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You need to get $p_4-p_3$. Taking the datum of elevation z as that of points 3 and 4, we have $$p_{atm}+(10)\rho g=p_2+\frac{1}{2}\rho v^2$$ $$p_3+\frac{1}{2}\rho v^2=p_2+\frac{1}{2}\rho v^2$$ $$p_{atm}+(120)\rho g=p_5+\frac{1}{2}\rho v^2+(120-h)\rho g$$where h is the depth of point 5 below the surface of the tank on the right. $$p_4+\frac{1}{2}\rho v^2=... 0 There are some small losses, but for saving electricity, turning off the appliances when not in use on a massive scale can be a great help for saving electricity. 2 Your error is to assume that only your red charges generate the heat, ie the red charges go through area A and they are not replaced by any other charges. If that were the case then the factor of \frac 12 would be correct. However as the red charges move through the resistor black charges to the left of the red charges would move into the resistor and ... 1 Since you equate W with NqU, that means W represents the amount of energy dissipated as heat during the time interval N charges passed through the cross section. That time interval is t / 2, so the resulting power is P = {{UIt / 2} \over {t / 2}} = UI. 0 I think it does matter where you place the pump. Regardless of how powerful the pump is, the minimum pressure it can create at the input is zero - a perfect vacuum. The pressure pushing the fluid up to the pump is then the atmospheric pressure. If the fluid is water, the maximum distance the atmosphere can pump it up to is about 10m. So the pump must ... 3 In the picture the pump is located ad height BB, but, if it was located at AA or CC, would something change? That is, would the fluid have different speed at the top when it flows out of the tube? In a nutshell: no. Bernoulli's equation, between the points 1 and 2, is as follows, where p is the pressure supplied by the pump:$$p_1+\frac{1}{2} \...

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This is my understanding: You already have $V = IR$, and take note that $R$ is constant because you are not adding any more resistor. When you lower the voltage by a factor of 2, you could lower the $I$ and keep the $R$ constant, or you can lower $R$ and keep $I$ constant. The latter is not possible, so the only way for us to decrease the voltage is to also ...

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I think there is a misunderstanding. What the article tells you is that if you change either the cuurent or the voltage by a factor of 2 then the power changes by a factor of 4.And if you change both of them by a factor of 2 then the power changes by a factor of 16.First equation is $P=VI$, if you change $I$ from $I$ to $2I$ the voltage $V$ also becomes $2V$...

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