Answer:

Explanation:



Distance between q1 and q2,d=20 cm

1m =100 cm
Electric force on charge q3 due to charge q1
(away from q1)
Electric force on charge q3 due to charge q2
(attract toward q2)
Net force=



Answer:
The gravitational potential energy (PE) of a particle or a system of particles is a form of energy that exists due to its position. Mathematically, it is expressed as the product of mass, gravitational acceleration, and the vertical displacement.
Explanation:
Given data:
Mass of the storage box, 
Height, 
The gravitational potential energy of the storage box can be expressed as,



In order to change the frictional force between two solid surfaces, it can be changed by shorter distances and by the amount of weight it has or the amount of force that is pushing that object to go however distance it can.
The gravitational potential energy of the mass in increased approximately by 490 J
Gravitational Potential energy formula is
P=mgh
where mass is given my m= 5 Kg
g is the gravitational force i.e. 9.8 
h is the height above the ground that is given by 10 m
P = 5*9.8*10
P = 490 J
What is gravitational potential energy?
The potential energy that a huge item has in relation to another massive object due to gravity is known as gravitational energy or gravitational potential energy. When two objects descend toward one another, the potential energy associated with the gravitational field is released (transformed into kinetic energy).
To learn more about gravitational potential energy
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Answer:
Fscos63
Explanation:
Given that a horizontal pole is attached to the side of a building. There is a pivot P at the wall and a chain is connected from the end of the pole to a point higher up the wall. There is a tension force F in the chain. What is the moment of the force F about the pivot P?
Taking the moment from the pivot point P, that means the moment at point p = 0
Then, if we consider the weight mg of the pole, according to the principle of equilibrium : sum of the upward forces equal to the sum of the downward forces.
Therefore, mg = Fsinø ....... (1)
Also, taking moment at point P
Let the length of the pole = s
The length of the weight of the pole = 1/2 S
Fscosø = mgs/2
The distance s will cancel out
2Fcosø = mg ...... (3)
Substitute mg in equation 1 into equation 3
2fcosø = fsinø
F will cancel out
Tanø = 2
Ø = tan^-1(2)
Ø = 63.4 degree
The moment of force F about pivot point P will be
Moment = force × distance
Moment = Fcos63 × S
Moment = Fscos63