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					Scientific Tools Unit
Topic #1 – Measurement


I start with the question “What is the distance from here to Truro?” Most students reply
“100km”. I then take two giant steps backward and ask the same question. I get the same
reply. This leads into a discussion about how precise the answer is. Students see that we
did not take the time to measure the distance carefully.
7
When measuring we need to measure accurately and precisely. If I measure a line with a
ruler and record the measurement as 3.641345343 cm I am lying. Most rulers measure
accurately to a millimetre. The most precise I can be is 3.64 cm. I did not measure all
those decimal places but I’m saying I did. I am positive about the 3 cm and about the 6
mm but I am doubtful about the 0.4 mm. I’m not positive that it’s a 4 or 3 or 5. I’m taking
my best estimation. This digit is called the doubtful digit. (It will be of vital importance
when we add/subtract measurements).

Let’s take another example. Measure the line below with the given ruler. (I draw this on
the board.)



The line measures 14.2 cm. This ruler is less precise than that of the students. I’m sure
about the 10 cm (the 1) and about the 4 cm but not about the 0.2 cm. It is my doubtful
digit. We can overline the doubtful digit (d.d.).

I get them to measure the height and width of the pictures on page 58 in their text. (One
measurement is 11.00 cm. They must include the zeros! They are significant!) We’ll use
these measurements later to find the area of the pictures.

I set up a classroom activity with different measuring tools. They include: a triple beam
balance, thermometer, spring scale (measures Newtons of force), metre stick, made-up
foot ruler (precise only to a foot), stopwatch, graduated cylinder. We go over the
measurements in the next class.



This leads into the idea of sig figs.
Scientific Tools Unit
Topic #2 – Significant Figures

Through measuring, students see and understand that there are reasons for a doubtful
digit. Now we will read other people’s measurements.

Ex. Who was more precise?
      Julia: 120.4cm                   Kevin: 120cm

This shows that Julia was doubtful about the 0.4 cm. Kevin was doubtful about the 2 in
the tens position. Julia is more precise.

When we are using measurements in different calculations in Chemistry and Physics and
even Biology, we need to account for the level of precision. To do so, scientists use
significant figures.

The last sig fig in a measurement is always the doubtful digit. But that is not always
clear. For example
12340cm or 12340.cm or 12340.0cm

To understand how precise a measurement is we need rules for sig figs.
   1) All non- zero digits are significant (34 cm --- 2 sig fig)
   2) Zeros in zero sandwiches are significant (2009 --- 4 sig fig)
   3) In order for zeros at the end of a measurement to be significant, there must be a
      decimal point. (23.000 -- 5 sig fig but 2300 ---2 sig fig)
   4) Zeros to the left of the first non-zero are NOT significant (0.00124 ---3 sig fig)

If Jenn measures a line to be 12.0 cm she is doubtful about the zero (it is the last sig fig
(see rule 3)).
If Darren measures a mass to be 1300 g, he is doubtful about the 3 (the last sig fig – rule
3)
If Lauren measures the temperature to be 20.1oC she is doubtful about the 1. (Rule 2)
Other examples
Measurement             # of Sig Figs
1) 1400.0               5
2) 300                  1
3) 0.0050               2
4) 6001.30              6
5) 11232.0              6
6) 5.00                 3
Scientific Tools Unit
Topic #3 – Scientific Notation

From the last lesson, we know how to count the number of sig figs and to indicate the
doubtful digit. What if I measured 300 km. What am I doubtful about? The 3. I really
measured to the 10 position. I want the first zero to count as significant. How do I show
that?
Scientists use scientific notation as a method of showing the proper amount of sig figs. It
also helps tremendously when we are dealing with huge or very small measurements in
Chemistry and Physics.
300 km can be written as 3 x 102 km. (102 is 100 – so 3 x 100 =300)
300 km (with 2 sig figs) can be written as 3.0 x 102 km. Notice we clearly have 2 sig figs.

When multiplying by 10+x we can move the decimal x times to make the number bigger.
4 x 105 N = 400000 N
3.00 x 104 kg = 30000 kg (3 sig figs –shown in sci notation)
 When multiplying by 10-x, we can move the decimal x times to make the number
smaller.                                            When using the calculator,
6.01 x 10-2 m = 0.0601 m                            use the “exp” or the “ee”
3.10 x 10-5 N = 0.0000310 N                         button for the “x 10”
More examples: Write in the following in another form. (Doubtful digits are underlined if
not obvious) Record the number of sig figs.

   1) 50112 km                               5.0112 x 104 km        5 sig figs

   2) 6000 m                                 6.00 x 103 m           3 sig figs

   3) 3.09 x 103 kg                          3090 kg                3 sig figs

   4) 8900 s                                 8.9 x 103 s            2 sig figs

   5) 1500000 N                              1.5000 x 106 N         5 sig figs

   6) 40900000 J                             4.09 x 107 J           3 sig figs

   7) 0.0000410 W                            4.10 x 10-5 W          3 sig figs

   8) 2.10 x 10-7 kg                         0.000000210 kg         3 sig figs

   9) 6.31 x 103 kg                          6310 kg                3 sig figs

   10) 1200000 m                             1.20 x 106 m           3 sig figs

   .
Scientific Tools Unit
Topic #4 – Mathematical Operations with Sig Figs

Remember, significant digits tell you how precise a measurement is. Now that we know
how to count the number of sig figs we can add, subtract, multiply and divide with them.
When calculating using measurements, we cannot increase our precision just by
calculating. We need to keep the appropriate amount of precision by keeping the
appropriate number of sig figs.

