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Bess [88]
3 years ago
7

A 1500 kg car originally moving at 10 m/s crashes into a wall and comes to rest in 0.25 s. What is the magnitude of the impulse

given to the car?
a. 37.5 kg m/s
b. 150 kg m/s
c. 15,000 kg m/s
d. 60,000 kg m/s
Physics
1 answer:
kirill [66]3 years ago
6 0

Answer:

Option C

Explanation:

Impulse is the change in momentum in an object hence given by

Impulse=F∆t

Impulse=∆p=m(v-u)

Here, F is force applied, t is rime, m is mass, v is final velocity and u is initial velocity

Given that the question has time but no magnitude of the force, we use the sexond equation.

Taking forward velocity as negative then u=-10 m/s, also since the car comes to rest, final velocity is 0 m/s. Substititing mass with 1500 kgs then

Impulse=1500(0--10)=15000 kg.m/s

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Answer:

\frac{a_{r,earth}}{a_{r,mars}} = 2.325

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v=\frac{2\pi\cdot r}{\Delta t}

Earth:

v_{earth} = \frac{2\pi\cdot (149.6\times 10^{9}\,m)}{(365\,days)\cdot(\frac{24\,hours}{1\,day} )\cdot(\frac{3600\,s}{1\,h} )}

v_{earth}=29806.079\,\frac{m}{s}

Mars:

v_{mars} = \frac{2\pi\cdot (227.9\times 10^{9}\,m)}{(687\,days)\cdot(\frac{24\,hours}{1\,day} )\cdot(\frac{3600\,s}{1\,h} )}

v_{mars}=24124.244\,\frac{m}{s}

Now, centripetal accelarations can be found:

Earth:

a_{r,earth} = \frac{(29806.079\,\frac{m}{s} )^{2}}{149.6\times 10^{9}\,m}

a_{r,earth} = 5.939\times 10^{-3}\,\frac{m}{s^{2}}

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a_{r,mars} = \frac{(24124.244\,\frac{m}{s} )^{2}}{227.9\times 10^{9}\,m}

a_{r,mars} = 2.554\times 10^{-3}\,\frac{m}{s^{2}}

The ratio of Earth's centripetal acceleration to Mars's centripetal acceleration is:

\frac{a_{r,earth}}{a_{r,mars}} = \frac{5.939}{2.554}

\frac{a_{r,earth}}{a_{r,mars}} = 2.325

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