Answer:
0.0078 N
Explanation:
The electrostatic force between two charges is given by Coulomb's law:
where:
is the Coulomb's constant
are the two charges
r is the separation between the two charges
The force is attractive if the two charges have opposite sign, and repulsive if the two charges have same sign.
In this problem, we have:
located at ![x_1=-3 m](https://tex.z-dn.net/?f=x_1%3D-3%20m)
located at ![x_2=0 m](https://tex.z-dn.net/?f=x_2%3D0%20m)
located at ![x_3=+3 m](https://tex.z-dn.net/?f=x_3%3D%2B3%20m)
The force between charge 1 and charge 2 is:
![F_{12}=k\frac{q_1 q_2}{(x_2-x_1)^2}=(8.99\cdot 10^9)\frac{(1\cdot 10^{-6})(+9\cdot 10^{-6})}{3^2}=0.0090 N](https://tex.z-dn.net/?f=F_%7B12%7D%3Dk%5Cfrac%7Bq_1%20q_2%7D%7B%28x_2-x_1%29%5E2%7D%3D%288.99%5Ccdot%2010%5E9%29%5Cfrac%7B%281%5Ccdot%2010%5E%7B-6%7D%29%28%2B9%5Ccdot%2010%5E%7B-6%7D%29%7D%7B3%5E2%7D%3D0.0090%20N)
And since the two charges have opposite sign, the force is attractive, so the force on charge 1 is towards the right.
The force between charge 1 and charge 3 is:
![F_{13}=k\frac{q_1 q_3}{(x_3-x_1)^2}=(8.99\cdot 10^9)\frac{(1\cdot 10^{-6})(5\cdot 10^{-6})}{6^2}=0.0012 N](https://tex.z-dn.net/?f=F_%7B13%7D%3Dk%5Cfrac%7Bq_1%20q_3%7D%7B%28x_3-x_1%29%5E2%7D%3D%288.99%5Ccdot%2010%5E9%29%5Cfrac%7B%281%5Ccdot%2010%5E%7B-6%7D%29%285%5Ccdot%2010%5E%7B-6%7D%29%7D%7B6%5E2%7D%3D0.0012%20N)
And since the two charges have same sign, the force is repulsive, so the force on charge 1 is towards the left.
Therefore, the net force on charge 1 is:
![F=F_{12}-F_{13}=0.0090-0.0012 = 0.0078 N](https://tex.z-dn.net/?f=F%3DF_%7B12%7D-F_%7B13%7D%3D0.0090-0.0012%20%3D%200.0078%20N)
towards the right.
Answer:
<h2>0.39m/s^2</h2>
Explanation:
Step one:
given data
mass m= 300kg
applied force F= 1000N
coefficient of friction μ= 0.3
Step two:
The net force Fn= applied force-friction force
Fn=F-F1
F1= limiting force
F1=μ*m*g
F1=0.3*300*9.81
F1=882.9N
the Net force= 1000-882.9
Fn=117.1N
Step three:
we know that
F=ma
Fnet=ma
a= Fnet/m
a=117.1/300
a=0.39m/s^2