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kenny6666 [7]
4 years ago
5

A 1500-kg car accelerates from 0 to 25 m/s in 7.0 s. What is the average power delivered by the engine? (1 hp = 746 W) A) 50 hp

B) 60 hp C) 70 hp D) 80 hp E) 90 hp
Physics
1 answer:
sergeinik [125]4 years ago
6 0

Answer:

power, P = 90 hp

Explanation:

It is given that,

Mass of the car, m = 1500 kg

Initial velocity of car, u = 0

Final velocity of car, v = 25 m/s

Time taken, t = 7 s

We need to find the average power delivered by the engine. Work done divided by total time taken is called power delivered by the engine. It is given by :

P=\dfrac{W}{t}

According to work- energy theorem, the change in kinetic energy of the energy is equal to work done i.e.

W=\Delta E=\dfrac{1}{2}m(v^2-u^2)

P=\dfrac{\dfrac{1}{2}m(v^2-u^2)}{t}

P=\dfrac{\dfrac{1}{2}\times 1500\ kg\times (25\ m/s)^2}{7\ s}

P = 66964.28 watts

Since, 1 hp = 746 W

So, P = 89.76 hp

or

P = 90 hp

So, the average power delivered by the engine is 90 hp. Hence, the correct option is (E) " 90 hp".                                                                                                  

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A uniform magnetic field of magnitude 0.23 T is directed perpendicular to the plane of a rectangular loop having dimensions 7.8
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Answer:

0.0025116weber/m²

Explanation:

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3 0
3 years ago
Read 2 more answers
A top fuel dragster with a mass of 500.0 kg starts from rest and completes a quarter mile (402 m) race in a time of 5.0 s. The d
blsea [12.9K]

The average power needed to produce this final speed is 1069.1 hp.

Mass of the dragster,  m = 500.0 kg,

Displacement travelled by the dragster,  s = 402 m,

Time taken in this travel,  t = 5.0 s,

Final velocity of the dragster,  v = 130 m/s.

Let the initial velocity of the dragster be u and acceleration be a.

Using kinematical equation,  s = ut + (1/2)at^2.

402  =  u*5  + (1/2)*a*5^2

10*u + 25*a  = 804.      ...........(1)

Using kinematical equation, v = u +at.

130 = u + 5*a

5*u + 25*a = 650.       .............(2)

Solving (1)and (2), we get,

u =  30.8 m/s.

According to work-energy theorem,

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W = (1/2)*500*(130^2 - 30.8^2)

W  =  3987840. J

Therefore power rating of the dragster is given by,

P  ⇒  W/t. =  3987840/5 = 797568 watt.

P  ⇒ 797568/746 =  1069.1 hp.

Learn more about Power rating here brainly.com/question/20137708

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