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Elina [12.6K]
3 years ago
9

two objects are connected by a light string that passes over a frictionless pulley as in the figure below, where m1 A) a1=g

B) a1>g
C) a1 D) a1>a2
E) a1
Physics
1 answer:
Shtirlitz [24]3 years ago
5 0
You didn't attach the figure. Your text is incomplete. And you never got around to asking a question. Other than that, we're on it.
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Mountain pull. A large mountain can slightly affect the direction of ""down"" as determined by a plumb line. Assume that we can
Nastasia [14]

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

3 0
3 years ago
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