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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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Suppose the velocity of an object moving along a line is positive. are position, displacement, and distance traveled equal? expl
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2 years ago
Water emerges straight down from a faucet with a 2.51-cm diameter at a speed of 3.04 m/s. (because of the construction of the fa
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This is a question on conservation of energy. That is,
mgh + KE1 = KE2
mgh +1/2mv1^2 = 1/2mv2^2
gh + 1/2v1^2 = 1/2v2^2

Where, h = 0.2 m, v1 =3.04 m/s
Therefore,
v2 = Sqrt [2(gh+1/2v1^2)] = Sqrt [2(9.81*0.2 + 1/2*3.04^2)] = 7.26 m/s

Now, Volumetric flow rate, V/time, t = Surface area, A*velocity, v
Where,
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At 0.2 m below,
V = 1.504*10^-3 m^3/s = A*7.26
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But, A = πr^2
Then,
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Diameter = 2r = 0.0162 m = 1.62 cm
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3 years ago
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