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alukav5142 [94]

2 years ago

14

A car traveling at 60km/h undergoes uniform acceleration at a rate of 2/ms^2 until is velocity reached 120km/h determine the dis

tance traveled and the time taken to make the distance

1 answer:

jenyasd209 [6]2 years ago

3 0

Explanation:

Given that,

Initial speed of a car, u = 60 km/h = 16.67 m/s

Acceleration, a = 2m/s²

Final speed, v = 120 km/h = 33.33 m/s

We need to find the distance traveled and the time taken to make the distance.

acceleration = rate of change of velocity

$a=\dfrac{v-u}{t}\\\\t=\dfrac{v-u}{a}\\\\t=\dfrac{33.33 -16.67 }{2}\\\\t=8.33\ s$

let the distance be d.

$d=\dfrac{v^2-u^2}{2a}\\\\d=\frac{33.33^{2}-16.67^{2}}{2(2)}\\\\d=208.25\ m$

Hence, the distance traveled and the time taken to make the distance is 208.25 m and 8.33 seconds respectively.

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Answer:

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Given

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Also $F\cos \theta =f_r$

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$f_r=\mu _k\cdot (mg-F\sin \theta )$

$F\cos \theta =\mu _kF_N$ -------1

$F\sin \theta =mg-F_N$ ---------2

Squaring 1 & 2 and then adding

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Substitute value of F in 1

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$\theta =cos^{-1}(\frac{\mu _KF_N}{\sqrt{(\mu _kF_N)^2+(mg-F_N)^2}})$

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How are particles in liquids, solids, and gases the same? How are they different?

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Explanation:

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$q_1$ and $q_2$ are the signed charge of the objects.
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For the two electrons:

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$r = 1\times 10^{-10}\;\text{m}$ .

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