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1 answer:
Answer:
about 10,652 balls
Step-by-step explanation:
The length of the court is 94 ft × 12 in/ft = 1128 in. That is 1128/8.5 ≈ 132.7 times the diameter of a ball.
The width of the court is 50 ft × 12 in/ft = 600 in. That is 600/8.5 ≈ 70.6 times the diameter of a ball.
When the balls are hexagonally packed, their row spacing is only (√3)/2 ≈ 0.866 times the ball diameter, so by cutting out one ball per row on alternate rows, about 15% more rows can be put in place. That is, there can be 81 rows of balls, alternating between 132 and 131 balls per row.
The number of balls that fit in that arrangement is ...
132×81 - 40 = 10,652
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Since there is some extra space beyond the last row, the balls can be "randomized" and a few more balls added, probably at least 2 or 3.
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General Idea:
The volume of cylinder is given by , where r is the radius and h is the height of the cylinder.
Applying the concept:
Step 1: We need to find the volume of full cylinder with the given dimensions using the formula.
Volume of full cylinder
Volume of half cylinder
Step 2: Let x be the number of minutes of filling the sand.
of sand filled every 15 seconds, there are four 15 seconds in a minute.
So volume of sand filled in 1 minute.
of sand taken out of cylindrical vase every minute.
Net volume of sand filled in 1 minute = Volume of sand filled in the vase in one minute - Volume of sand taken out in 1 minute
Net volume of sand filled in 1 minute
Volume of sand filled in x minutes .
We need to set up an equation to find the number of minutes need to fill half the volume in cylindrical vase.
Conclusion:
The number of minutes required for the base be half filled with sand is 57