Answer:
Explanation:
Energy of system of charges
= k q₁q₂ / r₁₂ + k q₁q₃ / r₁₃ + k q₃q₂ / r₃₂
q₁ , q₂ and q₃ are charges and r₁₂ , r₁₃ , r₃₂ are densities between them
9 x 10⁹ ( 2x2 x10⁻¹²/ .25 + 2x2 x10⁻¹²/ .25 + 2x2 x10⁻¹²/ .25 )
= 9 x 10⁹ x 3 x 16 x 10⁻¹²
= 432 x 10⁻³
= .432 J .
Answer:
N = 177843 sheets
Explanation:
We are given;
Mass;m = 0.0035 kg
Pressure; p = 101325 pa = 101325 N/m²
L = 0.279m
W = 0.216m
The weight of N sheets is N(mg)
Where;
m is the mass of one sheet
N is number of sheets
g is the acceleration due to gravity.
The pressure equals weight divided by the area on which the weight presses:
Thus,
p= F/A = Nmg/(L•W)
Therefore, making N the subject;
N = pLW/(mg)
N = 101325 x 0.279 x 0.216/ (0.0035 x 9.81)
N = 177843
The electrostatic force between two charges is given by Coulomb's law:

where
ke is the Coulomb's constant
q1 is the first charge
q2 is the second charge
r is the separation between the two charges
By substituting the data of the problem into the equation, we can find the magnitude of the force between the two charges:
Answer:
Part a)

Part b)

Explanation:
Since the ball and rod is an isolated system and there is no external force on it so by momentum conservation we will have

here we also use angular momentum conservation
so we have

also we know that the collision is elastic collision so we have

so we have

also we know

also we know

so we have


now we have


Part b)
Now we know that speed of the ball after collision is given as

so it is given as
