Physics, asked by monikachambial, 5 months ago


a sphere of mass 40 kg is attracted by another sphere of mass 80kg by a force of
2.5 x 10 ^-6 N when the centres are 3x10^-2 m apart
find the value of G​

Answers

Answered by SujalSirimilla
4

GIVEN:

  • Mass of the first sphere (m₁) = 40 Kg.
  • Mass of the second sphere (m₂) = 80 Kg.
  • Force (F) = 2.5 x 10⁻⁶ N.
  • Distance between the spheres (d) = 3x10⁻² m.

TO FIND:

  • Value of G.

SOLUTION:

Use this formula:

\mapsto \sf{\green{F=G\dfrac{m_1 \times m_2}{d^2} }}

Where:

  • F is force.
  • m₁ and m₂ are mass of the objects.
  • d is distance.
  • G is universal gravitation constant.

Substitute.

\to \sf 2.5 \times 10^{-6} =G \times \dfrac{40 \times 80}{\left(3 \times 10^{-2}\right)^2}

▣ Use the radicle rule : (a×b)ⁿ=aⁿ×bⁿ to simplify the denominator.

\to \sf 2.5 \times 10^{-6} =G \times \dfrac{40 \times 80}{3^2 \times (10^{-2})^2}

▣ Use the radicle rule = (mᵃ)ᵇ=mᵃᵇ. Moreover, 3²=9 and 40×80 = 3200.

\to \sf 2.5 \times 10^{-6} =G \times \dfrac{3200}{9 \times (10^{-2 \times 2})}

\to \sf 2.5 \times 10^{-6} =G \times \dfrac{3200}{9 \times (10^{-4})}

▣ Now transpose.

\to \sf 2.5 \times 10^{-6} \times \dfrac{9 \times (10^{-4})}{3200} =G

\to \sf \dfrac{22.5 \times 10^{-6}  \times (10^{-4})}{3200} =G

\to \sf \dfrac{22.5 \times 10^{-10}  }{3200} =G

\to \sf \dfrac{225 \times 10^{-10}  }{32\times 10^3} =G

\sf \to 7.03125 \times 10^{-13}N \cdot m^2/kg^2=G

G is 7.03×10⁻¹³ N·m²/kg².

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