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kondor19780726 [428]
3 years ago
5

An airplane has a mass of 3.1x10^4 kg and takes off under the influence of a constant net force of 3.7x10^4 N. What is the net f

orce that acts of the plane's 78-kg pilot?
Physics
1 answer:
ra1l [238]3 years ago
7 0

Answer:

Net force that acts on the pilot = 93 N (rounded off to nearest 1 N)

Explanation:

An airplane has a mass of 3.1 × 10^{4} Kg

The applied force for it to take off = 3.7 × 10^{4} N

According to Newton's second law of motion; applied force (F) is directly proportional to the product of mass (M) and acceleration (A) of the object.

i.e F ∝ MA

This means that the applied force (F) is also directly proportional to the mass (M) of the object.

i.e F ∝ M

Acceleration (A) in this case is constant.

If the airplane's mass = 3.1 × 10^{4} kg requires 3.7 × 10^{4} N to take off,

Then the pilot's mass of 78 kg will require;

\frac{78 * 3.7 * 10^4}{3.1 * 10^4} = 93.09677419 N = 93N (rounded off to nearest 1 N)

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Q=ne\\ =(1.50*10^9)(1.6*10^-^1^9C)\\ =2.4*10^-^1^0C

The potential difference V between the disks separated by a distance d is given by,

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Substitute 2.00×10⁵N/C for E and 0.50×10⁻³m for d.

V=Ed\\ =(2.00*10^5N/C)(0.50*10^-^3m)\\ =100V

The capacitance C of the capacitor is given by,

C=\frac{Q}{V} \\ =\frac{(2.4*10^-^1^0C)}{100V} \\ =2.4*10^-^1^2F

The capacitance of a parallel plate capacitor is given by,

C=\frac{\epsilon_0A}{d}

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Rewrite the expression for A.

C=\frac{\epsilon_0A}{d}\\ A=\frac{Cd}{\epsilon_0} \\ =\frac{(2.4*10^-^1^2F)(0.50*10^-^3m)}{(8.85*10^-^1^2C^2/Nm^2)} \\ =1.36*10^-^4m^3

the area A of the disks is given by,

A=\frac{\pi D^2}{4} \\ D=\sqrt{\frac{4A}{\pi } }

Here, D is the diameter of the disk.

D=\sqrt{\frac{4A}{\pi } }\\ =\sqrt{\frac{4(1.36*10^-^4m^2)}{3.14} } \\ =0.01316m\\ =1.32cm

The diameter of each disc is found to be 1.32 cm.



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