This can be done using integration and vector analysis. By considering a differential element of the line charge distribution dq = Q/a*dy, we could calculate the dFx and dFy and solve for Fx and Fy through integration.
dFx = -xkQdq / (y^2 + x^2)^3/2 = -xkQ^2/a/(y^2+x^2)^(3/2)*dy
Fx = integral from 0 to a -xkQ^2/a/(y^2+x^2)^(3/2)*dy
Fx = -xkQ^2/a*int(0, a, dy/(y^2+x^2)^(3/2))
Fx = -kQ^2/x(a^2+x^2)^(1/2)
dFy = kQy/(y^2+x^2)^(3/2) * dq
dFy = kQ^2/a*y/(y^2+x^2)^(3/2)*dy
Fy = int(0, a, kQ^2/a*y/(y^2+x^2)^(3/2)*dy)
Fy = kQ^2/a int(0, a, dy y/(y^2+x^2)^(3/2))
Fy = kQ^2/a * (1/x - 1/(a^2+x^2)^(1/2))
Answer:
c. Light energy to thermal energy
Explanation:
The energy from the sun comes in the form of light energy but is converted to thermal energy.
Answer: D. One-dimensional motion
Explanation:
D=m/v then v=m/d=17.5/2=8.75ml
