How can you measure the distance an object has moved?

1 answer:

8 0

You must observe the object twice.

-- Look at it the first time, and make a mark where it is.

-- After some time has passed, look at the object again, and
make another mark at the place where it is.

-- At your convenience, take out your ruler, and measure the
distance between the two marks.

What you'll have is the object's "displacement" during that period
of time ... the distance between the start-point and end-point.
Technically, you won't know the actual distance it has traveled
during that time, because you don't know the route it took.

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Answer:

the cochlea

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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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Answer:

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