Indiana Jones is in a temple searching for artifacts. He finds a gold sphere with a radius of 2 cm sitting on a pressure sensitive plate. To avoid triggering the pressure plate, he must replace the gold with something of equal mass. The density of gold is 19.3.103 kg/m3, and the volume of a sphere is V = 4/3 Ar3. Indy has a bag of sand with a density of 1,602 kg/m3. (A) What volume of sand must he replace the gold sphere with? If the sand was a sphere, what radius would it have?

Volume of Sand = 0.4 m³

Radius of Sand Sphere = 0.46 m

Explanation:

First we need to find the volume of gold sphere:

Vg = (4/3)πr³

where,

Vg = Volume of gold sphere = ?

r = radius of gold sphere = 2 cm = 0.02 m

Therefore,

Vg = (4/3)π(0.2 m)³

Vg = 0.0335 m³

Now, we find mass of the gold:

ρg = mg/Vg

where,

ρg = density of gold = 19300 kg/m³

mg = mass of gold = ?

Vg = Volume of gold sphere = 0.0335 m³

Therefore,

mg = (19300 kg/m³)(0.0335 m³)

mg = 646.75 kg

Now, the volume of sand required for equivalent mass of gold, will be given by:

ρs = mg/Vs

where,

ρs = density of sand = 1602 kg/m³

mg = mass of gold = 646.75 kg

Vs = Volume of sand = ?

Therefore,

1602 kg/m³ = 646.75 kg/Vs

Vs = (646.75 kg)/(1602 kg/m³)

Vs = 0.4 m³

Now, for the radius of sand sphere to give a volume of 0.4 m³, can be determined from the formula:

Vs = (4/3)πr³

0.4 m³ = (4/3)πr³

r³ = 3(0.4 m³)/4π

r³ = 0.095 m³

r = ∛(0.095 m³)

r = 0.46 m

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.