Scientists take the ideas of precision and accuracy very seriously.• You can actually take entire courses in University that show how to figure out the precision and
accuracy of measurements.• Guess what? I took 'em!... your opinion of me has probably slipped a few notches ; )
• We need to know that when another scientist reports a finding to us, we can trust the accuracy and precision of all the measurements that have been done.
• A set of guidelines is needed while we do calculations so that we get rid of all those “4.243956528452940472” kind of answers you see on your calculator.
• The guidelines are there so we will know how many digits we should round off the final answer to show the correct precision.
All of this boils down to something called “Significant Digits”, more commonly referred to as Sig Digs.
To determine how many significant (important) digits a number has, follow these rules:
1. The numbers 1 to 9 are always sig digs. Zero ("0") is a sig dig if it comes to the right of a number between 1 and 9.
Example 1:13.869 -> Five sig digs. All the numbers are digits between 1 and 9.1.304 -> Four sig digs. The zero counts because it appears to the right of the "3"576.00 -> Five sig digs. The zeros count because they appear to the right of the "6"0.0008 -> One sig dig. The zeros don’t count, because they are to the left of the non-zero digits.13 000 -> Five sig digs.
2. When you add or subtract numbers, always check which of the numbers is the least precise (least numbers after the decimal). Use that many decimals in your final answer.
Example 2:11.623 + 2.0 + 0.14 = ?If you type this on a calculator, you'll get 13.763. Round it off to a final answer of 13.8, since the number "2.0" is the least precise... it only has one sig dig after the decimal.
We do this because we can’t really trust how much rounding off might have been done for any decimals after this.
Warning!Some text books will tell you that this number only has two sig digs, and that the zeros are “ambiguous” (a fancy way of saying “iffy”). Sure, it is unlikely that your answer will come out to exactly 13000, but it could happen. And if those zeros were not significant we would have used a different way to write down the number (see the next lesson). We will simply follow the rule and say that it has five sig digs since those zeros are all appearing to the right of a non-zero digit.
3. When you multiply or divide numbers, check which number has the fewest sig digs. Round off your answer so it has that many sig digs.
Example 3:4.56 x 13.8973 = ?Putting this into a calculator you will get something like 63.371688. We round off our final answer to 63.4 (which has three sig digs), because "4.56" has the fewest sig digs... three sig digs.
Your Data Sheet...You'll see that there are a bunch of constants on the back of your data sheet. These are all given as three sig digs, so treat them appropriately.
You'll also notice that some formulas already have numbers on them. These numbers are considered perfect and can be considered to have an infinite number of sig digs.
There are also situations where you might need to convert a number from one set of units to another. For example, 1 minute = 60 seconds. This doesn't mean that 1 minute “sort of” equals 60 seconds... it is a perfect conversion, so it also has an infinite number of sig digs.
Student Worksheet Solutions LSM 2.6E Solutions for Predicting Chemical Reactions, Extra
Exercises
For each of the following questions, classify the reaction type (synthesis, decomposition, combustion, single replacement, double replacement, or other), and predict the balanced chemical equation. Provide a word equation as well. 1. 4 Al(s) + 3 O2(g) → 2 Al2O3(s)
formation or combustion aluminium + oxygen → aluminium oxide
4. A strip of zinc metal is placed into a copper(II) nitrate solution. Zn(s) + Cu(NO3)2(aq) → Cu(s) + Zn(NO3)2(aq) single replacement zinc + copper(II) nitrate → copper + zinc nitrate
Manipulating equations is probably one of the most important skills to master in a high school physics course.
• Although it is based on familiar (and fairly simple) math concepts, it is still a stumbling block for most new physics students.
• Manipulating an equation means that you rearrange the equation so that the unknown you are trying to calculate is on its own on one side of the equation.
• Later in the course it will also give you the power to combine formulas (which is necessary) to solve more complicated problems.
• Learning how to manipulate formulas now (while the formulas are still easy ones!) will pay off later.
At all times remember two basic rules from math...1. To move something to the other side, just do the opposite math operation to it.2. If you do it to one side, do it to the other.
Example 1: The basic formula for calculating the velocity of an object is v = d / t , where "v" is the velocity, "d" is the displacement, and "t" is the time. This formula is great "as-is" if we are going to calculate velocity, but what if I need to calculate the displacement and I've been given the velocity and time? Solve the formula to solve for "d".
