Two spheres each of mass $70 k g$ are $0.14 m$ apart. Find the force of attraction between them. Given $G = 6.67 x 10^{- 11} N m^{2} k g^{- 2} .$

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Solution

Step 1: Given data
Mass (m₁) = $70 K g$
Mass (m₂) = $70 space K g$
Distance of separation (d) = $0.14 space m$
$G space equals space 6.67 space x space 10 to the power of negative 11 end exponent space N space m squared space k g to the power of negative 2 end exponent.$
Step 2: The formula for force
$F = G \times \frac{m_{1} \times m_{2}}{d^{2}}$
Step 3: The calculation for the force of attraction
$F = \frac{6.67 \times 10^{- 11} \times 70 \times 70}{0.14 \times 0.14} F = 0.000016675 N F = 166.75 \times 10^{- 7} N$
Hence, the force of attraction is $166.75 \times 10^{- 7} N$ .

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Two spheres each of mass 70kg are 0.14m apart. Find the force of attraction between them. Given G=6.67x10-11Nm2kg-2.

Step 1: Given dataMass (m1) = 70KgMass (m2) = Distance of separation (d) = Step 2: The formula for forceF=G×m1×m2d2Step 3: The calculation for the force of attractionF=6.67×10-11×70×700.14×0.14F=0.000016675NF=166.75×10-7NHence, the force of attraction is 166.75×10-7N.

Two spheres each of mass $70 k g$ are $0.14 m$ apart. Find the force of attraction between them. Given $G = 6.67 x 10^{- 11} N m^{2} k g^{- 2} .$