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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 :

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

How did we get the value?

Calculate [tex]\(V_{\text{max}}\)[/tex]:

[tex]\[ R = 5 \, \text{inches} \][/tex]

[tex]\[ V_{\text{max}} = \frac{4}{3}\pi \times (5^3) \][/tex]

[tex]\[ V_{\text{max}} = \frac{4}{3}\pi \times 125 \][/tex]

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

Convert the flow rate to cubic inches per minute:

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

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

Calculate the time in minutes:

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

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

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

Convert the time to seconds:

[tex]\[ \text{Time (seconds)} = \text{Time (minutes)} \times 60 \][/tex]

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

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

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