Physics, asked by irfan8425, 4 months ago

If a body of mass 50 kg is kept at a distance of 1m from a body of mass 100kg , what is the mutual attraction between them?​

Answers

Answered by Itzranbowprince
1

=

(1)

2

6.67×10

−11

×50×50

=1.67×10

−7

N

Force of gravity F'=

r

2

GMm

Here r=R (the radius of earth)

and m

1

=m

2

=m

F

=

(6.4×10

6

)

2

6.67×10

−11

×6×10

24

×50

=0.48×10

3

N

F

is much greater than F so the persons will not move towards each other but each of them moves towards the earth.

Answered by SpreadLove
11

\huge\underline{\overline{\mid{\bold{\color{maroon}{\mathcal{Question:-}}\mid}}}}

If a body of mass 50 kg is kept at a distance of 1m from a body of mass 100kg , what is the mutual attraction between them?

\huge\bf\red{ \mid{ \overline{ \underline{ANSWER}}} \mid}

 \bf \: F=G×  \frac{m×M}{ {d}^{2} }  \\  \\  \bf \: m = 50 kg \\   \bf \: M = 120 kg \\  \bf \: Distance, d = 10 m \\  \\  \bf \: G=6.7× {10}^{ - 11}  \:  {Nm}^{2} / {kg}^{2}  \\  \\  \bf \: F=6.67× {10}^{ - 11}× \frac{50 \times 120}{ {10}^{2} } \\  \\  \bf \: F=6.67×60× {10}^{ - 11}  \\  \\   \red{ \boxed{ \bf \: F=4.02× {10}^{9}N}}

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