Question

a mass of 40 kg is attracted by a mass of 10 kg lying at a distance of 1 M with a force of 1.67 into 10 ^ -8 N. find the value of 'G' (note: it is capital G)

Answer 1

Answer:

F = GMm / d²

Here, M = 40 kg and m = 10 kg, d = 1 M

Substituting the values we get,

⇒ 1.67 × 10⁻⁸ = G × 40 × 10 / 1

⇒ 1.67 × 10⁻⁸ = G × 400

⇒ G = 1.67 × 10⁻⁸ / 400 = 1.67 × 10⁻⁸ / 4 × 10²

⇒ G = 0.4175 × 10⁻⁸⁻² = 0.4175 × 10⁻¹⁰ N

This is the required answer.

Hope it helped !!

Answer 2

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Using This Formula :-

$\bf\huge F=\frac{GMm}{d^{2} }$

According to the Question :-

Mass (First) = 40 kg

mass = 10 kg

Distance = 1 M

Put the values :-

⇒ 1.67 × 10⁻⁸ = $\bf\huge\frac{G\times40\times10}{1}$

⇒ 1.67 × 10⁻⁸ = G × 400

$\bf\huge G= \frac{1.67\times10 ^-8}{400}$

⇒ G = 0.4175 × 10⁻⁸⁺²

= 0.4175 × 10⁻⁶ N

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