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3 years ago

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If a bullet loses 1/nth of its velocity while passing through a plank,then no of planks required to stop the bullet is?

1 answer:

Novay_Z [31]3 years ago

8 0

Based on the answer provided, it seems the writer wanted you to assume that the energy loss per plank is constant. This is not the same as the bullet losing <span><span>1/nth</span><span>1/nth</span></span><span> of its velocity per plank (however, the fact that the question does not mention this assumption arguably makes the question ambiguous).

</span><span>With this assumption, the energy loss becomes
</span><span>
ΔE = <span>1/2 </span>m<span>v2 </span>− <span>1/2 </span>m <span><span>(<span>v−<span>v/n</span></span>) </span><span>2
</span></span></span>
and the number of planks <span>NN</span><span> becomes
</span>
N = <span><span><span>1/2</span>m<span>v2 /</span></span><span>ΔE </span></span>= <span><span>n2/ </span><span>2n−1
</span></span>
Otherwise, if you assume that the bullet loses <span><span>1/<span>nth</span></span><span>1/<span>nth</span></span></span><span> of its velocity per plank, then the answer is </span><span><span>N=∞</span></span><span><span>

</span>
</span>

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Oksanka [162]

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Let's convert the given speed (in cm/s) to m/s.

We know that:

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So, converting we have:

$14\frac{cm}{s}*\frac{1m}{100cm}=0.14\frac{m}{s}$

Then, calculating the time, we have:

$distance=speed*time\\\\3.6m=0.14\frac{m}{s}*time\\\\time=\frac{3.6m}{0.14\frac{m}{s}}=25.71s$

We have that the spider will cross the driveway in 25.71 seconds.

Have a nice day!

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