How do you get the answer to this problem? The Diameter of a typical pinhead is 1 milimeter. If a common virus measures 250 nanometers in diameter, and the virus fits together so well that there are no gaps or overlaps, how many viruses would fit on the pinhead? Express your answer to the nearest whole number.

Find the area of the virus (PI*(d/2)^2 and the area of the pinhead, then

number= areapinhead/areavirus

To get the answer to this problem, you need to compare the sizes of the pinhead and the virus, and then calculate how many viruses would fit on the pinhead.

Step 1: Convert the measurements to the same units
- The diameter of the pinhead is given as 1 millimeter.
- The diameter of the virus is given as 250 nanometers.

To compare these measurements, they need to be in the same unit. Since 1 millimeter is equal to 1,000,000 nanometers, we will convert the diameter of the pinhead to nanometers:
1 millimeter = 1,000,000 nanometers

So, the diameter of the pinhead is 1,000,000 nanometers.

Step 2: Calculate the number of viruses that would fit on the pinhead
To find out how many viruses can fit on the pinhead, we need to determine the ratio between the diameters of the pinhead and the virus.

Divide the diameter of the pinhead by the diameter of the virus:
1,000,000 nanometers / 250 nanometers = 4000

This means that the diameter of the pinhead is 4000 times larger than the diameter of the virus.

Since the question states that the viruses fit together so well that there are no gaps or overlaps, we can assume that the packing is close-packed, meaning that the viruses would arrange themselves in a hexagonal pattern.

In a close-packed hexagonal pattern, each virus touches six other viruses. Therefore, we can now calculate the number of viruses that can fit on the pinhead by dividing the surface area of the pinhead by the surface area of a single virus.

Surface area of the pinhead = π * (radius of the pinhead)^2
Surface area of a single virus = π * (radius of the virus)^2

Since we are given the diameter, we can find the radius by dividing the diameter by 2:
Radius of the pinhead = 1 millimeter / 2 = 0.5 millimeter = 0.5 * 1,000,000 nanometers = 500,000 nanometers
Radius of the virus = 250 nanometers / 2 = 125 nanometers

Now, we can calculate the number of viruses that fit on the pinhead:
Number of viruses = (Surface area of the pinhead) / (Surface area of a single virus)
= (π * (radius of the pinhead)^2) / (π * (radius of the virus)^2)
= (π * (500,000 nanometers)^2) / (π * (125 nanometers)^2)
= (π * 250,000,000,000 nanometers^2) / (π * 15,625 nanometers^2)
= (250,000,000,000 nanometers^2) / (15,625 nanometers^2)
= 16,000,000

Therefore, approximately 16,000,000 viruses can fit on the pinhead when arranged so well that there are no gaps or overlaps.

Expressing the answer to the nearest whole number, we can say that 16,000,000 viruses would fit on the pinhead.