An average human weighs about 600 n
if two such generic humans each carried 1.5 coulomb of excess charge, one positive and one negative, how far apart would they have to be for the electric attraction between them to equal their 600 n weight?

The distance between them, x = 5809.47 m

Explanation:

Given,

The average weight of a human, w = 600 N

The charge carried by the humans, q = 1.5 C

If 600 N of force acts between two opposite charges, the distance between the charges can be derived from the Coulomb's laws of force,

Where,

                          = 9 x 10⁹ N m² C⁻²

Therefore

                          x² = 9 x 10⁹ X 1.5² / 600

                              = 33750000

                           x = 5809.46 m

Hence, the distance between the humans should be, x = 5809.47 m

trig 15.3 cos

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.