5.0j as we know that work is derived as force into displacement so 5*1=5si force will be 5.0j
Q: Two charges q1 and q2, that are distance d apart , repel each other with a force of 6.40 N. what would be the force between two charges q1'=2q1 and q2'=3q2 that that are distance d apart?
Answer:
The force = 38.4 N
Explanation:
From coulombs law,
F = kq₁q₂/r² ............................ Equation 1
Where F = Force of attraction or repulsion between the charges, q₁ and q₂ = first and second charge respectively, r = distance between the charges, k = constant of proportionality.
When, F = 6.4 N, r = d m.
6.4 = kq₁q₂/d²......................... Equation 1
When q₁' = 2q₁, q₂' = 3q₂, r = d cm
F = k(2q₁)(3q₂)/d²
F = 6kq₁q₂/d².......................... Equation 2
Dividing Equation 1 by equation 2
6.4/F = kq₁q₂/d²/(6kq₁q₂/d²)
6.4/F = 1/6
F = 6.4×6
F = 38.4 N.
Thus the force = 38.4 N
The Volume of a Cube with a side(s) of 11.4 cm is
<span>V = s^3 </span>
<span> V = (11.4)^3 cm^3 </span>
<span> V = 1481.544 cm^3 </span>
<span> V = 1482 cm^3</span>
<span>however the Surface Area of a cube is the combined area of all the sides </span>
<span> A = 6s^2 </span>
<span> A = 6(11.4)^2 cm^2 </span>
<span> A = 779.76 cm^2 </span>