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Zepler [3.9K]
3 years ago
7

Four identical masses of 2.5 kg each are located at the corners of a square with 1.0-m sides. What is the net force on any one o

f the masses?

Physics
1 answer:
Ghella [55]3 years ago
8 0

Answer:

F=8.0*10^{-10}N

Explanation:

See the attached file for the masses distributions

The force between two masses at distance r is expressed as

F=\frac{Gm_{1}m_{2}  }{r^{2} }\\ G=Gravitional constant \\

since the masses are of the same value, the above formula can be reduce to

F=\frac{Gm^{2}}{r^{2} }\\

using vector notation,Let use consider the force on the lower left corner of the mass due to the upper left side of the mass is

F_{12} =\frac{Gm^{2}}{r^{2} }j\\

The force on the lower left corner of the mass due to the lower right side of the mass is

F_{14} =\frac{Gm^{2}}{r^{2} }i\\

The force on the lower left corner of the mass due to the upper right side of the mass is

F_{13} =\frac{Gm^{2}}{d^{2} }cos\alpha i +\frac{Gm^{2}}{d^{2} }sin\alpha j\\

The net force can be express as

F=\frac{Gm^{2}}{r^{2} }j +\frac{Gm^{2}}{r^{2} }i +\frac{Gm^{2}}{d^{2} }cos\alpha i +\frac{Gm^{2}}{d^{2} }sin\alpha j\\\\F=Gm^{2}[\frac{1}{r^{2}}+ \frac{1}{d^{2}cos\alpha }]i + Gm^{2}[\frac{1}{r^{2}}+ \frac{1}{d^{2}sin\alpha }]j\\\alpha=45^{0}, G=6.67*10^{-11}Nmkg^{-2}

if we insert values we arrive at

F=6.67*10^{-11}*2.5^{2}[\frac{1}{1^{2}}+ \frac{1}{\sqrt{2}^{2}cos45 }]i + 6.67*10^{-11}*2.5^{2}[\frac{1}{1^{2}}+ \frac{1}{\sqrt{2}^{2}sin45}]j\\F=5.643*10^{-10}i+5.643*10^{-10}j

if we solve for the magnitude, we arrive at

F=5.643*10^{-10}i+5.643*10^{-10}j \\F=\sqrt{(5.643*10^{-10})^{2} +(5.643*10^{-10})}^{2} \\F=8.0*10^{-10}

Hence the net force on one of the masses is

F=8.0*10^{-10}N

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Assume an axon has an internal diameter of 1μm and a myelin sheath 1μm thick. The internal specific resistance is 100 Ω cm. For
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The helicopter is moving south at 175 km/h, relative to the wind.

But the wind is moving east at 85 km/h, relative to the ground.

This means that the helicopter is moving south east relative to the ground.

Every hour, the helicopter will move 175 km to the south and 85 km to the east, relative to the ground.

This means that we can determine the speed and direction of the helicopter using a right triangle and simple trigonometry.

Refer to the triangle b1.

The distance traveled by the helicopter in 1 hour is denoted by d.

d is the hypotenuse of the right triangle.

Using the Pythagorean Theorem, we can calculate d to be 195 km (rounded to 3 s. f.)

Hence the helicopter is traveling at 195 km/h relative to the ground.

To calculate the direction we use,

tan (x) = opposite/adjacent = 85/175

So the angle x is,

x = arctan (\frac{85}{175} ) = 25.9°

Relative to the North, the helicopter is moving at 180° - 25.9° = 154° (rounded to 3 s. f.)

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