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2 years ago

The distance and direction between starting and ending positions

Physics

1 answer:

rodikova [14]2 years ago

8 0

Answer:

The length traveled by an object moving in any direction or even changing direction is called distance. The location of an object in a frame of reference is called position. For straight line motion, positions can be shown using a number line. The separation between original and final position is called displacement

Explanation:

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The idea that earth was the center of the universe is an example of ___ that has long been discarded

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2 years ago

Read 2 more answers

An eagle is flying horizontally at a speed of 3.10 m/s when the fish in her talons wiggles loose and falls into the lake 6.10 m

Alinara [238K]

Answer:

10.93m/s with the assumption that the water in the lake is still (the water has a speed of zero)

Explanation:

The velocity of the fish relative to the water when it hits the water surface is equal to the resultant velocity between the fish and the water when it hits it.

The fish drops on the water surface vertically with a vertical velocity v. Nothing was said about the velocity of the water, hence we can safely assume that the velocity if the water in the lake is zero, meaning that it is still. Therefore the relative velocity becomes equal to the velocity v with which the fish strikes the water surface.

We use the first equation of motion for a free-falling body to obtain v as follows;

v = u + gt....................(1)

where g is acceleration due to gravity taken as 9.8m/s/s

It should also be noted that the horizontal and vertical components of the motion are independent of each other, hence we take u = 0 as the fish falls vertically.

To obtain t, we use the second equation of motion as stated;

$h=ut+gt^2/2.................(2)$

Given; h = 6.10m.

since u = 0 for the vertical motion; equation (2) can be written as follows;

$h=\frac{1}{2}gt^2............(3)$

substituting;

$6.1=\frac{1}{2}*9.8*t^2\\6.1=4.9t^2\\hence\\t^2=6.1/4.9\\t^2=1.24\\t=\sqrt{1.24}=1.12s$

Putting this value of t in equation (1) we obtain the following;

v = 0 + 9.8*1.12

v = 10.93m/s

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Answer:

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Explanation:

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3 years ago

An object is dropped from an altitude of one Earth radiusabove Earth's surface. If M is the mass of Earth and R is itsradius the

Montano1993 [528]

Answer:

The speed of the object just before it hits Earth is $\sqrt{\dfrac{GM}{R}}$