28 times stronger than the force of gravity on the surface of the Earth.
Force on the particle is defined as the application of the force field of one particle on another particle. the electrical force between q₁ and q₃ will be –1. 1 × 10¹¹ N.
<h3>What is electric force?</h3>
Force on the particle is defined as the application of the force field of one particle on another particle. It is a type of virtual force.
The electric force in the second case will be the same as in the first case. Therefore the force on the particle will be the same.



Hence the electrical force between q₁ and q₃ will be –1. 1 × 10¹¹ N.
To learn more about the electric force refer to the link;
brainly.com/question/1076352
Answer:
acceleration a = 1.04 m/s2
Explanation:
Assume the train has a speed of 23m/s when the last car passes the railway workers. Once this happens the last car would have traveled a total distance of the 180m distance between the railway worker standing 180 m from where the front of the train started plus the 75m distance from the first car to the last car:
s = 75 + 180 = 255 m
We can use the following equation of motion to find out the distance traveled by the car:
where v = 23 m/s is the velocity of the car when it passes the worker,
= 0m/s is the initial velocity of the car when it starts, a m/s2 is the acceleration, which we are looking for.



Answer:
Explanation:
The relationship between angle and wavelength for maxima and minima in Young's double slit experiment is given by
For constructive interference

For Destructive interference

where 

m=order of maxima and minima
for second order maxima i.e. 
For smallest separation taking 


