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2 years ago

6

An airplane accelerat s down a runway at 3.20 m/s(squared) for 32.8s until is finally lifts off the ground. Presume to he plane

starts from the rest. Determine the distance traveled before takeoff.
The answer is 1721.3 but I want to know how to get it

1 answer:

rusak2 [61]2 years ago

5 0

SORRRY not this smart

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Explanation:in order to explain this result we have to use the Ohm law given by:

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Then we have

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3 years ago

On that system is<br> In an isolated system, the net __<br> zero.

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2 years ago

The _____ of an object is a measure of the distance an object moves in a given amount of time. *

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3 years ago

A worker on the roof of a house drops his 0.46 kg hammer, which slides down the roof at constant speed of 9.88 m/s. The roof mak

nekit [7.7K]

Answer:

The horizontal distance travelled in that time lapse is 12.94 m

Explanation:

In order to solve this problem, we'll need:

The horizontal speed
the time the hammer takes to fall from the roof to the ground

At the lowest point of the roof, the hammer has a 9.88 m/s speed that makes an angle of 27° with the horizontal, so we can calculate the horizontal and vertical speed with trigonometry. If we take right as x positive and down as y positive we get

$v_{x}=v*cos(27)=9.88 m/s *cos(27)=8.80 m/s \\v_{y}=v*sen(27)=9.88 m/s *sen(27)=4.49 m/s$

Now, we make two movement equation as we have a URM (no acceleration) in x and an ARM (gravity as acceleration) in y. We will wisely pick the lowest point of the roof as the origin of coordinates

$x(t)=8.8 m/s *t$

$y(t)=4.49m/s*t+\frac{1}{2}*9.8m/s^{2}*t^{2}$

Now we calculate the time the hammer takes to get to the floor

$17.1m=4.49m/s*t+\frac{1}{2}*9.8m/s^{2}*t^{2}\\t=1.47s$ or $t=-2.38s$

Now, we keep the positive time result and calculate the horizontal distance travelled

$x(1.47s)=8.8m/s*1.47s=12.94m$

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3 years ago