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Juliette [100K]
2 years ago
13

49.An object accelerates from rest to a speed of 24 m/s over a distance of 56 m. What

Physics
1 answer:
Nadusha1986 [10]2 years ago
3 0
<h3><u>Given</u><u>:</u></h3>

Initial velocity,u = 0 m/s

Final velocity,v = 24 m/s

Distance covered,s = 56 m

<h3><u>To</u><u> </u><u>be</u><u> </u><u>calculated:-</u><u> </u></h3>

Calculate the acceleration ( a ) .

<h3><u>Solution:-</u><u> </u></h3>

<u>According</u><u> </u><u>to</u><u> </u><u>the</u><u> </u><u>third</u><u> </u><u>equation</u><u> </u><u>of</u><u> </u><u>motion</u><u>, </u>

\bf \:  {v}^{2}  =  {u}^{2}  + 2as

★ Substituting the values in the third equation of motion:

\sf \implies \:  {(24)}^{2}  =  {(0)}^{2}  + 2 \times a \times 56

\sf \implies \: 576 = 112a

\sf \implies \: a =  \dfrac{576}{112}

\sf\implies \: a = 5.14 \: m {s}^{ - 2}

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B) Image height: 1.5 cm, upright

Explanation:

A)

In order to calculate the image position, we can use the lens equation:

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In this problem, we have:

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f = 40 cm (focal length, positive for a converging lens)

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B)

In order to calculate the image height, we use the magnification equation:

\frac{y'}{y}=-\frac{q}{p}

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p = 13 cm

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