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2 years ago

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A model rocket flies horizontally off the edge of a cliff at a velocity of 80.0m/s. If the canyon below is 128.0 m deep, how far

from the edge of the cliff does the model rocket land?
a. 112m
b. 225m
c. 337m
d. 409m

1 answer:

RSB [31]2 years ago

5 0

Answer:

c. 337

Explanation:

can someone answer my question

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An object with a mass m slides down a rough 370 inclined plane where the coefficient of kinetic friction is 0.20. If the plane i

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Answer:

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Explanation:

The Free Body Diagram of the system is presented in the image attached below. The final speed is determined by means of the Principle of Energy Conservation and the Work-Energy Theorem:

$K_{A} + U_{g,A} = K_{B} + U_{g,B} + W_{loss}$

$K_{B} = K_{A} + U_{g,A}-U_{g,B} - W_{loss}$

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$v = \sqrt{2\cdot g \cdot s \cdot (\sin \theta - \mu_{k}\cdot \cos \theta)}$

$v = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (10\,m)\cdot (\sin 37^{\textdegree} - 0.2\cdot \cos 37^{\textdegree})}$

$v \approx 9.312\,\frac{m}{s}$

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Explanation:

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