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3 years ago

10

An advertisement claims that a certain 1200kg can accelerate from rest to a speed of 25 m/s in a time of 8s. What average power

must the motor develop to produce this acceleration if the car is going up a 20 degree incline? Give your answer in Watts. Ignore friction.

1 answer:

expeople1 [14]3 years ago

3 0

Answer: 3MW

Explanation:

W = 1/2 mv^2

W = 0.5 x 1200 x 25^2 = 375 kW

T = 8s

P= 335 x 10^3 x 8 = 3,000 kW = 3MW

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Answer:

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a) Given data:

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a = 4.14 * 4.14

The moment of inertia is:

$I=mr^{2} \\r=\frac{a}{2} \\I=m(a/2)^{2} \\I=\frac{ma^{2} }{4}$

For 4 particles:

$I=4(\frac{ma^{2} }{4} )=ma^{2} =0.98*(4.14)^{2} =16.8kgm^{2}$

b) Distance from top left mass = x = a/2

Distance from bottom left mass = x = a/2

Distance from top right mass = x = √5 (a/2)

The total moment of inertia is:

$I=m(\frac{a}{2} )^{2} +m(\frac{a}{2} )^{2}+m(\frac{\sqrt{5a} }{2} )^{2}+m(\frac{\sqrt{5a} }{2} )^{2}=\frac{12ma^{2} }{4} =3ma^{2} =3*0.98*(4.14)^{2} =50.39kgm^{2}$

c)

$I=2m(\frac{m}{\sqrt{2} } )^{2} =ma^{2} =0.98*(4.14)^{2} =16.8kgm^{2}$

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