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Dennis_Churaev [7]

2 years ago

6

A car is moving 18 m/s to the eat. If it takes the car 5 seconds to reach a velocity of 19 m/s to the east, what is its accelera

tion?
A. 0.2 m/s^2 west
B. 0.2 m/s^2 east
C. 1 m/s^2 east
D. 5 m/s^2 east

1 answer:

defon2 years ago

3 0

<h3><u>Answer;</u></h3>

<u> 0.2 m/s²,east</u>

<h3><u>Explanation;</u></h3>

Using the equation;

v = u+at

Where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

making a the subject;

a= (v-u)/t

= (19 -18)/5

= 1/5

<u> = 0.2 m/s², east</u>

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Answer:

$q / q_{1} = 2$ , assuming that $q_{1}$ and $(q - q_{1})$ are point charges.

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$\begin{aligned}\frac{d}{d q_{1}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\right] \\ &= \frac{k}{r^{2}}\, \left[\frac{d}{d q_{1}} [q\, q_{1}] - \frac{d}{d q_{1}}[{q_{1}}^{2}]\right]\\ &= \frac{k}{r^{2}}\, (q - 2\, q_{1})\end{aligned}$ .

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