Adding/Subtracting
Ex 1                         The digits in bold are doubtful. The answer can
 11 m                        only have one d.d. therefore we round the number
+17.2 m                      at the first d.d. (when reading left to right).
 28.2 m

28 m is the answer.

Ex 2
1200 s
+ 413 s
1613s

Since we are doubtful about the 6, no digits after it can be significant. The answer is
1600s.

Multiplying/Dividing
Ex 1                                                 When multiplying or dividing the
120.4 m x 98.0 m = 11799.2 m2                        answer must have the same number
                                                     of sig figs as the least precise
4 sig fig     3 sig fig                              measurement.

The product must have 3 sig figs. The answer is 11800 m2.

Ex 2
59.501 m  3.9 s = 15.2566667 m/s

5 sig fig     2 sig fig

The answer is 15 m/s (2 sig figs)

When rounding, if the digit after the doubtful digit is greater than 5, it is rounded up. If it
is less than 5 it is rounded down. When the digit after the doubtful digit is a 5, then round
up if the d.d. is odd and round down if the d.d. is even.
Ex.
Round the following to 3 sig figs.
        a.   1459                        (1460)
        b.   653020                      (653000)
        c.   4540.041                    (4540)
        d.   0.00124720                  (1.25 x 10-3)
        e.   123.5                       (124)
        f.   124.5                       (124)
        g.   45.56                       (45.6)
        h.   129940                      (1.30 x 105)

        When doing a multi-step problem, take more than enough sig figs until the end of the
Note:
        calculation, then round to the least precise measurement.
        Ex. Pat walked 6.21m then 5.667m. The trip took 9.0 seconds in total. Calculate his
        average speed.

        (I know the answer must have 2 sig figs (like the 9.0s))
        total distance = 6.21m+5.667m = 11.877m (more than enough sig figs)
        v = 11.877m/9.0s
        v = 1.31967m/s (more than enough sig figs)
        v = 1.3 m/s (2 sig figs)

        See Sig fig Practice sheet.
Scientific Tools Unit
Topic #5 – Conversions

The precision in measurements is important and so is the unit of measure. If I say I
walked “100” today to school you don’t know if I mean metres, kilometres, or km/hr!!
In order to calculate with our measurements, they need to be of the proper unit.

       Measurement                       Unit                           Symbol
      length, distance                  metres                            m
           mass                          gram                              g
            time                       seconds                             s
          velocity                   metres/second                        m/s
        temperature                     kelvin                            K


   The metric system is easily used because it is based on factors of 10. There are 10
   mm in 1 cm, there are 100 cm in a metre. Each prefix means a certain factor of ten.

                   Prefix    Symbol         Factor          Example
                    kilo         k           1000          kJ, km, kg
                    centi        c           0.01            cm, cg
                    milli       m           0.001        mm, mg, meV
   There is a prefix and symbol for every factor of ten but we will use only these three.
   You must remember that:
    There are 1000 units in a kilounit
    There are 100 centiunits in a unit
    There are 1000 milliunits in a unit

Let’s do an example.
Ex. Convert the following measurements into the proper base unit.
    1) 23 kg
The base unit is grams (g). We convert by multiplying by the conversion factor WE
HAVE MEMORIZED. The unit we want to get rid of is on top so in the conversion
factor it must be on the bottom.

          1000 g
23kg x            = 23000g
            1kg
Notice that the kg’s divide out. We are left with only grams (what we wanted).

   2) 923 cm

            1m
923 cm x         = 9.23m
           100cm

   3) 23579mm
              1m
23579mm x          = 23.579m
            1000mm

This is called the factor label method and it is used ALL THE TIME in Physics and
Chemistry especially. These examples were easily done. Let’s look at some harder ones.

   3.90mm is how km?
   We will go to the base unit first.
                1m        1km
   3.90mm x            x          0.00000390km
             1000mm 1000m

   Notice that in the second conversion factor the metres are on the bottom. Why?

   Try these ones.

   1) 65 km  ? cm
   2) 272 min  ? hours
   3) 0.62 years  ? days
   4) 0.62 years  ? hours
   5) 8 m/s  ? km/s
   6) 15 m/s ? km/hr
   7) 50.0 km/hr  ? m/s
   8) 9.40 inches  ? cm (1 inch = 2.54 cm)
   9) 7.9 x 102 cm/hr ? km/s
   10) 9.92 x 10-1 kg  ? mg
Scientific Tools Unit
Topic #6 – Solving an equation: Undoing BEDMAS

In Science, particularly Physics and Math, we need to look closely at a relationship.
These relationships are generally expressed in mathematically relationships. When
solving an equation for a variable we need to undo BEDMAS (note: don’t say “bring it
over to the other side). When we undo something, we start from the end. SAMDEB. (The
last thing you put on when you get dress is your shoes. The first thing you take off is your
shoes.)

Example 1
Solve for F.

R  5TF  8E             Ask yourself “What is being done to F?”
                         We are multiplying by 5F and subtracting 8E.
                         Undo multiplying by dividing and subtracting by adding.
                         SAMDEB
     +8E              +8E
    R  5TF  8E

   R +8E = 5TF           Now undo the multiplying by dividing by 5T

    R  8E 5TF
                        The 5T’s divide out.
      5T    5T


         R  8E
    F                   Now we are ready plug and chug.
           5T


   Example 2
   Solve for s.

    v 2  s 2  2ad

    v 2  2ad  s 2      Undo the square now by taking the square root of both sides

     v 2  2ad  s 2

    s  v 2  2ad
Example 3

Solve for t
       
 d
v          Multiply both sides by t.
     t
        
    d
vt       t
      t
      
v t  d      Now divide by v
     
  d
t 
   v


Do worksheet

				
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posted:11/23/2011
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