Two charged particles are attracted to each other with a force of 7.6 newtons. what will happen if the distance between them shrinks to one third of what it started at?

The complete equation is as follows:

Force between two charge particles = k * Q1 * Q2 / r^2,

where k is constant, Q1 is charge on 1 particle, Q2 is charge on the other particle and r is the distance between the particles.

From the equation, if the distance between them shrinks to one third of what it started at, then the force will be 1/(1/3)^2 = 9 times larger.

It will become 7.6*9 = 68.4 Newtons.

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg

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