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amid [387]
3 years ago
13

Tom has a mass of 67.1 kg and Sally has a mass of 58.6 kg. Tom and Sally are standing 32.3 m apart on a massless dance floor. Sa

lly looks up and she sees Tom. She feels an attraction. If the attraction is gravitation, find its magnitude. Assume both can be replaced by point masses and that the gravitational constant is 6.67259 × 10−11 N · m2 /kg2 . Answer in units of N.
Physics
1 answer:
Gnom [1K]3 years ago
3 0

Answer:F=25.14\times 10^{-11} N

Explanation:

Given

mass of Tom(m_t)=67.1 kg

mass of sally(m_s)=58.6 kg

Distance between them(d)=32.3 m

Gravitational constant(G)=6.67\times 10^{-11} N.m^2/kg^2

Gravitational attraction is given by

F=\frac{Gm_1m_2}{d^2}

F=\frac{6.67\times 10^{-11}\times 67.1\times 58.6}{32.3^2}

F=25.14\times 10^{-11} N

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2 years ago
This same experiment is performed on the international space station. What is the primary issue with performing this experiment
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3 years ago
Mountain pull. A large mountain can slightly affect the direction of ""down"" as determined by a plumb line. Assume that we can
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Answer:

6.18 um

Explanation:

The plumb line will be pulled down by a combination of the gravitationall pull of Earth and of the mountain. The Earth pulls down and the mountain to the side. Because of this it will fall not in a straight line down, but slightly to the side. Since the plumb line will follow the compound gravity we can imagine a rectangle triangle formed by the plumb line, a vertical line that ends at the same height as the plumb line, and the sideways displacement.

The total gravity will be proportional to the plumb line lenght, the vertical line will be proportional to Earth's gravity and the sideways displacement to the mountain pull.

The gravity of Earth is 9.81 m/s^2

The pull of the mountain will be defined by Newton's law of universal gravitation:

F = G \frac{m1 * m2}{r^2}

Where

F: pull force

G: universal gravitational constant (6.67e-11 m^3/(kg * s)

m1: mass of the mountain

m2: mass of the plumb

r: distance between mountain and plumb (3 km in this case)

If we divide both sides by m2 we obtain the acceleration towards the mountain of the plumb

a = G \frac{m1}{r^2}

Now we need the mass of the mountain. This will be its volume times it's density. The volume depends on the radius (since we consider it as a sphere)

m1 = \delta * \frac{4}{3} * \pi * r^3

m1 = 2.6e3 * \frac{4}{3} * \pi * 1000^3 = 1.09e13 kg

So, the acceleration on the plumb will be

a = 6.67e-11 \frac{1.09e13}{3000^2} = 8.08e-5 m/s^2

This is very small compared to the pull of Earth, so we can make an approximation that the length of the plumb line is equal to vertical line.

We can use the principle of similar triangles to say that:

\frac{\Delta x}{L} = \frac{a}{g}

So:

\Delta x = \frac{a * L}{g} = \frac{8.08e-5 * 0.75}{9.81} = 6.18e-6 m = 6.18 um

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