Question: The force between a pair of 0.005 C is 750 N. What is the distance between them?
Answer:
17.32 m
Explanation:
From coulomb's Law,
F = kqq'/r²........................... Equation 1
Where F = Force between the force, q' and q = both charges respectively, k = coulomb's constant, r = distance between both charges.
make r the subject of the equation above
r = √(kqq'/F)..................... Equation 2
From the question,
Given: q = q' = 0.005 C, F = 750 N
Constant: k = 9.0×10⁹ Nm²/C².
Substitute these values into equation 2
r = √(9.0×10⁹×0.005×0.005/750)
r = √(300)
r = 17.32 m.
Hence the distance between the pair of charges = 17.32 m
K.E.= 1/2 x MV^2 = 1/2 x 40(kg) x (25x25) =12500J
Weight is force.
The metric unit of force is the Newton.
1 Newton is the force needed to accelerate 1 kilogram at the rate of 1 m/s² .
To solve this problem we will use the kinematic equations of descriptive motion of a projectile for which both the height reached and the distance traveled are defined. From this type of movement the lion reaches a height (H) of 3m and travels a horizontal distance (R) of 10 m. Mathematically the equations that describe this movement are given as,


Dividing the two equation we have that





Substituting values of H and R, we get



Substituting the value of \theta in equation we get,






Therefore the speed of the mountain lion just as it leaves the ground is 9.98m/s at an angle of 50.2°