Answer:
3.036×10⁻¹⁰ N
Explanation:
From newton's law of universal gravitation,
F = Gm1m2/r² .............................. Equation 1
Where F = Gravitational force between the balls, m1 = mass of the first ball, m2 = mass of the second ball, r = distance between their centers.
G = gravitational constant
Given: m1 = 7.9 kg, m2 = 6.1 kg, r = 2.0 m, G = 6.67×10⁻¹¹ Nm²/C²
Substituting into equation 1
F = 6.67×10⁻¹¹×7.9×6.1/2²
F = 321.427×10⁻¹¹/4
F = 30.36×10⁻¹¹
F = 3.036×10⁻¹⁰ N
Hence the force between the balls = 3.036×10⁻¹⁰ N
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Answer:
trail mix, soup, and gold
hope this helps
have a good day :)
Explanation:
The total work <em>W</em> done by the spring on the object as it pushes the object from 6 cm from equilibrium to 1.9 cm from equilibrium is
<em>W</em> = 1/2 (19.3 N/m) ((0.060 m)² - (0.019 m)²) ≈ 0.031 J
That is,
• the spring would perform 1/2 (19.3 N/m) (0.060 m)² ≈ 0.035 J by pushing the object from the 6 cm position to the equilibrium point
• the spring would perform 1/2 (19.3 N/m) (0.019 m)² ≈ 0.0035 J by pushing the object from the 1.9 cm position to equilbrium
so the work done in pushing the object from the 6 cm position to the 1.9 cm position is the difference between these.
By the work-energy theorem,
<em>W</em> = ∆<em>K</em> = <em>K</em>
where <em>K</em> is the kinetic energy of the object at the 1.9 cm position. Initial kinetic energy is zero because the object starts at rest. So
<em>W</em> = 1/2 <em>mv</em> ²
where <em>m</em> is the mass of the object and <em>v</em> is the speed you want to find. Solving for <em>v</em>, you get
<em>v</em> = √(2<em>W</em>/<em>m</em>) ≈ 0.46 m/s
I know it’s the Coulomb’s law and that I’m pretty sure the answer would be C.Inverse Square.
