A solid copper sphere hangs at the bottom of a steel wire of negligible mass. The top end of the wire is fixed. When the wire is struck, it vibrates with a fundamental frequency of 300 Hz. The copper object is then submerged in water so that half its volume is below the water line. Determine the new fundamental frequency. The density of water, copper, and steel is 1000kg/m³, 8960kg/m³, and 8050kg/m³ respectively

Explanation:

The problem is based on vibration in string .For it the formula is as follows :

n = k √T , n is frequency of fundamental tone , k is constant for a string of given length and T is tension in the string .

In the first case weight of sphere is equal to tension .

weight = volume of sphere x density of copper x g

= v x 8960 x g

Putting the values

300 = k √ (v x 8960 x g)

300 = k √ 87808v

When the sphere is half submerged

tension is equal to apparent weight of sphere in water

=  v x 8960 x g - .5  v x 1000 x 9.8 , v is volume

= 82908v

for second case

n =  k √ 82908v

dividing the two equations

n / 300 =

= .9716

n = 291.5 Hz .

i think the answer is a, hope this is right, that would be my best guess.

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.