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RSB [31]
3 years ago
14

A 1451 kg car is traveling at 48.0 km/h. Compared to 48.0 km/h, what would be the fraction of energy the car would have if it we

re traveling at 24.0 km/h.
Physics
2 answers:
Rzqust [24]3 years ago
8 0
Ke = (1/2)(m)(v^2)
Ke = (1/2)(1451kg)(48^2)
Ke = 1,671,552 joules

Ke = (1/2)(1451)(24^2)
Ke = 417,888

1/4th the energy




Marat540 [252]3 years ago
3 0
We know, kinetic energy is directly proportional to the square of velocity, So, if kinetic energy is halved it's velocity will reduce to its 1/4th value

In short, Your Answer would be 1/4

Hope this helps!
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Answer:

32 turns

Explanation:

From the expression for the induced emf,

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E = emf = 24 V

N = number of turns = ?

B = magnetic field strength = 0.10 T

A = Cross sectional Area of the loop = 100 cm² = 0.01 m²

w = Angular speed = (2πf) = (2π × 120) = 754.3 rad/s

24 = N (0.1)(0.01)(754.3)

N = (24/0.7543)

N = 31.8 ≈ 32 turns.

Hope this Helps!!!

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