Multiplication

Subtraction

Let us now perform the subtraction operation.

4.56 cm ± 0.02 cm

1.45 cm ± 0.01 cm

The resulting difference is: 3.11 cm ± 0.03 cm

Here you subtract the values, but add the errors.

Using the same approach, subtract the following numbers:

1. 1.45 cm ± 0.02 cm – 1.22 cm ± 0.01 cm.

0.230 cm ± 0.03 cm

2. 5.41 feet ± 0.01 feet – 5.210 feet ± 0.001 feet.

0.20 feet ± 0.01 feet

3. 1.23456 cm ± 0.00001cm – 1.23469 cm ± 0.00001cm. How many significant digits are there in the two values being subtracted? How many significant digits are there in the result?

-0.000130000 cm ± 0.00002 cm

4. Can you compute 1.33 feet ± 0.01 feet – 6.25 inches ± 0.25 inches? If yes, show the result and explain. If not, explain why not.

first of all you do have to convert them to uniform measuring unit. Then do operations.

Multiplication

From your reading about multiplication in the Precision handout, you should have learned that you cannot compute the error term by just adding the error terms of the operands as you do with addition or subtraction. The following is the algorithm that is recommended:

To compute the result's error term (r_errorTerm) we first need to compute the result, r:

r = a × b where a, b are the values of the input operands.

Then we need to compute the relative fraction of “a” (a_{errorFraction}) that is uncertain:

a_{errorFraction} = a_errorTerm / ⎢ a ⎢

Next we compute the relative fraction of “b” (b_{errorFraction}) that that is uncertain:

b_{errorFraction} = b_errorTerm / ⎢ b ⎢

From these two, we then compute the result's error (r_errorTerm) and round it to one or two significant digits:

r_errorTerm = (a_{errorFraction} + b_{errorFraction}) × ⎢ r ⎢

Given the error term, we now can determine how many significant decimal places the result should have.

Using this algorithm for computing the error term, multiply the following numbers and provide the product, the error term, and the units:

1. 1.23 cm ± 0.01cm × 1.26 cm ± 0.01cm

aerrorFraction= 0.01 cm / 1.23 cm= 0.0008 cm

berrorFraction= 0.01 cm / 1.26 cm= 0.0079 cm

rerrorTerm = (0.0008 + 0.0079) x (1.23 x 1.26) = 0.0087 x 1.55 = 0.0135 cm

2. 5.41 feet ± 0.03 feet × 2.21 feet ± 0.01feet

aerrorFraction= 0.03 feet / 5.41 feet= 0.0055 feet

berrorFraction= 0.01 feet / 2.21 feet= 0.0045 feet

rerrorTerm= (0.0055 + 0.0045) x (5.41 x 2.21) = 0.01 x 12.0 = 0.12 cm

3. Can you multiply 1.211 kg ± 0.001 kg × 1.34 m ± 0.02 m? If yes, give an explanation. If not, explain why not.

Yes, you could multiply kilograms and meters if then you intend to divide it to seconds, which give us a momentum formula, or in cases like that, when there is something additional, that

gives that multiplication a meaning. But if you just want to multiply kg and meter, and then leave it, it is not acceptable, cause it is meaningless. So therefore you couldn’t do such a thing.

4. 3.12 m/sec ± 0.01 m/sec × 2.15 sec ± 0.02 sec? Is it possible to do this multiplication? If yes, do the multiplication and explain what you did and why. If not, explain why not.

aerrorFraction= 0.01 m/sec / 3.12 m/sec= 0.003 m/sec

berrorFraction= 0.02 sec / 2.15 sec= 0.009 sec

rerrorTerm= (0.003 + 0.009) x (3.12 x 2.15) = 0.012 x 6.71 = 0.08 m

5. 9.52 m/sec² ± 0.01 m/sec² × 3.50 sec ± 0.02 sec? Is possible to do this multiplication? If yes, explain what you did and why. If not, explain why not.

aerrorFraction= 0.01 m/sec2 / 9.52 m/sec2= 0.001 m/sec2

berrorFraction= 0.02 sec / 3.50 sec= 0.006 sec

rerrorTerm= (0.001 + 0.006) x (9.52 x 3.50) = 0.007 x 33.3 = 0.2 cm