Answer:
The maximum theoretical height that the pump can be placed above liquid level is 
Explanation:
To pump the water, we need to avoid cavitation. Cavitation is a phenomenon in which liquid experiences a phase transition into the vapour phase because pressure drops below the liquid's vapour pressure at that temperature. As a liquid is pumped upwards, it's pressure drops. to see why, let's look at Bernoulli's equation:

(
stands here for density,
for height)
Now, we are assuming that there aren't friction losses here. If we assume further that the fluid is pumped out at a very small rate, the velocity term would be negligible, and we get:


This means that pressure drop is proportional to the suction lift's height.
We want the pressure drop to be small enough for the fluid's pressure to be always above vapour pressure, in the extreme the fluid's pressure will be almost equal to vapour pressure.
That means:

We insert that into our last equation and get:

And that is the absolute highest height that the pump could bear. This, assuming that there isn't friction on the suction pipe's walls, in reality the height might be much less, depending on the system's pipes and pump.
You can find air pods that look real on letgo. or you can go to wish.com but if you want a good pair jus get the real ones
Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem
Answer:
il(t) = e^(-100t)
Explanation:
The current from the source when the switch is closed is the current through an equivalent load of 15 + 50║50 = 15+25 = 40 ohms. That is, it is 80/40 = 2 amperes. That current is split evenly between the two parallel 50-ohm resistors, so the initial inductor current is 2/2 = 1 ampere.
The time constant is L/R = 0.20/20 = 0.01 seconds. Then the decaying current is described by ...
il(t) = e^(-t/.01)
il(t) = e^(-100t) . . . amperes