# Differences

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 phy127:hw5 [2011/03/01 21:24]mvfernandezserra phy127:hw5 [2011/03/01 22:37] (current)mvfernandezserra Both sides previous revision Previous revision 2011/03/01 22:37 mvfernandezserra 2011/03/01 21:24 mvfernandezserra 2011/03/01 21:22 mvfernandezserra created 2011/03/01 22:37 mvfernandezserra 2011/03/01 21:24 mvfernandezserra 2011/03/01 21:22 mvfernandezserra created Line 4: Line 4: ====19==== ====19==== + ====20==== + ====38==== + Hint: You need to evaluate in which situation will the current *I* be larger. Use P=V*I + ====41==== + ====45==== + 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*! + ====83==== + 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.
phy127/hw5.txt · Last modified: 2011/03/01 22:37 by mvfernandezserra