 Teacher resources and professional development across the curriculum

Teacher professional development and classroom resources across the curriculum            Solutions for Session 3, Part B

See solutions for Problems: B1 | B2 | B3 | B4 | B5 | B6 | B7 | B8 | B9 | B10 | B11 | B12 | B13 | B14 | B15    Problem B1 Answers will vary.   Problem B2

 a. Answers will vary. Distance A is approximately 1 cm. Distance B is approximately 8 or 9 cm. Distance C is approximately 1 dm. Distances D and E are approximately 1 m. b. Answers will vary. Measure the lengths in terms of your own referent measures and then approximate.   Problem B3

 a. One way to estimate distances is to know the length of one's individual pace. First determine your average walking pace, and then use this information to approximate 100 m. b. One way to do this is to walk the 100 m, then multiply the time it took to walk 100 m by 10. Your average walking pace would be measured in meters per second, and could be found by dividing 1,000 by the total time taken (in seconds) to walk the kilometer.   Problem B4

 a. You will need 12 metersticks to do this, one for each edge of the cube. b. One cubic meter is equivalent to 1,000 dm3, since there are 10 dm in each dimension and three dimensions: 10 • 10 • 10 = 1,000. Similarly, 1 m3 is equivalent to a 1 million cm3 and 1 billion mm3. Since 1 L is equivalent to 1 dm3, there are 1,000 L in 1 m3.   Problem B5

Answers will vary, but here are a few examples:

 • A teacup holds about 200 to 250 mL (or 2 to 2.5 dL). • A thimble will probably be about a tenth of a teacup, or 20 mL. • A gas tank of a car holds anywhere from 40 to 70 L.   Problem B6 The phrase 3 cc refers to cubic centimeters, which is equivalent to milliliters; "3 cc" of medicine is 3 mL.   Problem B7 If you measure how much liquid is in a 1 L or 2 L bottle, you'd most likely find that there is a small amount of extra liquid in each, probably as a result of a particular bottling procedure.   Problem B8 A cubic decimeter is the same volume measure as a liter. So a woman's lungs hold about 4.4 dm3 of air, and a man's lungs about 5.8 dm3 of air.   Problem B9

 a. The dimensions of the cubic decimeter are 10 • 10 • 10 cm. The capacity is 1 L, and the weight of 1 L of water is 1 kg. b. One cubic centimeter of water weighs 1 g. The capacity of 1 cm3 (or cc) is 1 mL.   Problem B10 A balance scale compares two different masses. It typically only tells us when one side of the scale is larger in mass than the other. A spring scale uses the force of gravity to determine the weight of an object placed under the spring. The biggest difference between scale types is that some scales rely on mass calculations (typically balance scales), while others rely on weight calculations (typically spring scales). One interesting thought is whether these scales would report different answers on the Moon; the spring scale would because it depends on gravity, and the balance scale would not.   Problem B11

Answers will vary. Here are some possibilities:

 a. A large paper clip weighs about 1 g. b. A chocolate bar is generally 100 g. c. A small bag of flour is 500 g. d. Four medium apples together will weigh approximately 1 kg.   Problem B12 Depending on the temperature of the water and the level of precision of the instrument you used, you should notice that the measured mass came close to 1 kg. This confirms what we know already: One liter of water has a mass of 1 kg when the water is at 4 degrees Celsius.   Problem B13 Depending on the object used, the weight may not be 1 kg. For example, sand is heavier than water, so a cubic decimeter of sand (a liter of sand) should weigh more than a cubic decimeter of water. In general, a substance heavier than water but with the same volume weighs more than 1 kg. A substance lighter than water weighs less than the same volume of water.   Problem B14 A newborn baby has a mass of approximately 3 to 4 kg. A fifth grader has a mass of approximately 35 to 40 kg. An average adult woman has a mass of approximately 60 to 70 kg. An average adult male has a mass of approximately 75 to 90 kg.   Problem B15

 a. It takes only one weighing. Put one coin on one side of the balance and a second coin on the other side. If one is heavier than the other, the heavier coin is the fake. If they balance, the third coin is the fake. b. It takes up to two weighings. Weigh two coins against each other. If one is heavier, it is the fake one. If they balance, weigh the other two coins against each other. Again, the heavier coin is the fake one. Alternatively, you could start by putting two coins on each side of the balance. The side with the fake coin will be heavier. Weigh those two coins against each other to determine which one is heavier, and thus fake. c. The minimum number of weighings is still two. Put three coins on each side of the balance. The heavier side will contain the fake coin. This reduces the number of possibly heavy coins to the three on that side. Then the method from part (a) can be applied to find the heavy coin. Alternatively, you could divide the coins into three groups of two. Weigh two of the groups against each other. If one side is heavier, the fake coin is there. If they balance, the fake coin is in the group that was left out. Weigh the two coins in the "bad pair" against each other; the heavier is the fake one. d. You can do this with three weighings. Put four coins on each side for the first weighing. The heavier side will contain the fake coin. This reduces the number of possibly heavy coins to the four on that side. Then divide the heavier group into two groups, and weigh them; finally, weigh the remaining two coins from the heavier group. You can also do this with two weighings: Place three coins on each side of the balance. If they're the same weight, proceed to weigh the two remaining coins that haven't been weighed. If not, use the method in part (a) to decide which is the fake coin. e. It will take three weighings. First, divide the 12 coins in half, weigh them, and determine which half the heavier coin is in. Then with the heavier side, follow the six-coin strategy from part (c).   