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serg [7]
3 years ago
9

A crow drops .11 -kg clam onto a rocky beach from a height of 9.8 m. What is the kinetic energy of the clam when it is 5.0 m abo

ve the ground? What is it's speed at that point?
Physics
1 answer:
Alchen [17]3 years ago
5 0

Answer:

Kinetic Energy=5.17J

Speed=9.7m/sec

Explanation:

Given the mass of the clam is m=0.11kg

The height from which the rock was thrown is h=9.8m

Given the rock is at 5m above the ground.

The distance travelled by the rock is (s)=9.8-5=4.8m

The initial velocity is u=0m/sec

let the final velocity be v m/sec

We know that the acceleration of the particle is a=9.8m/sec^{2}

We know that v^{2} -u^{2} =2as

v^{2} -0^{2} =2\times 9.8\times 4.8

v^{2} =94.08

v=9.7m/sec

Now the kinetic energy=\frac{1}{2}mv^{2}

KE=\frac{1}{2}\times 0.11\times 9.7^{2}=5.17J

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Answer:

the correct solution is 13 s

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      a₁/a₂ t² = t² -2 4 t + 16

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the correct solution is 13 s, if you have to select one the nearest 12s

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