In the formula v = d / t, "d" is being divided by "t". To get "t" to the other side, we need to do the opposite... multiply by "t"!
v= dt
v=dt t
but what we do to one side, we do to the other...
t v= dtt
the "t" on the right side cancel each other out leaving...(t) v = d
the last step (and it's basically just a tradition) is to put our unknown on the left side of the equation. So let's flip flop the whole thing to get our final equation!
d = t v
And that's it! This is not a new formula you need to memorize; it is a formula already on your data sheet that you have manipulated to use more easily for a particular question.
Warning!Do not use any methods like "the triangle" that you learned in Junior High for the formulav = d/t. This works great for an easy formula, but try using it for vf
2 = vi2 + 2ad...
it does not work!
Now try to solve the same formula for "t". You should get...
t= dv
Be careful with formulas with addition, subtraction, square roots and squares.• You basically need to follow the BEDMAS (Brackets, Exponents, Division, Multiplication,
Addition, Subtraction) rule from math, but backwards. • Usually take care of any addition and subtraction first, then multiplication and division, and
finally exponents (remember, square root is just an exponent).
Example 2: Solve the formula “vf2 = vi
2 +2ad” for vi
Before doing anything else, take care of anything being added or subtracted to vi by doing the opposite...
vf2 - (2ad) = vi
2 +2ad – (2ad)
which leaves us with...vf
2 - 2ad = vi2
flip the whole formula (so vi is on the left) and take the square root of both sides...v i= v f2−2ad
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Equations Warm-up: Rules for manipulating equations
Learning objectives:
3.A.3. to be able to rearrange equations using the following rules:
add or subtract the same thing to both sides
if a = b then a + c = b + c
multiply or divide both sides by the same thing
if a = b then a x c = b x c
replace any term or expression by another equal expression
if a + b = c and b = d x e then a + (d x e) = c
square or square root both sides
if a + b = c then (a + b)2 = c2 also
if c
ba =2
then c
ba =
expand out an equation
y(a + x) = 1 becomes ya + yx = 1
simplify (factorise)
ab + ac = a(b + c)
3A3: Manipulating equations
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Now use these rules to answer the following questions. You may want to think about some of these tips.
When rearranging an equation, don’t be afraid to use a lot of small steps and write down every step.
Sometimes it isn’t at all clear how best to proceed – just start, remembering what it is that you need to make the subject of the equation – and eventually you will get there. There can be a lot of different ways of doing it.
Brackets are useful because you can move the whole term (ie what is inside the brackets) around as if it is a single item.
Questions: Q1. Consider the equation v = u + at. Make a the subject of the equation. Q2. Rearrange s = ut + ½ at2 to make a the subject of the equation.
Q3. Rearrange p
mv = to make p the subject
Q4. Rearrange 24 d
LF
π= to make d the subject.
Q5. Rearrange x
y+
=1
1 to make x the subject.
Q6. If k
CV = and
tk
69.0= , write an equation for V in terms of C and t.
Q7. Drugs in the blood can be bound to plasma proteins and/or free in solution, in practice there is an equilibrium whereby Cfree + protein <--> Cbound.
The percentage of drug bound is given by total
bound
C
Cb 100= where Ctotal = Cbound + Cfree
The fraction of drug in plasma that is free is given by total
free
C
Cf = .
Express f as a function of b. Q8. Rewrite the following so that the brackets are removed. (a – 2)(b – 3) = 0 Q9. Rewrite the following so that the brackets are removed. (x + 3y + 2)(x – 3) = 0 Q10. Factorise 3x2 – x
3A3: Manipulating equations
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Q11. Factorise the following expression: 25 – y2 Q12. In pharmacology, the proportion of receptors bound with drug D is given by eqn 1.
eqn 1. 1+
=DK
DKboundproportion (K is the affinity constant.)
when a competing drug B is added, a higher concentration of drug D1 is given to get the same number of receptors bound with drug D.
eqn 2. 11
1
++=
BBKKD
KDboundproportion (KB is the affinity constant for drug B)
since the proportion bound is the same in eqn 1 and eqn 2 we can make the right hand side of eqn 1 equal to the right hand side of eqn 2.
11 1
1
++=
+BBKKD
KD
DK
DK
Simplify this as much as possible, getting D1 as a function of B.
