Question

Ross will make a water balloon that can be modeled with a sphere. A constraint he must consider is that when the radius of the balloon exceeds 5 inches, the balloon will pop. If he uses a garden hose with a flow rate of 12 gallons per minute to fill up the balloon, for how many seconds can he fill it before it pops? Round to the nearest tenth of a second. (1 gallon = 231 cubic inches)

Enter your answer in the box.

Answer 1

Answer: no

Explanation:

Answer 2

Ross can fill the water balloon for approximately 11.34 seconds before it pops, given the constraints.

How did we get the value?

Calculate
`\(V_{\text{max}}\)`:

`\[ R = 5 \, \text{inches} \]`

`\[ V_{\text{max}} = (4)/(3)\pi * (5^3) \]`

`\[ V_{\text{max}} = (4)/(3)\pi * 125 \]`

`\[ V_{\text{max}} \approx 523.8 \, \text{cubic inches} \]`

Convert the flow rate to cubic inches per minute:

`\[ \text{Flow rate} = 12 \, \text{gallons/minute} * 231 \, \text{cubic inches/gallon} \]`

`\[ \text{Flow rate} \approx 2772 \, \text{cubic inches/minute} \]`

Calculate the time in minutes:

`\[ \text{Time (minutes)} = \frac{V_{\text{max}}}{\text{flow rate}} \]`

`\[ \text{Time (minutes)} = (523.8)/(2772) \]`

`\[ \text{Time (minutes)} \approx 0.1890 \, \text{minutes} \]`

Convert the time to seconds:

`\[ \text{Time (seconds)} = \text{Time (minutes)} * 60 \]`

`\[ \text{Time (seconds)} \approx 11.34 \, \text{seconds} \]`

So, Ross can fill the water balloon for approximately 11.34 seconds before it pops, given the constraints.