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uranmaximum [27]
3 years ago
9

Assume the Earth is a ball of perimeter 40, 000 kilometers. There is a building 20 meters tall at point a. A robot with a camera

placed at 1.75 m. above the surface of the Earth starts walking away from the building. At what distance from a, computed on the surface, does the robot cease to see the (top of the) building?

Physics
1 answer:
torisob [31]3 years ago
5 0

Answer:

Approximately 21 km.

Explanation:

Refer to the not-to-scale diagram attached. The circle is the cross-section of the sphere that goes through the center C. Draw a line that connects the top of the building (point B) and the camera on the robot (point D.) Consider: at how many points might the line intersects the outer rim of this circle? There are three possible cases:

  • No intersection: There's nothing that blocks the camera's view of the top of the building.
  • Two intersections: The planet blocks the camera's view of the top of the building.
  • One intersection: The point at which the top of the building appears or disappears.

There's only one such line that goes through the top of the building and intersects the outer rim of the circle only once. That line is a tangent to this circle. In other words, it is perpendicular to the radius of the circle at the point A where it touches the circle.

The camera needs to be on this tangent line when the building starts to disappear. To find the length of the arc that the robot has travelled, start by finding the angle \angle \mathrm{B\hat{C}D} which corresponds to this minor arc.

This angle comes can be split into two parts:

\angle \mathrm{B\hat{C}D} = \angle \mathrm{B\hat{C}A} + \angle \mathrm{A\hat{C}D}.

Also,

\angle \mathrm{B\hat{A}C} = \angle \mathrm{D\hat{A}C} = 90^{\circ}.

The radius of this circle is:

\displaystyle r = \frac{c}{2\pi} = \rm \frac{4\times 10^{7}\; m}{2\pi}.

The lengths of segment DC, AC, BC can all be found:

  • \rm DC = \rm \left(1.75 \displaystyle + \frac{4\times 10^{7}\; m}{2\pi}\right)\; m;
  • \rm AC = \rm \displaystyle \frac{4\times 10^{7}}{2\pi}\; m;
  • \rm BC = \rm \left(20\; m\displaystyle +\frac{4\times 10^{7}}{2\pi} \right)\; m.

In the two right triangles \triangle\mathrm{DAC} and \triangle \rm BAC, the value of \angle \mathrm{B\hat{C}A} and \angle \mathrm{A\hat{C}D} can be found using the inverse cosine function:

\displaystyle \angle \mathrm{B\hat{C}A} = \cos^{-1}{\rm \frac{AC}{BC}}

\displaystyle \angle \mathrm{D\hat{C}A} = \cos^{-1}{\rm \frac{AC}{DC}}

\displaystyle \angle \mathrm{B\hat{C}D} = \cos^{-1}{\rm \frac{AC}{BC}} + \cos^{-1}{\rm \frac{AC}{DC}}.

The length of the minor arc will be:

\displaystyle r \theta = \frac{4\times 10^{7}\; \rm m}{2\pi} \cdot (\cos^{-1}{\rm \frac{AC}{BC}} + \cos^{-1}{\rm \frac{AC}{DC}}) \approx 20667 \; m \approx 21 \; km.

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Explanation:

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Assuming that the wires are cylindrical in shape

As radius of the first wire is half that of the second wire and therefore the area of cross-section of the first wire will be one-fourth of the second wire( ∵ wire is cylindrical, the cross-sectional part will be circle and the area of the circle = π × r² )

As the volume is same for both wires

∴ π × (r_{1})² × l_{1} = π × (r_{2})² × l_{2}

Here

r_{1} is the radius of the first wire

r_{2} is the radius of the second wire

l_{1} is the length of the first wire

l_{2} is the length of the second wire

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By cancelling the same terms on both sides

we get

l_{1} = 4 × l_{2}

⇒ Length of first wire will be four times of the length of second wire

<h3>Strain is defined as the elongation per unit length</h3>

Strain in first wire = ΔL ÷ l_{1} = ΔL ÷ (4 × l_{2})

where ΔL is the elongation of the wire which in this case is same in both wires

Strain in second wire = ΔL ÷ l_{2}

∴ Strain in second wire is four times of strain in first wire

<h3>Stress = F ÷ A</h3>

where F is the force perpendicular to the cross-sectional area

A is the area of cross-section

Force in first wire = m_{1} × g

where m_{1} is the mass hanged to the first wire

g is the acceleration due to gravity

Force in second wire = m_{2} × g

where m_{2} is the mass hanged to the second wire

g is the acceleration due to gravity

Let A_{1} be the cross-sectional area of first wire

A_{2} be the cross-sectional area of second wire

A_{2} = 4 × A_{1} (∵ cross=sectional area of the wire = π × (radius of the wire)² )

Stress in first wire = (m_{1} × g) ÷ (A_{1})

Stress in second wire = (m_{2} × g) ÷ (A_{2}) = (m_{2} × g) ÷ (4 × A_{1})

<h3>Young's modulus is defined as Stress per unit strain</h3>

As Young's modulus is same for both wires, Stress per unit strain must be same for both wires

Stress per unit strain of first wire = ((m_{1} × g) ÷ (A_{1})) ÷ (ΔL ÷ (4 × l_{2}))

Stress per unit strain of second wire = ((m_{2} × g) ÷ (4 × A_{1})) ÷ (ΔL ÷ l_{2})

By equating them we get

m_{2} = 16 × m_{1}

⇒ m_{2} = 16 × 3 = 48 kg

∴ m_{2} = 48 kg

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