Physically inactive (sitting down a lot during free time)

Physics

1 answer:

alekssr [168]3 years ago

6 0

Yes, that’s physically inactive. You should be moving your arms and legs if you’re going to be sitting down.

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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

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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

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