Answer:
Explanation:
To find out how much plastic and aluminum is needed for the cylindrical tube of tennis balls, we first need to calculate the dimensions of the tube. Since the tube is just large enough to fit all three balls, we can calculate its height by multiplying the diameter of a tennis ball (6.9 cm) by 3 (the number of balls that fit in the tube) to get 20.7 cm.
The volume of a cylinder is given by the formula V = πr^2h, where r is the radius of the cylinder and h is its height. Since we know that three tennis balls fit in the tube, we can calculate its radius by dividing the diameter of a tennis ball (6.9 cm) by 2 to get 3.45 cm. Therefore, the volume of the cylindrical tube is:
V = π(3.45 cm)^2(20.7 cm) ≈ 2,260.8 cubic centimeters
The bottom and side of the tube are made of plastic, so we can calculate the amount of plastic needed by subtracting the volume of the lid (which is made of aluminum) from the total volume of the tube. The lid is a circular disk with a diameter equal to that of the cylinder and a height equal to its thickness (which we don't know). Therefore, its volume can be calculated using the formula for the volume of a cylinder:
V = πr^2h
where r is half the diameter of the lid and h is its thickness.
Since we know that three tennis balls fit in the tube, we can calculate its diameter by multiplying the diameter of a tennis ball (6.9 cm) by 3 to get 20.7 cm (the same as the height). Therefore, r = 10.35 cm.
We don't know how thick the lid is, so let's call its thickness t. Then its height is h = t.
The volume of aluminum needed for the lid is:
V = π(10.35 cm)^2t ≈ 3375t cubic centimeters
Therefore, the amount of plastic needed for the bottom and side of the tube is:
2,260.8 - 3375t cubic centimeters
To find out how much unfilled space there is in the tube, we need to subtract from its total volume (2,260.8 cubic centimeters) the volume occupied by three tennis balls:
V = 3(4/3π(3.45 cm)^3) ≈ 131.9 cubic centimeters
Therefore, there are approximately 2,128.9 cubic centimeters of unfilled space in the tube.
To express this as a percentage of the total volume of the tube:
(2,128.9 / 2,260.8) x 100% ≈ **94%**
So approximately **94%** of the total volume of the tube is unfilled space.
I hope this helps! Let me know if you have any other questions.