Question

Three balls A, B and C are kept in a straight line. The separation between A and C is 1 m, and B is placed at the midpoint between them. The masses of A, B, C are 100 g, 200 g and 300 g respectively. Find the net gravitational force on (a) A, (b) B, and (c) C.​

Answer 1

Answer:

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A ------ 0.50 m -------- B -------- 0.50 -------- C

Ma = 0.100 kg Mb = 0.200 kg Mc = 0.300 kg

Net gravitational force Fa on A is towards right side.

Fa = G Ma Mb / 0.50² + G Ma Mc / 1.0²

= 6.67 * 10⁻¹¹ * 0.100 [ 0.200/ 0.50² + 0.300 /1 ] N

= 7.34 * 10⁻¹² N

Net gravitational force Fc on C is towards B, left side and its magnitude is equal to Fa ie., 7.34 * 10⁻¹² N

Net gravitational force on B is Fb & is towards C

Fb = G Mb [ Mc - Ma ]/ 0.50²

= 6.67 * 10⁻¹¹ * 0.200 * 0.200 * 4 = 1.07 * 10⁻¹¹ N

Answer 2

Answer:

पुढील वाक्य वाच व अधोरेखित शब्दाचा समानार्थी शब्द लिही व शब्दासह वाक्यात करून वाक्य परत लिही . ( नीलमला तिच्या मैत्रिणीला मदत करता न आल्याचा खेद वाटला) .