What happens to the air pressure inside a balloon when the balloon is squeezed to half its volume at constant temperature?

Calculate the force of gravity between a comet with a mass of 500kg and a small asteroid with a mass of 20kg that is separated b

givi [52]

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The equivalent gravitational force is ~

$F \approx1.48\times 10 {}^{ - 7} \: \: N$

$\large \boxed{ \mathfrak{Step\:\: By\:\:Step\:\:Explanation}}$

We know that ~

$\huge\boxed{\mathrm{F = \dfrac{ Gm_1m_2}{ r²}}}$

where,

F = gravitational force

$m_1$ = mass of 1st object = 500 kg

$m_2$ = mass of 2nd object = 20kg

G = gravitational constant = $6.674 × {10}^ {-11}$

r = distance between the objects = 2.12 m

Let's calculate the force ~

$F = \dfrac{6.674 \times 10 {}^{ - 11} \times 500 \times 20}{(2.12) {}^{2} }$

$F = \dfrac{6.674 \times 10 {}^{ - 11} \times 10 {}^{4} }{4.4944}$

$F = \dfrac{6.674}{4.4944} \times 10 {}^{ - 7}$

$F =1.484 \times 10 {}^{ - 7} \: \: newtons$

7 0

2 years ago

The force of gravity on Jupiter is much stronger than the force of gravity on earth what is the answer?

Pachacha [2.7K]

This would be true. On Jupiter you would weigh 234 pounds if you were 100 pounds on Earth.

7 0

3 years ago

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What materials in a torch are conductors and insulators?

Alisiya [41]

Are you sure you weren't given any options? Basically the electrical components, e.g wires and bulbs will be conductors, and the part comprising the outer casing, e.g plastic, glass, will be insulators

8 0

4 years ago

At the county fair, Chris throws a 0.12kg baseball at a 2.4kg wooden milk bottle, hoping to knock it off its stand and win a pri

viva [34]

Answer:

$v_{f2}$ =6.5% $v_{i1}$

Explanation:

Mass of the ball: $m_{1} =0.12kg$ ]

Initial velocity of the ball: $v_{i1}$

final velocity of the ball: $v_{f1}$ which is -30/100 of $v_{i1}$ = $-0.3v_{i1}$

Mass of the bottle: $m_{2} =2.4kg$

Initial velocity of the bottle: $v_{i2}=0m/s$

final velocity of the bottle: $v_{f2}$ is unknown (to find)

by using conservation momentum, which stated that the initial momentum is equal to the final momentum.

$m_{1} v_{i1} +m_{2} v_{i2} =m_{1} v_{f1} +m_{2} v_{f2}$

so since the bottle is at rest firstly, therefore $v_{i2} =0$

$m_{1} v_{i1} +m_{2} (0) =m_{1} v_{f1} +m_{2} v_{f2}$

$m_{1} v_{i1} =m_{1} v_{f1} +m_{2} v_{f2}$ equation 1

so now substitute $v_{f1}$ into equation 1

$m_{1} v_{i1} =m_{1} (-0.3v_{i1} ) +m_{2} v_{f2}$

$m_{1} v_{i1} = -0.3m_{1}v_{i1} +m_{2} v_{f2}$

collect the like terms

$m_{1} v_{i1} +0.3m_{1}v_{i1} =m_{2} v_{f2}$

$1.3m_{1} v_{i1} =m_{2} v_{f2}$

divide both side by $m_{2}$

$v_{f2}=\frac{1.3m_{1} v_{i1}}{m_{2} }$

Now substitute

$v_{f2} =\frac{1.3*0.12*v_{i1}}{2.4}\\v_{f2} =\frac{0.156v_{i1} }{2.4} \\v_{f2} =0.065v_{i1}$

$v_{f2} =$ 6.5% $v_{i1}$

6 0

3 years ago

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How can you tell that this paragraph uses a quotation? Check all that apply?

Debora [2.8K]

Answer- There are two reasons that we know quotations have been used first is the use of of name of the person who quoted it and secondly the quotation is written inside the quotation marks.

Explanation- Quotation is nothing but using a line that has been already quoted by someone somewhere. Such sentences are normally written inside quotation marks. In the above given paragraph the name of the person who quotes the sentence is also given hence we know that our quotation has been used.

8 0

3 years ago