A245-kg object and a 545-kg object are separated by 3.30 m. (a) find the magnitude of the net gravitational force exerted by these objects on a 59.0-kg object placed midway between them. n (b) at what position (other than an infinitely remote one) can the 59.0-kg object be placed so as to experience a net force of zero from the other two objects? m from the 545 kg mass toward the 245 kg mass

a) The magnitude of the net gravitational force exerted by these objects on a 59.0 kg is .

b) By placing object of 59.0 kg 1.976 m from 245 kg object and 1.324 m from 545 kg object.

Explanation:

Mass of an object-1=

Mass of an object-2=

Distance between   = 3.30 m

Mass of an object-3 =

If placed midway between them .

Force on due to

Distance between  

Force on due to

Distance between  

Net force on :

The magnitude of the net gravitational force exerted by these objects on a 59.0 kg is .

b) Distance between   = r = 3.30 m

Let the distanced between be r

Force experienced by due to :

Let the distanced between be r'

Force experienced by due to :

Net force equal to zero, then :

R = r+r'

3.30 m = 1.4914\times r' + r'

r' = 1.324 m

r = 1.976 m

By placing object of 59.0 kg 1.976 m from 245 kg object and 1.324 m from 545 kg object.

b.transmitters send radio waves, and receivers capture radio waves.

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.