Answers: A1.
start v = u + at
Step 1. You want to get a on its own on the left. So start by reversing it.
u + at = v
Step 2. You want a to be on its own so start by subtracting u from both sides
u – u + at = v – u at = v – u
Step 3. To get a on its own , you have to divide both sides by t.
( )t
uv
t
u
t
va
−=−=
A2.
Step 1. You want to get a on its own on the left. So start by reversing it.
ut +½at2 = s
Step 2. You want a to be on its own so start by subtracting ut from both sides
ut – ut + ½at2 = s – ut ½at2 = s – ut
Step 3. To get a on its own , you have to multiply both sides by 2 then divide both sides by t2
at2 = 2(s - ut)
( )22
2 )2
t
uts
t
at −=
( )2
)2
t
utsa
−=
3A3: Manipulating equations
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A3.
Step 1. Start by squaring both sides.
p
mv =2
Step 2. You want p to be on the left, so multiply both sides by p.
pp
mpv ×=2
mpv =2
Step 3. Now divide both sides by v2
2v
mp =
A4.
Step 1. You need to get d2 off the bottom. To do this multiply both sides by 4πd2
LFd
dd
LFd
=
×=
2
2
2
2
4
44
4
π
ππ
π
Step 2. Now to leave d2 on its own, divide both sides by 4πF
F
Ld
F
L
F
Fd
π
ππ
π
4
44
4
2
2
=
=
Step 3. Now square-root both sides.
F
Ld
π4=
A5.
Step 1. You need to get (1+x) off the bottom. To do this multiply both sides by (1+x) Here you are treating what’s inside the brackets (1+x) as a single term.
( )( )
( )
( ) 11
11
11
=+
+×+
=+
xy
xx
xy
Step 2. You want to get x on its own, so expand out the brackets.
y+xy = 1
Step 3. Now subtract y from both sides to leave xy on the left on its own.
xy = 1 – y
Step 4. Now to get x on its own, divide both sides by y.
y
yx
or
yx
y
y
yy
xy
−=
−=
−=
1
11
1
these last two expressions are equivalent.
3A3: Manipulating equations
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A10. The term “factorise” means to find the terms which were multiplied together to give this. In this case you can take x out of both terms
3x2 – x = x(3x – 1)
A11. Here you have to remember that the difference of two square numbers is the same as the product of their sum and difference i.e. 25 – y2 = (5 – y)(5 + y)
3A3: Manipulating equations
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A12.
Step 1. This is our starting point. Our aim is to get all terms with D1 on the left and all terms with B in them on the right.
11 1
1
++=
+BBKKD
KD
DK
DK
Step 2. In order to move things around we need to get them off the bottom (denominator to use the technical). Start by multiplying both sides by (DK+1). (OK so it doesn’t look a lot better but just wait...)
( )( ) ( )
1
)())((
11
11
1
11
1
1
++
+=
+×++
=+×+
B
B
BKKD
KDDKKDDK
DKBKKD
KDDK
DK
DK
Step 3. Now multiply both sides by (D1K + BKB+1) I’ve used brackets strategically so that I can see it more easily – otherwise it can look like a real mess. Setting things out clearly is really important here.
( )( )( ) ( )( ) )())((
)())((1
111
111
KDDKKDDKBKDKKDDK
KDDKKDBKKDDK
B
B
+=++
+=++
Step 4. Now have a look and see what terms appear on both sides. See that (DK)(D1K) appears on both sides, so you can subtract (DK)(D1K) from both sides leaving... Which looks much better.
(DK)(D1K) + (DK)(BKB) + DK = (DK)(D1K) + (D1K)
( )( ) )( 1KDDKBKDK B =+
Step 5. Now you have (DK) appearing in both terms on the left hand side. Try simplifying this...
DK(BKB + 1) = D1K
Step 6. Now you can see that you can divide both sides by K which will get rid of the K.
DK(BKB + 1) = D1K D(BKB + 1) = D1
Step 7. If you want to you can divide both sides by D so you have both D and D1 on the same side but that is a bit cosmetic.
BKB + 1 = D1/D
3A3: Manipulating equations
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looking back, you started with something not too big, then went through something that looked really quite horrible, then ended up with something quite simple.
