Day 4 after-lunch quiz:

1.  Look at the plastic cup in front of you.  Without measuring anything:

• Estimate its inside width at the top, as a percent of its inside width at the bottom.

• Estimate the cross-sectional area at the top of the cup, as a percent of the cross-sectional area at the bottom.

2.  Again using that plastic cup:

• Using the paper rulers make your best estimate of the inside volume of the cup.

• If there was a hole in the side of the cup, right next to the bottom of the cup, how fast would the water flowing out of the cup be moving?

If the hole had cross-sectional area .1 cm^2, then

• How much water would flow out of the cup in 10 seconds, assuming the cup is initially full? 

• How long would it take the water level to fall by 2 centimeters? 

• How long would it then take the water level to fall by 2 more centimeters?

• How long do you estimate it would take the cup to empty?

3.  On a 10 x 10 grid of square tiles:

• How many corners are there to measure between?

• Is there any distance between corners that can't be duplicated by the distance between the lower left-hand corner and some other corner of the grid?

• Place the lower upper limit you can on how many distances are possible between corners of a 10 x 10 grid.  That is, find the smallest number you can that you're sure is at least as great as the number of total distances.

• Place the largest lower limit you can on the number of possible distances.

• Could the fact that 50 is equal to 1^2 + 7^2, and also to 5^2 + 5^2, have anything to do with your estimates?

4.  Consider a 10 x 10 x 10 stack of small cubes.  Can you place upper and lower limits on the possible number of distances that could be measured between corners of the small cubes?

5.  How many intact spheres of diameter 1 centimeter could be fit into a 1-meter cube?

You're unlikely to be able to give the actual number, so place the lowest upper limit you can on the number, and the greatest lower limit you can be sure of.

6.  How many of the tiles on this floor would it take to cover, as nearly as possible, a hemispherical dome 600 feet in diameter?  What percent of the area of the dome do you estimate would have to be be left uncovered because of the uneven cracks between the tiles?