Question

A spherical helium balloon that is 8 feet in diameter can lift about 17 pounds. What does the diameter of a balloon need to be to lift a person who weighs 136 pounds? Explain.

Answer 1

Answer:

The diameter of the balloon that can lift a person weighing 136 pounds is

16 feet

Step-by-step explanation:

To get the diameter of the balloon that can lift a person weighing 136 pounds, we need to know the volume which this diameter will occupy

Firstly, we must know that the spherical balloon will have same density irrespective of its diameter since it is made of same material.

Thus since, density is constant, what will change will be the volume and the weight. Hence, what we are saying is that the ratio of the volume to the weight at any point in time irrespective of the weight and volume will be the same. This basis will provide a solution to our question.

With an 8 feet diameter, the volume that can be lifted would be the volume of a sphere with diameter 8. If the diameter is 8, then the radius is 4.

The volume of a sphere = 4/3 × π× $r^{3}$ = 4/3 × π × $4^{3}$ = 256π/3 $ft^{3}$

So, what this means is that, a spherical helium balloon of size 256π/3 $ft^{3}$ can lift a person weighing 17 pounds

Now, let the radius of the balloon that can lift a person of weight 136 pounds be x feet

The needed volume is thus 4/3 × π× $x^{3}$ = 4π$x^{3}$/3 $ft^{3}$

Now let's make a relationship;

A volume of  256π/3 $ft^{3}$  lifts 17 pounds

A volume of  4π$x^{3}$/3 $ft^{3}$    lifts 136 pounds

To get x, we simply use a cross-multiplication;

256π/3 $ft^{3}$ × 136 = 4π$x^{3}$/3 $ft^{3}$ × 17

4π$x^{3}$/3 $ft^{3}$ / 256π/3 $ft^{3}$ = 136/17

$x^{3}$/64 = 8

$x^{3}$ = 64 × 8

$x^{3}$ = 512

x = $\sqrt[3]{512}$

x = 8 feet

Since the radius is 8 feet, the diameter will be 2 × 8 = 16 feet