Calculate the force of attraction between two balls m1 and M2 are 1kg

1.	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» $\blue{\bold{\underline{\underline{Answer:}}}}$ $\green{\tt{\therefore{F=6.67\times 10^{-9}\:N}}}$ $\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$ $\green{\underline \bold{<klux>GIVEN</klux> :}} \\ \tt: \implies Mass \: of \: ball( M_{1}) = 1 \: kg \\ \\ \tt: \implies Mass \: of \: ball( M_{2}) = 1 \: kg \\ \\ \tt: \implies Distance \: between \: radius \: of \: balls (R)= 10 \: cm \\ \\ \tt: \implies Value \: of \: G = 6.67 \times {10}^{ - 11} N {m}^{2} {kg}^{ - 2} \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Force \: of \: attraction(F)= ?$ • ACCORDING to given QUESTION : $\bold{As \: we \: know \: that} \\ \tt: \implies F= \frac{G M_{1} M_{2} }{ {R}^{2} } \\ \\ \tt: \implies F= \frac{6.67 \times {10}^{ - 11} \times 1 \times 1 \times }{(0.1)^{2} } \times \frac{N \times {m}^{2} \times {kg}^{ - 2} \times kg \times kg }{ {m}^{2} } \\ \\ \tt: \implies F = \frac{6.67 \times {10}^{ - 11} }{0.01} \times N \times {kg}^{ - 2} \times {kg}^{2} \\ \\ \tt: \implies F= \frac{6.67 \times {10}^{ - 11} }{1^{ - 2} } \times N \\ \\ \tt: \implies F= 6.67 \times {10}^{ - 11 + 2} \: N \\ \\ \green{\tt: \implies F= 6.67 \times 10^{ - 9} \: N} \\ \\ \green{\tt \therefore Attraction \: force \: between \: two \: balls} \\ \: \: \: \: \green{\tt is \: 6.67 \times {10}^{ - 9} \: N}$

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»

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

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

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

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

• ACCORDING to given QUESTION :

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

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

• ACCORDING to given QUESTION :

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