Aparticle with charge 3 is located on the x0axis at the point 4cm, and a second aprtucle with charge, what is the magnitude of the total electrostatic force on a third particle with charge 6?

179.7 N

Explanation:

Notice: check attached file/picture for the diagram showing the arrangement of charged particles.

The mathematical expression for coulumb's law is given below;

F= Kq1q2/ r^2 ---------------------------(1).

Slotting in the given parameters;

F(1)= 8.9875× 10^9 × 3× 6/(0.12)^2.

F(1)= 161,775,000,000/ 0.144.

F(1)= 11,234,375,000,000

F(1)= 11.2 N towards left.

Also, the force due to -5 uC charge placed at x = 6 cm.

Using coulumb's law equation in equation (1), we have;

F(2)= 8.9875×10^9 × 5 × 6/ (0.042).

F(2)= 168.5 N.

Therefore, net force on 6 uC charge will be;

=F(1)+ F(2)

= 168.5 N + 11.2 N

= 179.7N.

Hence, the magnitude of the total electrostatic force on a third particle with charge 6 is 179.7 N

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

according to question, we have said that if the population has twice its initial population, then it becomes

so, time taken to reach this above mentioned population is given by

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.