Can someone help with this question :)

A negative charge of -0.550 m exerts an upward <a href="/cdn-cgi/l/email-protection" class="__cf_email__" data-cfemail="5868766e

almond37 [142]

Answer:

a. +10.9μC

b. 0.600N and downward

Explanation:

To determine the magnitude of the charge, we use the force rule that exist between two charges which us expressed as

F=(kq₁q₂)/r²

since q₁=-0.55μC and the force it applied on the charge above it is upward,we can conclude that the second charge is +ve, hence we calculate its magnitude as

q₂=Fr²/kq₁

q₂=(0.6N*0.3²)/(9*10⁹*0.55*10⁻⁶)

q₂=0.054/4950

q₂=1.09*10⁻⁵c

q₂=10.9μC.

Hence the second charge is +10.9μC

b. From the rule of charges which state that like charges repel and unlike charges attract, we can conclude that the two above charges will attract since they are unlike charges. Hence the direction of the force will be downward into the second charge and the magnitude of the force will remain the same as 0.600N

8 0

4 years ago

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

Ghella [55]

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

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