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LUCKY_DIMON [66]
3 years ago
14

what would be the value of 'G' on the surface of earth if it's mass was twice & its radius half of what it is how?​

Chemistry
2 answers:
Katen [24]3 years ago
8 0

Answer:

What would be the value of g on the surface of the earth if its mass was twice and its radius half of what it is now ? g2=8g1 Thus , the value of g on the surface of the earth would be eight times the present value.

Explanation:

Hope this helps! :)

son4ous [18]3 years ago
8 0

Explanation:

What could be the value of g on the surface of Earth if its mass were twice as large, and the radius also twice what it is now?

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Andreas Xi

Updated 1 year ago

G would be half of what it is now, since it is proportional to mass and inversely proportional to distance squared. Such a planet would be only a little more dense than water, quite similar to Neptune, but smaller.

See below for the answer to the (completely different) original question:

What would be the value of g on the surface of the earth if its mass was twice as large and its radius half of what it is now?

G would be eight times larger, since it is proportional to mass and inversely proportional to distance squared. [Thanks to Niels for pointing out an earlier mistake]

The Earth would need to be 16 times as dense, though, which is quite impossible. The density of rock is around 3 g/cm^3, of iron around 8 and the Earth overall (consisting mostly of those two materials, including compression) around 6. The densest known material is around 22 g/cm^3 (Osmium), which is only roughly 4 times as much. Even when you consider increased compression, no planet could possibly satisfy your conditions, even if it were made of pure Osmium.

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The golden rule is to head for cover.

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The temperature of a sample of gas in steel container at 30.3 Atm is
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Answer:

74.09 atm

Explanation:

Using the gas laws ( Charles and Boyle's law). We have the formula ,

P1/T1 = P2/T2

Where P1 = 30.3atm

T1 = -100 degree Celsius

to kelvin = -100+ 273 = 173K

T2 = 150 degree Celsius

To Kelvin = 150 = 150+273 = 423K

Imputing values

P1/T1 = P2/T2

30.3/173 = P2/ 423

Cross multiply

173×P2 = 30.3 ×423

173P2 = 12816.9

Divide both sides by 173

P2 = 12816.9/173

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6 0
2 years ago
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Any one know the answer?
quester [9]
The last one, answer is D
8 0
3 years ago
Which of the following changes will decrease the rate of collisions between gaseous molecules oftype A and B in a closed contain
Leni [432]
D - take away B molecules
4 0
3 years ago
Hydrogen and oxygen react chemically to form water how much water would form if 14.8grams of hydrogen reacted with 34.8 grams of
pishuonlain [190]

Answer:

There will be formed 39.1935 grams H2O formed

Explanation:

<u>Step 1:</u> The balanced equation

2H2 + 02 → 2H20

<u>Step 2</u>: Given data

mass of hydrogen = 14.8 grams

Molar mass of hydrogen = 2.02 g/mole

mass of oxygen = 34.8 grams

Molar mass of oxygen = 32 g/mole

<u>Step 3: </u>Calculate moles

moles = mass / Molar mass

moles of hydrogen = 14.8g/ 2.02 g/mole = 7.33 moles

moles of oxygen = 34.8g / 32g/mole = 1.0875 moles

For 2 moles hydrogen consumed, we need 1 mole of oxygen.

This means oxygen is the limiting reagens and will be consumed completely. Hydrogen is the reactant in excess, there will remain 5.155 moles of hydrogen

<u>Step 4:</u> Calculate moles of H2O

We see that for 2 moles of H2 consumed, there is needed 1 mole of O2, to produce 2 moles of H2O.

For 1.0875 moles of oxygen consumed, there will be produced 2.175 moles of H2O

<u>Step 5:</u> Calculate mass of water

Mass of H2O = moles of H2O * Molar mass of H2O

Mass of H2O = 2.175 moles * 18.02 g/moles 39.1935 grams

There will be formed 39.1935 grams H2O formed

4 0
3 years ago
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