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phy127:hw5 [2011/03/01 21:24] mvfernandezserra |
phy127:hw5 [2011/03/01 22:37] (current) mvfernandezserra |
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+ | Hint: You need to evaluate in which situation will the current *I* be larger. Use P=V*I | ||

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+ | Hint: You need to use $I=P/V$, $P=I^2/R$. In this equation you see that the larger the voltage, the smaller the current. | ||

+ | A smaller current involves less dissipation in heat, and therefore less waste of power. Evaluate P1 and P2 for the two different currents. | ||

+ | ====80==== | ||

+ | Hint: You do not need to use the mass. It is only for distraction! | ||

+ | For part a you need to evaluate the horsepower of the car. Use P=F*V and convert the result to HP. | ||

+ | For part b you can do many things. What I would do is obtain the total charge of the battery.\\ | ||

+ | Qtotal = number of batteries* Q of 1 battery (note that 1 A h=3600 C).\\ | ||

+ | The total energy that the battery can deliver is Qtotal*V. Note that V is 12 (batteries in parallel).\\ | ||

+ | Then evaluate how long will the car run with this energy using that this energy is totally used in friction energy\\ | ||

+ | U$_{friction}$=F*X. And obtain *X*! | ||

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+ | You need to use $P=V^2/R$ in the two situations. | ||

+ | ====85==== | ||

+ | You need to use I=Q/t and l=v*t . $l=2*\pi*R$ . | ||

+ | From here you can obtain the total charge and then dive this total charge by the charge of 1 proton. |