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64. Integrated Concepts(a) Assuming 95.0$\%$ efficiency for the conversion of electrical power by the motor, what current must the $12.0-\mathrm{V}$ batteries of a 750 -kg electric car be able to supply: (a) To accelerate from rest to 25.0 $\mathrm{m} / \mathrm{s}$ in 1.00 $\mathrm{min}$ (b) To climb a $2.00 \times 10^{2}-\mathrm{m}$-high hill in 2.00 $\mathrm{min}$ at a constant $25.0-\mathrm{m} / \mathrm{s}$ speed while exerting $5.00 \times 10^{2} \mathrm{N}$ of force to overcome air resistance and friction? (c) To travel at a constant $25.0-\mathrm{m} / \mathrm{s}$ speed, exerting a $5.00 \times 10^{2} \mathrm{N}$ force to overcome air resistance and friction? See Figure 20.47.

(a) $I=343 \mathrm{A}$(b) $I=2.17 \mathrm{kA}$(c) $I=1.10 \mathrm{kA}$

Physics 102 Electricity and Magnetism

Chapter 20

Electric Current, Resistance, and Ohm's Law

Current, Resistance, and Electromotive Force

University of Washington

Hope College

University of Winnipeg

McMaster University

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So in this problem we have Ah, we have known that the battery power P B is i times v. Right? And, uh, because of the the transfer inefficiency, the actual the extra power acting on the car PC a poor PC p b times 95% No, 5% is the efficiency. And, uh, so we camped in that Ah, this single 0.95 PB so or not for 95 I time to be. So we camped in that I call pc be bye bye 0.95 times feet. Right. So, uh, we have three scenarios in this problem in this number one. So we accelerate the car from zero meter per second to 25 meters per second. Ah, a one minutes. So we want to determine the current in this case. So, uh, in this case, the total energy equal PC. The total energy that the car acquired is P C. Times T, and it is also equal to half. And the squid, right, So have m iss. M B square is the is the kind of energy increments of this car. So we have known the mass of the car and the V is 25 meters per second and the tea is one minute in one minutes so we can sew for PC in the after we got PC, We kempton the current, right? So from this equation, we know that peace equal half and this squared over tea. And so that's the current we can sew for it. And the current I got ihsaa 3 42 amperes. Okay. And for the second scenario Ah, oui. The car is plumbing on the incline and the hide only implies that 200 meters and uh so the car travel from the bottom to top in two minutes and the speed is constant. It's 25 meters per second and we also have 500 newtons to overcome the friction and air resistance. Okay, so the total the total Ah, the total energy on the car simply equal to P C. Times t and it also equal to ah ngh. Ngh is the, uh, the gravitational potential difference from the bottom to the top in the plus f times, uh, little distance, which is let's say l so f is the air resistance and the friction. So here every kwo 500 Newtons. And how is the total distance from the from the bottom to top. So to figure out this cell, we can utilize plus half times v times t right. So because because the velocity of this car is as constant is 25 meters per second and we also in the total time so we can figure out the lens. So in this equation, we console for PC peace equal MGH plus halftimes v times T Bye bye. Uh, t right. So, uh, in this solution, we already know the mass of the car and we also know h which is 200 meters and the friction is 500 of these 25 tease to two minutes. So this tea also is two minutes so yourself a pc Ah, and the current so that the current, because it becomes ski muc the current PC, the rabbi 0.95 Times week, right? So the final solution, I don't I obtain sa 2170 peers and in the third scenario, the car is simply moving on the cars on the plan and with the constant will also TV equal 25 meters per second and the friction air resistance F it's Ah, it's my smile. Face of 500 mutants. So, in this case, the power PC simply equal to at halftime, Zvi. Right. So, uh, we already know f and we also know be so we can sell for this PC. So the current equal PC over 0.95 time. See? And we can see that this sequel, uh, turn 96 amperes. Okay.

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