Question

Calculate the force of attraction between two balls m1 and M2 are 1kg then their centers are 10cm apart.given G=6.67×10-¹¹ Nm² kg-²

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{F=6.67\times 10^{-9}\:N}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Mass \: of \: ball( M_{1}) = 1 \: kg \\ \\ \tt: \implies Mass \: of \: ball( M_{2}) = 1 \: kg \\ \\ \tt: \implies Distance \: between \: radius \: of \: balls (R)= 10 \: cm \\ \\ \tt: \implies Value \: of \: G = 6.67 \times {10}^{ - 11} N {m}^{2} {kg}^{ - 2} \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Force \: of \: attraction(F)= ?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies F= \frac{G M_{1} M_{2} }{ {R}^{2} } \\ \\ \tt: \implies F= \frac{6.67 \times {10}^{ - 11} \times 1 \times 1 \times }{(0.1)^{2} } \times \frac{N \times {m}^{2} \times {kg}^{ - 2} \times kg \times kg }{ {m}^{2} } \\ \\ \tt: \implies F = \frac{6.67 \times {10}^{ - 11} }{0.01} \times N \times {kg}^{ - 2} \times {kg}^{2} \\ \\ \tt: \implies F= \frac{6.67 \times {10}^{ - 11} }{1^{ - 2} } \times N \\ \\ \tt: \implies F= 6.67 \times {10}^{ - 11 + 2} \: N \\ \\ \green{\tt: \implies F= 6.67 \times 10^{ - 9} \: N} \\ \\ \green{\tt \therefore Attraction \: force \: between \: two \: balls} \\ \: \: \: \: \green{\tt is \: 6.67 \times {10}^{ - 9} \: N}$

Answer 2

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Question :

Calculate the force of attraction between two balls m1 and M2 are 1kg then their centers are 10cm apart.given G=6.67×10-¹¹ Nm² kg-²

Solution :

Formula :

$\tt{ \purple{ \implies{F \: = \dfrac{k \: q_{1}q_{2}}{ {r}^{2} } }}}$

Equating we get :

F = 6.67 × 10^-9 N.