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erastova [34]
3 years ago
7

Two objects attract each other gravitationally. If the distance between their centers decreases by a factor of 2, how does the g

ravitational force between them change? Two objects attract each other gravitationally. If the distance between their centers decreases by a factor of 2, how does the gravitational force between them change? The gravitational force decreases by a factor of 2. The gravitational force increases by a factor of 4. The gravitational force decreases by a factor of 4. The gravitational force remains unchanged. The gravitational force increases by a factor of 2.
Physics
1 answer:
kramer3 years ago
6 0

Answer:

The gravitational force between them increases by a factor of 4

Explanation:

Gravitational force is a force of attraction between two objects with masses M and m which are separated by a distance R. It is given mathematically as:

Fg = GMm/R²

Where G = Gravitational constant.

If the distance between their centers, R, decreases by a factor of 2, then it means the new distance between their centers is:

r = R/2

Hence,the gravitational force becomes:

Fg = GMm/r²

Fg = GMm/(R/2)²

Fg = GMm/(R²/4)

Fg = 4GMm/R²

Hence,the gravitational force increases by a factor of 4.

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Answer:

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Now, using Length contraction formula of relativity theory:

L'' = L'\sqrt{1 - \frac{v^{2}}{c^{2}}}                           (1)

time taken = 5 years

We know that :

time = \frac{distance}{speed}

5 = \frac{L''}{v}                                                      (2)

Dividing eqn (1) by v on both the sides and substituting eqn (2) in eqn (1):

\frac{L'\sqrt{1 - \frac{v^{2}}{c^{2}}}}{v} = 5

\frac{20'\sqrt{1 - \frac{v^{2}}{c^{2}}}}{v} = 5

Squaring both the sides and Solving above eqution, we get:

v = 0.97 c

(b) Time observed from Earth:

Using time dilation:

t'' = \frac{t'}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}

t'' = \frac{5}{\sqrt{1 - \frac{(0.97c)^{2}}{c^{2}}}}

Solving the above eqn:

t'' = 20.56 years

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calculate the mass of potassium chlorate (kcio3) required to obtain 10g of oxygen in the following reaction:kclO3-kcl+O2​
igor_vitrenko [27]

First, balance the reaction:

_ KClO₃   ==>   _ KCl + _ O₂

As is, there are 3 O's on the left and 2 O's on the right, so there needs to be a 2:3 ratio of KClO₃ to O₂. Then there are 2 K's and 2 Cl's among the reactants, so we have a 1:1 ratio of KClO₃ to KCl :

2 KClO₃   ==>   2 KCl + 3 O₂

Since we start with a known quantity of O₂, let's divide each coefficient by 3.

2/3 KClO₃   ==>   2/3 KCl + O₂

Next, look up the molar masses of each element involved:

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• Cl: 35.453 g/mol

• O: 15.999 g/mol

Convert 10 g of O₂ to moles:

(10 g) / (31.998 g/mol) ≈ 0.31252 mol

The balanced reaction shows that we need 2/3 mol KClO₃ for every mole of O₂. So to produce 10 g of O₂, we need

(2/3 (mol KClO₃)/(mol O₂)) × (0.31252 mol O₂) ≈ 0.20835 mol KClO₃

KClO₃ has a total molar mass of about 122.549 g/mol. Then the reaction requires a mass of

(0.20835 mol) × (122.549 g/mol) ≈ 25.532 g

of KClO₃.

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2 years ago
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