The new magnitude of the force of attraction will be 6 times the original force of attraction
<h3>How to determine the initial force </h3>
- Mass 1 = m₁
- Mass 2 = m₂
- Gravitational constant = G
- Distance apart = r
- Initial force (F₁) = ?
F = Gm₁m₂ / r²
F₁ = Gm₁m₂ / r²
<h3>How to determine the new force </h3>
- Mass 1 = 2m₁
- Mass 2 = 3m₂
- Gravitational constant = G
- Distance apart (r) = r
- New force (F₂) =?
F = Gm₁m₂ / r²
F₂ = G × 2m₁ × 3m₂ / r²
F₂ = 6Gm₁m₂ / r²
But
F₁ = Gm₁m₂ / r²
Therefore
F₂ = 6Gm₁m₂ / r²
F₂ = 6F₁
Thus, the new magnitude of the force of attraction will be 6 times the original force of attraction
Learn more about gravitational force:
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I know i did part a correctly. heres what i did: momentum is conserved: m1 * u - m2 * u = m2 * v or (m1 - m2) * u = m2 * v Also, for an elastic head-on collision, we know that the relative velocity of approach = relative velocity of separation (from conservation of energy), or, for this problem, 2u = v Then (m1 - m2) * u = m2 * 2u m1 - m2 = 2 * m2 m1 = 3 * m2 m1 is the sphere that remained at rest (hence its absence from the RHS), so m2 = 0.3kg / 3 m2 = 0.1 kg b) this part confuses me, heres what i did (m1 - m2) * u = m2 * v (.3kg - .1kg)(2.0m/s) = .1kg * v .4 kg = .1 v v = 4 m/s What my teacher did: (.3g - .1g) * 2.0m/s = (.3g + .1g) * v I understand the left hand side but i dont get the right hand side. Why is m1 added to m2 when m1 is at rest which makes its v = zero?? v = +1.00m/s since the answer is positive, what does that mean? Also, if v was -1.00m/s what would that mean? thanks!
<span>Reference https://www.physicsforums.com/threads/elastic-collision-with-conservation-of-momentum-problem.651261...</span>
When is at the end of the runway the velocity of the plane is given by the equation
where s=1800 m is the runway length. Thus
At half runway the velocity of the plane is
Therefore at midpoint of runway the percentage of takeoff velocity is
‰
I'm not completely sure, but I believe the answer is B. Solid
