Assuming 95.0% efficiency for the conversion of electrical power by the motor, what current must the 12.0-V batteries of a 728-kg electric car be able to supply to do the following?
(a) accelerate from rest to 25.0 m/s in 1.00 min
A
(b) climb a 200-m high hill in 2.00 min at a constant 25.0 m/s
speed while exerting 443 N of force to overcome air resistance and
friction
A
3)gain in energy in 1 min ( 60 sec ) = mv^2 /2
Also, Power required = E/t
= 0.5 mv^2 / t
= 0.5*728*25^2/60
= 3791.67 W
Since efficiency for the conversion of electrical power by the motor = 95.0%
Therefore , P = 0.95 VI
3791.67 = 0.95*12*I
I = 332.6 A
so the current is 332.6 A
b)
height,h = 200 m
L = 120 sec x 25 = 3000 m
Work done,W = 3000 x 443 + 728 x 9.81 x 200 = 2757336 J
P = rate of doing work = W/t = 22977.8
12 * I * 0.95 =22977.8
I = 2015.596 A
Assuming 95.0% efficiency for the conversion of electrical power by the motor, what current must the...
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