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3 years ago

15

A certain corner of a room is selected as the origin of a rectangular coordinate system. If a fly is crawling on an adjacent wal

l at a point having coordinates (3.1, 0.5), where the units are meters, what is the distance of the fly from the corner of the room? Answer needs to be in appropriate significant figures.

1 answer:

asambeis [7]3 years ago

5 0

Answer:

The distance is 3.1 m

Explanation:

The position vector of the fly relative to the corner of the wall is

r = (3.1, 0.5).

The distance of the fly from the corner will be calculated as the magnitude of the vector "r"

magnitude of vector $r = \sqrt{(3.1 m)^{2} + (0.5 m)^{2}} = 3.1 m$

Since the numbers to be added have only one decimal place 3.<u>1</u> and 0.<u>5</u>, the result of the sum will have to have one decimal place. The result of the square root will also have one decimal place.

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4 years ago

Use the drop-down menus to complete the statements.

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Answer:

Use the drop-down menus to complete the statements.

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3 years ago

A 75 kg baseball player runs at a velocity of 6 m/s before sliding to a stop at second base. a. What is the kinetic energy of th

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Answer:

a. $\displaystyle k_o=1350\ J$

b. $\displaystyle k_1=0\ J$

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d. $W=-1350\ J$

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Explanation:

<u>Work and Kinetic Energy </u>

When an object moves at a certain velocity v0 and changes it to v1, a change in its kinetic energy is achieved:

$\Delta k=k_1-k_0$

Knowing that

$\displaystyle k=\frac{mv^2}{2}$

We have

$\displaystyle \Delta k=\frac{mv_1^2}{2}-\frac{mv_0^2}{2}$

The work done by the force who caused the change of velocity (acceleration) is

$\displaystyle W=\frac{mv_1^2}{2}-\frac{mv_0^2}{2}$

If we know the distance x traveled by the object, the work can also be calculated by

$W=F.x$

Being F the force responsible for the change of velocity

The 75 kg baseball player has an initial velocity of 6 m/s, then he slides and stops

a. Before the slide, his initial kinetic energy is

$\displaystyle k_o=\frac{mv_0^2}{2}$

$\displaystyle k_o=\frac{(75)6^2}{2}$

$\boxed{\displaystyle k_o=1350\ J}$

b. Once he reaches the base, the player is at rest, thus his final kinetic energy is

$\displaystyle k_1=\frac{(75)0^2}{2}$

$\boxed{\displaystyle k_1=0\ J}$

c. The change of kinetic energy is

$\Delta k=k_1-k_0=0\ J-1350\ J$

$\boxed{\Delta k=-1350\ J}$

d. The work done by friction to stop the player is

$W=\Delta k=k_1-k_0$

$\boxed{W=-1350\ J}$

e. We compute the force of friction by using

$W=F.x$

and solving for x

$\displaystyle F=\frac{W}{x}$

$\displaystyle F=\frac{-1350\ J}{2\ m}$

$\boxed{F=-675\ N}$

The negative sign indicates the force is against movement

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3 years ago

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