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grin007 [14]
3 years ago
11

how can there be so many different substances in the world if there are only a few elements that are common? 

Physics
1 answer:
LuckyWell [14K]3 years ago
3 0
-- If there are only  <em>10</em>  elements in the universe that can make compound molecules, and a compound molecule can be formed by combining  1,  2,  3, or  4  different elements, then that's already the possibility of  at least 400 different molecules.

-- There are many more than  10  elements that can combine to form compound molecules.

-- Every single "<em>organic</em>" molecule, of which there are thousands, is the combination of <em>carbon</em> with other elements.

-- Most all of the substances that can be distilled out of oil, including the paraffin waxes, the alcohols, gasoline, kerosene, butane, propane, octane, and natural gas, are made of just carbon, hydrogen, and oxygen, only with different numbers of each one. 

-- Plastics, drugs, rubber, and DNA are examples of molecules that are made of <em>hundreds</em> of atoms.
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For which pair of objects would adding the same amount of electrons to each object result in a decrease in the electric force?
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A bubble, located 0.200 m beneath the surface in a glass of beer, rises to the top. The air pressure at the top is 1.01 ???? 10
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Answer:

the ratio of the bubble’s volume at the top to its volume at the bottom is 1.019

Explanation:

given information

h = 0.2 m

P_{0} = 1.01  x 10^{5} Pa

P_{1} V_{1} = P_{2} V_{2}

\frac{V_{2} }{V_{1}} = \frac{P_{1} }{P_{2}}

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8 0
3 years ago
A 6.00 kg ball is dropped from a height of 12.0 m above one end of a uniform bar that pivots at its center. The bar has mass 5.0
Andre45 [30]

Answer:

The height of the other ball go after the collision is 2.304 m.

Explanation:

Given that,

Mass of ball = 6.00 kg

Height = 12.0 m

Mass of bar =5.00 kg

Length = 4.00 m

Suppose we need to calculate how high will the other ball go after the collision

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v=\sqrt{2gh}

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Using formula of angular momentum

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Put the value into the formula

l_{before}=6.00\times15.33\times2.0

l_{before}=183.96\ kgm^2/s

We need to calculate the angular momentum

Using formula of angular momentum

l_{after}=I_{t}\omega

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Put the value into the formula

l_{after}=(\dfrac{5\times4.00^2}{12}+6.00\times2.0^2+6.00\times2.0^2)\omega

183.96=54.66\omega

\omega=\dfrac{183.96}{54.66}

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After collision the ball leaves with velocity

We need to calculate the velocity after collision

Using formula of the velocity

v= r\omega

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We need to calculate the height

Using formula of height

h=\dfrac{v^2}{2g}

Put the value into the formula

h=\dfrac{(6.72)^2}{2\times9.8}

h=2.304\ m

Hence, The height of the other ball go after the collision is 2.304 m.

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