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2 years ago

11

A car is traveling at 30 m/s in a straight line. The drive applies the brakes for 3 seconds and the car slows down to 12 m/s. Wh

at is the acceleration of the car?

1 answer:

jeyben [28]2 years ago

3 0

Answer:

The answer to your question is: -6m/s²

Explanation:

Data

vo = 30 m/s

vf = 12 m/s

t = 3 s

a = ?

Formula

vf = vo + at

a = (vt - vo) / t

Process

a = (12 - 30) / 3 substitution

a = -18 / 3 simplify

a = -6 m/s² result, is negative because the car

is slowing down.

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inna [77]

Answer:

$\displaystyle m=\frac{2}{3},\ n=\frac{4}{3}$

Explanation:

<u>Dimensional Analysis</u>

It's given the relation between quantities A, B, and C as follows:

$\displaystyle A=\frac{3}{2}B^mC^n$

and the dimensions of each variable is:

$A=L^2T^2$

$B=LT^{-1}$

$C=LT^2$

Substituting the dimensions into the relation (the coefficient is not important in dimension analysis):

$\displaystyle L^2T^2=\left(LT^{-1}\right)^m\left(LT^2\right)^n$

Operating:

$L^2T^2=\left(L^mT^{-m}\right)\left(L^nT^{2n}\right)$

$L^2T^2=L^{m+m}T^{-m+2n}$

Equating the exponents:

$m+n=2$

$-m+2n=2$

Adding both equations:

$3n=4$

Solving:

$n=4/3$

$m=2-4/3=2/3$

Answer:

$\mathbf{\displaystyle m=\frac{2}{3},\ n=\frac{4}{3}}$

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2 years ago

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Answer:

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2 years ago

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Answer:

a)F=3 x 10⁻⁷ N

b)x=2.405 m

Explanation:

Given that

m₁=295 kg

m₂=595 kg

d= 4.1 m

a)

m₃=63 kg

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The force between the mass m₁ and m₃

$F_{13}=\dfrac{Gm_1m_3}{r^2}$

by putting the values

$F_{13}=\dfrac{Gm_1m_3}{r^2}$

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F₁₃=2.94 x 10⁻⁷ N

The force between the mass m₂ and m₃

by putting the values

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The net force F

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b)

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x=1.42 (4.1-x)

x=5.82 - 1.42x

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