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An all-electric car (not a hybrid) is designed to run from a bank of 12.0 V batteries with total energy storage of 2.30 ✕ 107 J. (a) If the electric motor draws 8.80 kW as the car moves at a steady speed of 20.0 m/s, what is the current (in A) delivered to the motor? A (b) How far (in km) can the car travel before it is "out of juice"? km (c) What If? The headlights of the car each have a 55.0 W halogen bulb. If the car is driven with both headlights on, how much less will its range be (in m)? m

Respuesta :

Answer:

a) [tex]I=733.33\ A[/tex]

b) [tex]d=52272.7273\ m[/tex]

c) [tex]d'=51948.0519\ m[/tex]

Explanation:

Given:

  • voltage of the battery, [tex]V=12\ V[/tex]
  • energy storage capacity of the battery, [tex]E=2.3\times 10^7\ J[/tex]
  • speed of the car, [tex]v=20\ m.s^{-1}[/tex]

a)

power drawn by the car, [tex]P=8.8\ kW[/tex]

Now the Current delivered to the motor:

we the relation between the power and electrical current,

[tex]P=V.I[/tex]

[tex]8800=12\times I[/tex]

[tex]I=733.33\ A[/tex]

b)

Distance travelled before battery is out of juice:

we first find the time before the battery runs out,

[tex]t=\frac{E}{P}[/tex]

[tex]t=\frac{2.3\times 10^7}{8800}[/tex]

[tex]t=2613.636\ s[/tex]

Now the distance:

[tex]d=v.t[/tex]

[tex]d=20\times 2613.636[/tex]

[tex]d=52272.7273\ m[/tex]

c)

When the head light of 55 W power is kept on while moving then the power   consumption of the car is:

[tex]P'=P+55[/tex]

[tex]P'=8800+55[/tex]

[tex]P'=8855\ W[/tex]

Now the time of operation of the car:

[tex]t'=\frac{E}{P'}[/tex]

[tex]t'=\frac{2.3\times 10^7}{8855}[/tex]

[tex]t'=2597.4026\ s[/tex]

Now the distance travelled:

[tex]d'=v.t'[/tex]

[tex]d'=20\times 2597.4025[/tex]

[tex]d'=51948.0519\ m[/tex]

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