Question

Calculate the attraction force of two 150kg mass objects which lies at 6 meter distance.Pls explain the steps.​

Answer 1

Answer:

8.18 × 10^-8 N

Explanation:

m₁ = 150 kg

m₂ = 150 kg

d = 6 m

F(g) = Gm₁m₂/ d²

      = 6.67 × 10^-11 × 150 × 150 / 6²

      = 6.67 × 10^-11 × 35 × 35          (Canceling 150 by 6 twice)

      = 667 × 1225 × 10^-13

      = 817075 × 10^-13

      ≈ 8.18 × 10^-8 N                   (ANS.)

Answer 2

To Find :

The of Attraction between two objects .

Given :

• Mass $\rightarrow \sf{m_{1} = m_{2} = 150 kg}$

• Distance between two objects = 6 m

• Value of "G" = $6.67 × 10^{-11}$

We Know :

Force :

$\purple{\sf{\underline{\boxed{F = G\dfrac{m_{1}m_{2}}{r^{2}}}}}$

Where ,

• F = Force

• G = Universal Gravitational Constant

• m = mass of the object

• r = distance

Solution :

• Mass $\rightarrow \sf{m_{1} = m_{2} = 150 kg}$
• Distance between two objects = 6 m

Using the formula, and Substituting the values in it , we get :

$\purple{\sf{F = G\dfrac{m_{1}m_{2}}{r^{2}}}} \\ \\ \\ \implies \sf{F = 6.67 \times 10^{-11}\bigg(\dfrac{150 \times 150}{6^{2}}\bigg)} \\ \\ \\ \implies \sf{F = 6.67 \times 10^{-11}\bigg(\dfrac{22500}{36}\bigg)} \\ \\ \\ \implies \sf{F = 6.67 \times 10^{-11}\bigg(\dfrac{2.25 \times 10^{4}}{36}\bigg)} \\ \\ \\ \implies \sf{F = 6.67 \times \dfrac{2.25 \times 10^{-7}}{3.6 \times 10}} \\ \\ \\ \implies \sf{F = 6.67 \times \dfrac{2.25 \times 10^{-8}}{3.6}} \\ \\ \\ \implies \sf{F = \dfrac{15 \times 10^{-8}}{3.6}} \\ \\ \\ \implies \sf{F = 4.17 \times 10^{-8} N}$

Hence the Force of attraction is 4.17 × 10^(-8) N