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Dima020 [189]
3 years ago
10

An evacuated long tube contains a coin and a feather. If both objects fall together starting from the top of the tube, it is exp

ected that: If this experiment is repeated at a place 2000 kilometers above the sea-level, the acceleration due to gravity gexp is expected to :
Physics
1 answer:
Natasha_Volkova [10]3 years ago
5 0

Answer:

gexp = 3.65 m/s²

Explanation:

The value of acceleration due to gravity changes with the altitude. The following formula gives the value of acceleration due to gravity at some altitude from the sea level:

gexp = g(1 - 2h/Re)

where,

gexp = expected value of g at altitude = ?

g = acceleration due to gravity at sea level = 9.8 m/s²

h = altitude = 2000 km = 2 x 10⁶ m

Re = Radius of Earth = 6.37 x 10⁶ m

Therefore,

gexp = (9.8 m/s²)(1 - 2*2 x 10⁶ m/6.37 x 10⁶ m)

<u>gexp = 3.65 m/s²</u>

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Answer:

The third charged particle must be placed at x = 0.458 m = 45.8 cm

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F: Electric force in Newtons (N)

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d: distance between the charges in meters (m)

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1m = 100 cm

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q₂= +7.00 μC = +7.00 * 10⁻⁶ C

d₁ = x (m)

d₂ = 1-x (m)

Problem development

Look at the attached graphic.

We assume a positive charge q₃ so F₁₃ and F₂₃ are repulsive forces and must be equal so that the net force is zero:

We use formula (1) to calculate the forces F₁₃ and F₂₃

F_{13} = \frac{k*q_1*q_3}{d_1^2}

F_{23} = \frac{k*q_2*q_3}{d_2^2}

F₁₃ = F₂₃

\frac{k*q_1*q_3}{d_1^2} = \frac{k*q_2*q_3}{d_2^2} We eliminate k and q₃ on both sides

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5-10x+5x^2=7 x^2

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We solve the quadratic equation:

x_1 = \frac{-b+\sqrt{b^2-4ac} }{2a} = \frac{-10+\sqrt{10^2-4*2*(-5)} }{2*2} = 0.458m

x_2 = \frac{-b-\sqrt{b^2-4ac} }{2a} = \frac{-10-\sqrt{10^2-4*2*(-5)} }{2*2} = -5.458m

In the option x₂, F₁₃ and F₂₃ will go in the same direction and will not be canceled, therefore we take x₁ as the correct option since at that point the forces are in  opposite way .

x = 0.458m = 45.8cm

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