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3 years ago

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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:

4.6 years

Explanation:

This is solved using Kepler's third law which says:

$T^2=\frac{4\pi ^2}{GM} a^2$

Where

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a = Distance from the star (in meters)

G = Gravitational constant

M = Mass of the parent star (in kg)

From the information given

$M = 3.5M_{sun} = 6.96*10^{30} kg$

$a=4.2AU = 6.28*10^{11} meters$

$G = 6.67*10^{-11}m^3kg^{-1}s^{-2}$

We put this into Kepler's law and get:

$T=\sqrt{\frac{4\pi ^2}{6.67*10^{-11}*6.96*10^{30}} (6.28*10^{11})^3}=145,128,196 seconds.$

This when converted to years is 4.6 years.

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Multiple-Concept Example 13 presents useful background for this problem. The cheetah is one of the fastest accelerating animals,

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Answer:

9241.6 W or 12.39318 hp

Explanation:

u = Initial velocity = 0

v = Final velocity

m = Mass

t = Time taken

Energy

$KE=\frac{1}{2}m(v^2-u^2)\\\Rightarrow KE=\frac{1}{2}108(30.4^2-0^2)\\\Rightarrow KE=49904.64\ Joules$

Power

$P=\frac{KE}{t}\\\Rightarrow P=\frac{49904.64}{5.4}\\\Rightarrow P=9241.6\ W$

Converting to hp

$1\ W=\frac{1}{745.7}\ hp$

$\\\Rightarrow 9241.6\ W=\frac{9241.6}{745.7}\ hp=12.39318\ hp$

The power developed by the cheetah is 9241.6 W or 12.39318 hp

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