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OLEGan [10]
3 years ago
13

Converting 15 miles to kilometer

Physics
2 answers:
Dmitry [639]3 years ago
8 0
15 miles to kilometers would be: 24.14 kilometers
dimaraw [331]3 years ago
5 0

1 mile = 1609.344... meters

1,000 meters = 1 km

(15 miles) x (1609 m/mile) x (1 km/1000 m) =

(15 x 1609 / 1000) (mile-meter-km / mile-meter) =  <em>24.14 km</em>


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Answer:

To convert 400 mm to m you can apply the formula [m] = [mm] / 1000; use 400 for mm. Thus, the conversion 400 mm m is the result of dividing 400 by 1000. 0.4

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2 years ago
The wave shown below is produced in a rope.
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<h3>B) A dot vector drawn up and vector drawn down.</h3>

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A geneticist looks through a microscope to determine the phenotype of a fruit fly. The microscope is set to an overall magnifica
guajiro [1.7K]

Answer:

f_{e} = 1.7 cm

Explanation:

The magnification of the compound microscope is given by the product of the magnification of each lens

        M = M₀ m_{e}

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Where f₀ and f_{e} are the focal lengths of the lens and eyepiece, respectively, all values ​​in centimeters

In this exercise they give us the magnification (M = 400X), the focal length of the lens (f₀ = 0.6 cm), the distance of the tube (L = 16 cm), let's look for the focal length of the eyepiece (f_{e})

         f_{e} = - L / f₀ 25 / M

Let's calculate

        f_{e} = - 16 / 0.6 25 / (-400)

        f_{e} = 1.67 cm

The minus sign in the magnification is because the image is inverted.

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<em>The final speed of the second package is twice as much as the final speed of the first package.</em>

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<u>Free Fall Motion</u>

If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:

v=gt

And the distance traveled downwards is:

\displaystyle y=\frac{gt^2}{2}

If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:

\displaystyle t=\sqrt{\frac{2y}{g}}

Replacing into the first equation:

\displaystyle v=g\sqrt{\frac{2y}{g}}

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Let's call v1 the final speed of the package dropped from a height H. Thus:

\displaystyle v_1=\sqrt{2gH}

Let v2 be the final speed of the package dropped from a height 4H. Thus:

\displaystyle v_2=\sqrt{2g(4H)}

Taking out the square root of 4:

\displaystyle v_2=2\sqrt{2gH}

Dividing v2/v1 we can compare the final speeds:

\displaystyle v_2/v_1=\frac{2\sqrt{2gH}}{\sqrt{2gH}}

Simplifying:

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The final speed of the second package is twice as much as the final speed of the first package.

5 0
3 years ago
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Answer:

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This shows the compass if following its direction properly.

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3 years ago
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