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svetlana [45]
3 years ago
10

A car with a mass of 1.50x10^3 kg starts from rest and accelerates to a speed of 18.0m/s in 12.0 s. assume that the force of res

istance remains constant 400.0n during this time. what is the average power developed by the car's engine.
Physics
1 answer:
Luden [163]3 years ago
8 0
The first thing you should know for this case is the definition of distance.
 d = v * t
 Where,
 v = speed
 t = time
 We have then:
 d = v * t
 d = 9 * 12 = 108 m
 The kinetic energy is:
 K = ½mv²
 Where,
 m: mass
 v: speed
 K = ½ * 1500 * (18) ² = 2.43 * 10 ^ 5 J
 The work due to friction is
 w = F * d
 Where,
 F = Force
 d = distance:
 w = 400 * 108 = 4.32 * 10 ^ 4
 The power will be:
 P = (K + work) / t
 Where,
 t: time
 P = 2.86 * 10 ^ 5/12 = 23.9 kW
 answer:
 the average power developed by the engine is 23.9 kW
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Answer:

ax = 2.60m/s^{2}, t = 26.92s

Explanation:

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(vx)f^{2} = (vx)i^{2} + 2ax \Lambda x (1)

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2ax \Lambda x = (vx)f^{2} - (vx)i^{2}

ax = \frac{(vx)f^{2} - (vx)i^{2}}{2 \Lambda x}  (2)

Since the plane starts from rest, its initial velocity will be zero ((vx) = 0):

Replacing the values given in equation 2, it is gotten:

ax = \frac{(70m/s)^{2} - (0m/s)^{2}}{2(940m)}

ax = \frac{4900m/s}{2(940m)}

ax = \frac{4900m/s}{1880m}

ax = 2.60m/s^{2}

So, The acceleration of the plane is 2.60m/s^{2}    

Now that the acceleration is known, the next equation can be used to find out the time:

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Rewritten equation (3) in terms of t:

t = \frac{(vx)f - (vx)i}{ax}

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t = 26.92s

<u>Hence, the plane takes 26.92 seconds to reach its take-off speed.</u>

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