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jok3333 [9.3K]
3 years ago
13

If kerosene has a specific gravity of 0.820, what force will be exerted on the circular bottom of a cylindrical kerosene tank th

at has a diameter of 12-ft and a height of 30 ft?
Chemistry
1 answer:
Svet_ta [14]3 years ago
4 0

Answer:

F = 774146.534\,N

Explanation:

The pressure at the bottom of the tank is:

P_{bottom} = (0.820)\cdot (1000\,\frac{kg}{m^{3}})\cdot (9.807\,\frac{m}{s^{2}})\cdot (30\,ft)\cdot (\frac{0.305\,m}{1\,ft} )

P_{bottom} = 73581.921\,Pa

The force exerted on the circular bottom is:

F=(73581.921\,Pa)\cdot (\frac{\pi}{4} )\cdot [(12\,ft)\cdot (\frac{0.305\,m}{1\,ft} )]^{2}

F = 774146.534\,N

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3 years ago
13. Which statement best describes an element? *
Natasha2012 [34]

Explanation:

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A mixture is composed of different types of atoms or molecules that are not chemically bonded.

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4 0
3 years ago
How do you solve this ??
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Here, we have the value of particle in terms of Joules which is 3.2 X 10^{-19}

So, on substituting we get,
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E_{eV} = 1.99 eV so, it can be rounded off to 2.00 eV.
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