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2 years ago

write the formula of gravitational force between two bodies of mass 'x'and 'y' kept at a distance 'r'

Mathematics

1 answer:

Rudik [331]2 years ago

5 0

$\huge \orange{F =G.\frac{xy}{r^2}}$

Step-by-step explanation:

Masses of the bodies are x and y

Distance at which they are kept is r

When two bodies are of masses x and y are placed at a distance r then the gravitational force acting on them is given as:

$\huge \purple{F =G.\frac{xy}{r^2}}$
Where G is the gravitational constant.

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Answer:

The weight of the parcel is approximately 6 kilograms.

Step-by-step explanation:

The weight of the parcel is 12 lbs and 8 oz.

1 lb = 16 oz

So, 8 oz = lb.

That means, 12 lbs and 8 oz = (12 + 0.5) lbs = 12.5 lbs

Now, 1 lb = 454 g

So, 12.5 lbs = (12.5 × 454) g = 5675 g

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So, 5675 g kilograms. (Rounded to the nearest whole kilogram)

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3 years ago

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Stella [2.4K]

Answer: $-\frac{\sqrt{2a}}{8a}$

=======================================================

Explanation:

The (x-a) in the denominator causes a problem if we tried to simply directly substitute in x = a. This is because we get a division by zero error.

The trick often used for problems like this is to rationalize the numerator as shown in the steps below.

$\displaystyle \lim_{x\to a} \frac{\sqrt{3a-x}-\sqrt{x+a}}{4(x-a)}\\\\\\\lim_{x\to a} \frac{(\sqrt{3a-x}-\sqrt{x+a})(\sqrt{3a-x}+\sqrt{x+a})}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{(\sqrt{3a-x})^2-(\sqrt{x+a})^2}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{3a-x-(x+a)}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{3a-x-x-a}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\$

$\displaystyle \lim_{x\to a} \frac{2a-2x}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{-2(-a+x)}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{-2(x-a)}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{-2}{4(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\$

At this point, the (x-a) in the denominator has been canceled out. We can now plug in x = a to see what happens

$\displaystyle L = \lim_{x\to a} \frac{-2}{4(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\L = \frac{-2}{4(\sqrt{3a-a}+\sqrt{a+a})}\\\\\\L = \frac{-2}{4(\sqrt{2a}+\sqrt{2a})}\\\\\\L = \frac{-2}{4(2\sqrt{2a})}\\\\\\L = \frac{-2}{8\sqrt{2a}}\\\\\\L = \frac{-1}{4\sqrt{2a}}\\\\\\L = \frac{-1*\sqrt{2a}}{4\sqrt{2a}*\sqrt{2a}}\\\\\\L = \frac{-\sqrt{2a}}{4\sqrt{2a*2a}}\\\\\\L = \frac{-\sqrt{2a}}{4\sqrt{(2a)^2}}\\\\\\L = \frac{-\sqrt{2a}}{4*2a}\\\\\\L = -\frac{\sqrt{2a}}{8a}\\\\\\$

There's not much else to say from here since we don't know the value of 'a'. So we can stop here.

Therefore,

$\displaystyle \lim_{x\to a} \frac{\sqrt{3a-x}-\sqrt{x+a}}{4(x-a)} = -\frac{\sqrt{2a}}{8a}\\\\\\$

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3 years ago

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Kruka [31]

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3 years ago

write the formula of gravitational force between two bodies of mass 'x'and 'y' kept at a distance 'r'​

write the formula of gravitational force between two bodies of mass 'x'and 'y' kept at a distance 'r'