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patriot [66]
2 years ago
8

A particular electric car is supplied with 300 kJ of chemical energy by the battery. Of this, a total of 70.5 kJ of energy is wa

sted as heat.
Calculate the overall efficiency of the electric car.
Physics
1 answer:
Lorico [155]2 years ago
3 0

Supplied energy=300kJ

  • Wasted energy=70.5J

Used energy:-

\\ \sf\longmapsto 300-70.5=229.5kJ

We know

\boxed{\sf Efficiency=\dfrac{Used\:Energy}{Supplied\:Energy}\times 100}

\\ \sf\longmapsto Efficiency=\dfrac{229.5}{300}\times 100

\\ \sf\longmapsto Efficiency=\dfrac{229.5}{3}

\\ \sf\longmapsto Efficiency=76.5\%

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Answer:

Part a)

v = -(8.33\hat j + 9.33\hat i)\times 10^6 m/s

Part b)

E = 4.4 \times 10^{-13} J

Explanation:

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So we will have

m_1v_1 + m_2v_2 + m_3v_3 = 0

(5 \times 10^{-27})(6 \times 10^6\hat j) + (8.4 \times 10^{-27})(4 \times 10^6\hat i) + (3.6 \times 10^{-27}) v = 0

(30\hat j + 33.6\hat i)\times 10^6 + 3.6 v = 0

v = -(8.33\hat j + 9.33\hat i)\times 10^6 m/s

Part b)

By equation of kinetic energy we have

E = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 + \frac{1}{2}m_3v_3^2

E = \frac{1}{2}(5 \times 10^{-27})(6\times 10^6)^2 + \frac{1}{2}(8.4 \times 10^{-27})(4 \times 10^6)^2 + \frac{1}{2}(3.6 \times 10^{-27})(8.33^2 + 9.33^2) \times 10^{12}

E = 9\times 10^{-14} + 6.72 \times 10^{-14} + 2.82\times 10^{-13}

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3 years ago
An electric current in a metal consists of moving
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A car drives around a curve with radius 400 m at a speed of 32 m/s. The road is banked at 7.0 degree. The mass of the car is 150
Ronch [10]

Answer:

The magnitude of the centripetal force to make the turn is 3,840 N.

Explanation:

Given;

radius of the cured road, r = 400 m

speed of the car, v = 32 m/s

mass of the car, m = 1500 kg

The magnitude of the centripetal force to make the turn is given as;

F_c = \frac{mv^2}{r}

where;

Fc is the centripetal force

F_c = \frac{mv^2}{r} \\\\F_c = \frac{(1500)(32)^2}{400}\\\\F_c = 3,840 \ N

Therefore, the magnitude of the centripetal force to make the turn is 3,840 N.

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