Lecture, Monday, August 26, 2002

Today in class we practiced applying significant figures and rounding when doing mathematical calculations: addition, subtraction, multiplication and division.

The rules for reporting numbers for addition and subtraction operations differ from the rules for multiplication and division.

We'd discussed these rules in class on Friday, August 23, 2002 and in the Lecture Notes for that day.

So we practiced doing some problems:

Problem #1:

Determine the result to the correct number of significant figures.

Answer 1

Answer 2

Answer 3

Answer 4

Answer 5



7.29 x 103

7.29 x 107

7.29 x 10-5

This calculation is only a division so after doing the calculation the number of significant figures must be determined. Depending on your calculator setting the answer is equal to Answer 2.

However, the number of sig figs in the answer is three. How do we get that? The numerator (3.6675) has 5 sig figs and the denominator (5.03 x 10-4) has 3 sig figs. Following the rule for multiplication and division:

When multiplying and dividing measured quantities, there should be as many significant figures in the answer as there are in the measurement with the least number of significant figures.

The answer can only have 3 sig figs. So answers 1 and 2 are acceptable. Answer 2 has too many sig figs and answer 4 has the incorrect exponent, probably from an error entering one or the other number. Finally answer 5 is the result of using the TI calculator incorrectly.

To obtain Answer 5 students solved the division by entering into their calculator the following:

3.6675 divided by 5.03 times 10^-4

TI calculators interpret this entry by dividing 3.6675 by 5.03 then multplying the result (0.729) by 10-4. Giving 7.29 x 10-5.

The EE function on the calculator is the correct way to enter numbers in scientific notation. So the entry would go;

3.6675 divided by 5.03 EE-4

That result is the correct answer, rounded to the correct number of sig figs!

Moral..?? Watch out. I will try to get you on this deliberately on the next exam....warning Will Robinson!

Problem #2:

Determine the result to the correct number of significant figures.

We then tried the problem from our Lecture Notes for Monday, August 26, 2002.

Problem #3

Report the result of the following calculation to the correct number of significant figures.

c) Answer


After practicing these calculations everyone should try some problems on their own. Check out the InClass Exercises 1 and 2, the PLE3 for Monday's class, as well as Problems 1.35 - 1.40 in BLB and some of the other problems in our lecture notes.

Next we discussed conversions. Again we did several examples to see how to approach solving these kinds of problems.

The problems we did were from PLE3.

Problem #4

Determine the number of centimeters in 3.00 feet.

This is a straight forward example of a conversion of feet to centimeters. Beginning with 3.00 feet we end up with 91.4 cm. We've not changed the length in any way we have re-expressed the length in a different unit. The two conversion factors are used to re-express the length in a different unit. Essentially we have multipled 3.00 feet by 1 twice. The first factor 12 inches = 1 foot when expressed as a ratio with 12 in in the numerator and 1 foot in the denominator converts feet to inches. The second conversion factor 2.54 cm = 1 in converts inches to centimeters. We started with a number with 3 sig figs so our answer must have 3 sig figs.

NOTE: Conversion factors NEVER change the number of significant figures in your answer.

Problem #5

Determine the number of ft2 in 144 in2.

This problem hasw proved to be a little challenging for many students, so lets take it apart and look at it a little more carefully. We are asked to convert 144 in2 to ft2. We know a conversion between inches and feet, 12 inches = 1 foot, but how do we handle in2? As you can see in the solution we set up the factor so inches cancel, but we must square the factor. This is very important. The square means that all units and all numbers within the bracket must be squared. So the answer could also be written as,

Problem #6

Convert 3.513 g cm-3 to pounds ft-3.

This is a little more challenging problem because we have a unit in the numerator (grams) that must be converted to pounds, and a unit in the denominator (cm3) that needs to be converted to ft3.

So here is the approach we would use;