Lesson 7: Graphing
Graphing is an essential skill for both Physics 20 and 30. You MUST be able to follow all of the rules of properly drawing a graph, and also be able to do basic interpretation of graphs.
When you are presented with a chart of numbers that you are going to graph, you should start by identifying which variable is the manipulated variable and which is the dependent variable.
● In a lab, you are usually watching to see one thing change on its own, while letting something just plod along in an expected way.
● The one that you're watching for changes in is called the dependent or responding variable, drawn on the y-axis.
● The one that is changing in a regular expected pattern is the independent or manipulated variable, drawn on the x-axis.
● As a rule of thumb, time almost always goes on the x-axis.
Graphing RulesGraphs you draw must have the following five basic characteristics. If you miss any part you will lose marks.
Title
● Your title should be short, but still clearly tell what you have graphed.● The most common and recommended way to name your graph is to say what your y-axis and
x-axis are.● One way to say it is “Y-Axis versus X-Axis”● The other common way is to say “Y-Axis as a function of X-Axis”
Labeled Axis
● Make sure to write out the full name of what you have graphed on each axis, along with the units you used.
● If you are using any sort of scientific notation for the numbers, make sure you show it here also.
A Well Chosen Scale
● The information you plot should always cover at least 75% of the area on your graph.● Start by looking at the maximum values you have for both the x and y axis. ● Then check out how many major “ticks” you have on each axis of the actual graph.● Divide your maximums by the ticks to find out roughly what to label each tick as.
Data Plotted Correctly
● It’s too bad when a person does all this work, and then does a sloppy job of plotting their information.
● Make sure you are as careful as possible when marking your points on your graph, otherwise everything else is a waste of time.
● You should always put little circles around each dot, since they might be hard to see on the graph paper. It also shows that each data point is a bit “iffy.”
Notice that on the graph I have marked off two points on the best fit line...
(x2 , y
2) = (4.2, 40)
(x1 , y
1) = (2, 18)
I use these to calculate the slope.
Line Straightening
A special skill involving graphs is a technique called line straightening, sometimes called an averaging technique.
● Essentially, you will sometimes have data to graph that would give you a curved line.○ Analyzing this line is very difficult... calculating a single slope
or area under the line is impossible.● What we need to do is come up with a way to manipulate the info
you have so that, when plotted, it will give you a straight line of best fit (a linear relationship).
You'll need to do some thinking to get these done.1. Identify the main formula/concept that has something to do with the data you're looking at.2. Get rid of the variables in the formula that are not part of the data you're looking at... this will
leave you with a relationship, instead of a formula.3. Look at what is being done mathematically to the variables. Most often one of them will either
be squared or square rooted. Create a new column of values where you do this to the values.4. Plot your new data as a straight line graph.
Quite often, the resulting graph will have a slope or an area under the line that has some sort of meaning.
Example 2: A student performs an experiment to measure the how centripetal force changes as the velocity is altered. He collects the data shown on the table below. Using a suitable averaging technique, determine a new set of values that, if graphed, would give a linear graph. Identify the meaning of the slope of this graph, and determine the mass of the object if the radius of the circle is 1.15 m.
Fc (N) v (m/s)
20.0 1.0
80.0 2.0
180 3.0
320 4.0
500 5.0
First we need to identify the main formula (which we haven't studied yet!).● Don't worry, questions you'll do will be based on formulas you've seen before.
F c=m v2
r
Next get rid of the unneeded parts of the formula.● The only variables we are concerned about according to the chart of values given are Fc and v.
F cv 2
Now evaluate what this relationship is telling us.● Nothing has been done to Fc , but v is being squared.
Warning!Do not make the mistake of calculating some sort of average based on the numbers given if you are asked to use an “averaging technique” for a graphing problem.
● Come up with a new column of values with v squared. No, really, just square v. Be sure to make the units squared as well.
Fc (N) v (m/s) v2 (m2/s2)
20.0 1.0 1.0
80.0 2.0 4.0
180 3.0 9.0
320 4.0 16
500 5.0 25
● The graph would be Fc vs v2.○ Try graphing it yourself and then calculate the slope... you should get about 20 N/m2/s2
● To figure out what the slope actually means in this case, you would look at your graph this way...
slope=riserun
=F cv2
and from the original formula F c=mv2
r
F cv2
=mr
slope=F cv2
=mr
slope=mr
● So if we want to determine the mass of the object...
