An unstrained horizontal spring has a length of 0.34 m and a spring constant of 180 N/m. Two small charged objects are attached to this spring, one at each end. The charges on the objects have equal magnitudes. Because of these charges, the spring stretches by 0.024 m relative to its unstrained length. Determine (a) the possible algebraic signs and (b) the magnitude of the charges. g

a

            ,

OR

              ,    

b

and

Explanation:

Generally the force exerted on the string is mathematically represented as

substituting values 180 N/m for k and 0.024 m for e

This force can also equivalent to the electrostatic force between the charges i.e

substituting for k and ( 0.34 + 0.024 = 0.364 m) for r we have

Given the spring was stretched it means that the force between the charges is a repulsive for which tell us that both charge are of the same sign thus the possible  algebraic signs  of the charges are

            ,

OR

,

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.