Two point charges are located on the y axis as follows: charge q1 = -2.30 nc at y1 = -0.600 m , and charge q2 = 2.80 nc at the origin (y = 0). what is the magnitude of the net force exerted by these two charges on a third charge q3 = 7.50 nc located at y3 = -0.300 m ?

3.825*10^-6 N

Explanation:

As particle 1 and particle 3 has opposite types of charge, particle 3 will be attracted to particle 1. And as particle 2 and 3 has the same sign, they will repel each other. Due to the position of the particles, both the Force that 1 exerts on 3 and the force that 2 exerts on 3, will have the same direction. Now, you need the magnitude. You can use the following expression:

K is the Coulomb constant equal to 9*10^9 N*m^2/C^2, q1 and q2 is the charge of the particles, and r is the distance.

Force 1 on 3 is equal to:

Force 2 on 3:

The magnitude of the resultant force is the addition of both forces:

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explanation:

physics means: "the branch of science concerned with the nature and properties of matter and energy. the subject matter of physics, distinguished from that of chemistry and biology, includes mechanics, heat, light and other radiation, sound, electricity, magnetism, and the structure of atoms."

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm