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3 years ago

You can see lightning before you can hear the thunder. Which is the best explanation for this?

2 answers:

8 0

Salutations!

You can see lightning before you can hear the thunder. Which is the best explanation for this?

The best explanation for this statement is the transmission of light in air is faster than the transmission of sound in the air. Light is faster than sound, thus you can see the thunder then you can hear the sound.

Hope I helped :D

ser-zykov [4K]3 years ago

8 0

The transmission of light in air is faster than the transmission of sound in air

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What is the force needed to move an 225 kg car a distance of 150 meters

Kobotan [32]

Physics

mass = 225 kg

weight = m × a = 2250 N

distance = 150 m

force : ______?

Answers :

W = F × s

W = 2250 × 150

W = 337500 ✅

4 0

3 years ago

Read 2 more answers

A package is dropped from an air balloon two times. In the first trial the distance between the balloon and the surface is Hand

enyata [817]

Answer:

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

Explanation:

Free Fall Motion

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:

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

Rationalizing:

$\displaystyle v=\sqrt{2gy}$

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:

$\displaystyle v_2/v_1=2$

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

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3 years ago

In a electrically natural atom of any element there are equal numbers of

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Answer:

I hope it's helpful for you ☺️