Answer:
Explanation:
Given that:
length of side , a = 0.5 m
charge , q = 6.65 mC
length of diagonal , d = 0.5 * sqrt(2)
d = 0.707 m
F is the force due to adjacent particle ,
F1 is the force due to diagonal particle
Now , for the net charge on a particle
Fnet = 2 * F * cos(45) + F1
Fnet = 2*cos(45) * k * q^2/a^2 + k * q^2/d^2
Fnet = 9*10^9 * 0.00665^2 * (2* cos(45)/.5^2 + 1/.707^2)
Fnet = 3.05 *10^6 N
the magnitude of net force acting on each particle is 3.05 *10^6 N
part B)
for the direction of particle
d) along the line between the charge and the center of the square outward of the center
The force of static friction acting on the boy is 333 Newtons.
Hence, option (d) 333N is the correct answer.
Given the data in the question;
- Weight of boy;

- Coefficient of static friction;

Force of static friction acting on the boy; 
<h3>Friction</h3>
Friction is simply referred as the resistance an object experience when moving or trying to move over another object. It can either by dynamic or static.
Static friction is the resistance experienced by a body at rest.
Coefficient of friction is the measure of how easily a body moves in relation to the other.
Static friction is expressed as;

Where
is the normal force and
is the coefficient of static friction.
We substitute our given values into the equation above

The force of static friction acting on the boy is 333 Newtons.
Hence, option (d) 333N is the correct answer.
Learn more friction: brainly.com/question/17237604
Answer:
When light is scattered, It reflects in many different directions, think about sunlight reflecting off the clouds.
Explanation:
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