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myrzilka [38]
3 years ago
8

The gravitational constant G was first measured accurately by Henry Cavendish in 1798. He used an exquisitely sensitive balance

to measure the force between two lead spheres whose centers were 0.27 m apart. One of the spheres had a mass of 188 kg, while the mass of the other sphere was 0.93 kg.
What was the ratio of the gravitational force between these spheres to the weight of the lighter sphere?
Physics
1 answer:
belka [17]3 years ago
7 0

The gravitational force between the spheres is

F_{\rm g}=\dfrac{G(188\,\mathrm{kg})(0.93\,\mathrm{kg})}{(0.27\,\mathrm m)^2}\approx1.6\times10^{-7}\,\mathrm N

where <em>G</em> = 6.674 x 10⁻¹¹ N m²/kg².

The weight of the lighter sphere is

F_{\rm w}=(0.93\,\mathrm{kg})g\approx9.1\,\mathrm N

where <em>g</em> = 9.80 m/s².

The ratio between the two forces is then

\dfrac{F_{\rm g}}{F_{\rm w}}\approx1.8\times10^{-8}

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The total momentum of the players after collision is 130 kgm/s.

The given parameters:

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F = kx\\\\k =\frac{F}{x}\\\\where \ F = mg\\\\k = \frac{mg}{x}\\\\given \ that:\\\\mass \ of \ each \ wheel = 425 \ kg\\\\x = 7.7cm = 0.077 m\\\\g = 9.8 \ m/s^2\\\\Then;\\\\k = \frac{425 \ kg * 9.8 \ m/s^2}{0.077 \ m}\\\\k = 5.409*10^4 \ N/m

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