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Mila [183]
3 years ago
7

A boy 12.0 m above the ground in a tree throws a ball for his dog, who is standing right below the tree and starts running the i

nstant the ball is thrown. If the boy throws the ball horizontally at 8.50 m/s, (a) how fast must the dog run to catch the ball just as it reaches the ground, and (b) how far from the tree will the dog catch the ball?
Physics
1 answer:
miss Akunina [59]3 years ago
4 0

Answer:

Explanation:

Height covered = 12m

time to fall by 12 m

s = 1/2 gt²

12 = 1/2 g t²

t = 1.565 s

Horizontal distance of throw

= 8.5 x 1.565

= 13.3 m

This distance is to be covered by dog during the time ball falls ie 1.565 s

Speed of dog required = 13.3 / 1.565

= 8.5 m /s

b ) dog will catch the ball at a distance of 13.3 m .

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Answer:

L - h = 12.3672 in

Explanation:

Given

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P*V = n*R*T

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V₁ = A*L = (3.00 in²)*(16.8 in)   ⇒   V₁ = 50.4 in³

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P₂ = P₁ + P = (14.6959 lb/in²) + (41.0 lb/in²) = 55.6959 lb/in²

Inserting various values we get

V₂ = (14.6959  P.S.I)*(50.4 in³) / (55.6959 lb/in²)

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⇒  h = 4.4328 in

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