thread ball valves
technician a uses a current output test to check ac generator output. technician b uses a voltage output test to check output. w
1 answer:
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Answer:
Explanation:
For pressure gage we can determine this by saying:
The closed tank with oil and air has a pressure of P₁ and the pressure of oil at a certain height in the U-tube on mercury is p₁gh₁. The pressure of mercury on the air in pressure gauge is p₂gh₂. The pressure of the gage is P₂.
We want to work out P₁-P₂: Heights aren't given so we can solve it in terms of height: assuming h₁=h₂=h
Answer:
a) 2,945 mC
b) P(t) = -720*e^(-4t) uW
c) -180 uJ
Explanation:
Given:
i (t) = 6*e^(-2*t)
v (t) = 10*di / dt
Find:
( a) Find the charge delivered to the device between t=0 and t=2 s.
( b) Calculate the power absorbed.
( c) Determine the energy absorbed in 3 s.
Solution:
- The amount of charge Q delivered can be determined by:
dQ = i(t) . dt
- Integrate and evaluate the on the interval:
- The power can be calculated by using v(t) and i(t) as follows:
v(t) = 10* di / dt = 10*d(6*e^(-2*t)) /dt
v(t) = 10*(-12*e^(-2*t)) = -120*e^-2*t mV
P(t) = v(t)*i(t) = (-120*e^-2*t) * 6*e^(-2*t)
P(t) = -720*e^(-4t) uW
- The amount of energy W absorbed can be evaluated using P(t) as follows:
- Integrate and evaluate the on the interval: