AP Physics 1 Homework Packet Orientation
1.
5.5 kilogram(s) contains this many grams (show a unit conversion)
Always start by writing down what you are given, in this case, 5.5 kg. Then think—I want to
create a conversion factor that gets rid of kg and creates g so I will put kg on the bottom so it
cancels and g on the top so it stays. I know there are 1000 g in a kg so I add the numbers in
and perform the multiplication. There are 2 sig figs in the answer. Answer 5.5 x 10 g
3
5.5 kg x
2.
1000g
= 5.5 x 103 g
1 kg
Convert 0.2974 m to mm (show a unit conversion).
Always start by writing down what you are given, in this case, .2974m. Then think—I want to
create a conversion factor that gets rid of m and creates mm so I will put m on the bottom so
it cancels and mm on the top so it stays. I know there are 1000 mm in 1 m so I add the
numbers in and perform the multiplication. There are 4 sig figs in the answer. Answer
297.4mm
3.
1000mm
= 297.4mm
1m
3.2 seconds contain this many picoseconds (show a unit conversion).
.2974 m x
Always start by writing down what you are given, in this case, 3.2s. Then think—I want to
create a conversion factor that gets rid of s and creates ps so I will put s on the bottom so it
cancels and ps on the top so it stays. I know there are 1012 ps in 1 second so I add the
numbers in and perform the multiplication. Conversely, I could also say in 1 ps, there are
10-12 seconds. There are 2 sig figs in the answer. Answer = 3.2 x 10
12
10 ps
12
3.2 s x
4.
s
= 3.2 x 1012 ps or
3.2 s x
1ps
10
-12
s
ps
= 3.2 x 1012 ps
9.16 seconds contain this many nanoseconds (show a unit conversion).
Always start by writing down what you are given, in this case, 9.16 s. Then think—I want to
create a conversion factor that gets rid of s and creates ns so I will put s on the bottom so it
cancels and ns on the top so it stays. I know there are 109 ns in 1 s so I add the numbers in
and perform the multiplication. Conversely, I could also say in 1 ns, there are 10-9 seconds.
There are 3 sig figs in the answer. Answer = 9.16 x 10 ps
9
9.16 s x
5.
109 ns
= 9.16 x 109 ns or 9.16 s x
109 ns
= 9.16 x 109 ns
s
s
The distance of 54 km equals how many m (show a unit conversion).
Always start by writing down what you are given, in this case, 54 km. Then think—I want to
create a conversion factor that gets rid of km and creates m so I will put km on the bottom so
it cancels and m on the top so it stays. I know there are 1000m in 1 km so I add the numbers
in and perform the multiplication. There are 2 sig figs in the answer. Answer 5.4x10 m
(The answer 54000 m also contains 2 sig figs).
4
54 km x
1000m
1km
= 5.4x104 m (could also be 54,000 m)
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6.
2
Convert 4450.4 g to mg.
Always start by writing down what you are given, in this case, 4450.4 g. Then think—I want
to create a conversion factor that gets rid of g and creates mg so I will put g on the bottom so
it cancels and mg on the top so it stays. I know there are 1000 mg in 1 g (or 1 mg is .001 g) so
I add the numbers in and perform the multiplication. There are 5 sig figs in the answer.
Answer 4.4504x106 mg
4450.4 g x
7.
1000mg
= 4.4504x106 mg or 4450.4 g x
1mg
= 4.4504x106 mg
1g
.001 g
Express the volume 622.6 cm3 in liters (show a unit conversion).
Always start by writing down what you are given, in this case, 622.6 cm3. Remember that
cm3 is the same thing as mL. Change cm3 to mL and then think—I
want to create a conversion factor that gets rid of mL and creates L so I will put mL on the
bottom so it cancels, and L on top so it stays. I know there are 1000 mL in 1 L (or 1 mL is .001
L) so I add the numbers in and perform the calculation. There are 4 sig figs in the answer.
Answer .6626 L
622.6cm3 = 622.6 mL x
8.
1L
1000 mL
= .6626L or 622.6cm3 = 622.6 mL x
.001L
1 mL
= .6626L
Convert 10.6 m3 to mm3 (show a unit conversion).
Always start by writing down what you are given, in this case, 10.6 m3. Then think—I want to
create a conversion factor that gets rid of m3 and creates mm3 so I put m3 in the denominator
of the conversion factor so it cancels, and mm3 so it stays. I know that there are 1000 mm in
1 m so I add these numbers in. However, because the units are cubed, the values also need to
be cubed. There are 3 sig figs in the answer. Answer 1.06 x 1010 mm3
10.6 m3 x
10003 mm3
1m
3
= 1.06x1010 mm3
9.
The pressure of the earth's atmosphere at sea level is 14.7 lb/in2. What is the
pressure when expressed in g/m2? (2.54 cm = 1 in., 2.205 lb = 1 kg) (show a unit
conversion).
Always start by writing down what you are given, in this case, 14.7 lb/in2. Then think—I
need a set of conversion factors that is going to convert pounds to grams and square inches
to square meters. Start by writing down the given value with lb in the numerator and in2 in
the denominator. I always begin by cancelling out the value in the numerator, so we know
we want to get rid of lb and create g. But, the conversion factor we are given converts lb to
kg, so we will have to do this first and then convert kg to g. Because we want to cancel lb put
this in the denominator of the conversion factor; because we want to create kg put this in the
numerator; then add in the corresponding values of the conversion factor that are given.
Then, in the next factor we want to get rid of kg and create g so put kg in the denominator
and g in the numerator. We know that 1 kg equals 1000 g so put these numbers in.
Now that we have created g, go back and get rid of in2, and create m2. Unfortunately, the
conversion factor we are given only changes cm to in, so we will first have to change in2 to
cm2, and then cm2 to m2. Because we want to cancel in2, in the conversion factor put in2 in the
numerator, and then put cm2 in the denominator. Then put the values given for the
conversion factor, remembering that because the units are squared, the values must also be
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3
squared. Then, in the next factor we want to get rid of the cm2, so put it in the numerator,
and we want to create m2 so put it in the denominator. We know that there are 100 cm in 1
m so put these values in, remembering to square the values. Finally, perform the calculation.
There are 3 sig figs in the answer. Answer 1.03x107 g/m2
14.7 lb
in2
10.
x
1 kg
2.205 lb
x
1000g
1 kg
x
1 in2
2.54 2 cm2
x
1002 cm2 1.03x107 g
=
1 m2
m2
Convert 4315 mL to qts. (1 L = 1.06 qt) (show a unit conversion)
Always start by writing down what you are given, in this case, 4315 mL. Then think—I need
a conversion factor that gets rid of mL and creates quarts so I put mL in the denominator of
conversion factor so it cancels. But, I can’t put qt in the numerator because I haven’t been
told how many mL is in a quart. But I have been told how many L are in a quart and I know I
can convert mL to L. So, I put L in the numerator. I know there are 1000 mL in 1 L so I add
these numbers to the conversion factor. Then, I want to get rid of L so I put this in the
denominator of the next conversion factor so it cancels. I want to create qt so I put this in the
numerator. I have been given that 1 L = 1.06 qt, so I add these numbers to the conversion
factor. Then I perform the multiplications. There are 3 sig figs in the answer.
Answer 4.57 qt
4315 mL x
11.
1L
1000 mL
x
1.06qt
1L
= 4.57 qt
Convert 24.3 lb to g. (1 lb = 453.6 g) (show a unit conversion)
Always start by writing down what you are given, in this case, 24.3 lb. Then think—I need a
conversion factor that gets rid of pounds and creates g so I put lb in the denominator of the
conversion factor so it cancels. Then I put g in the numerator so it stays. I have been given a
conversion factor that converts pounds directly grams so I just add the corresponding
numbers to the conversion factor. Then perform the multiplication to complete the problem.
There are 3 sig figs in the answer. Answer 1.10 x 104 g
24.3 lb x
453.6g
1lb
= 1.10x104 g
12.
Convert 72.3 mi to km. (1 m = 1.094 yds, 1 mi = 1760 yds) (show a unit
conversion)
Always start by writing down what you are given, in this case, 72.3 mi. Then think—I need a
conversion factor that gets rid of mi and creates km so I put mi in the denominator of
conversion factor so it cancels. But, I can’t put km in the numerator because I haven’t been
told how many mi is in a km. But I have been told how many 1760 yd are in 1 mi and I have
also been told how many yds are in 1 m. So, I put yd in the numerator and add the
corresponding numbers to the conversion factor. Then, I want to get rid of yd so I put this in
the denominator of the next conversion factor so it cancels. I want to create m so I put this in
the numerator. I have been given that 1 m = 1.094 yd so I add these numbers to the
conversion factor. But, we still have to get to km. For the last conversion factor we need to
cancel m so we put this in the denominator. We want to create km so we put this in the
numerator. We know that 1 km equals 1000 m so we add these numbers to the conversion
factor. Then I perform the calculations. There are 3 sig figs in the answer.
Answer 1.16 x 102 km
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72.3 mi x
1760 yd
1 mi
x
1m
1.094 yd
x
1km
1000 m
4
= 1.16 x 102 km
13.
The density of liquid chloroform is 1.48 g/mL. What is its density in units of
3
lb/in ? (2.54 cm = 1 in., 2.205 lb = 1 kg) (show a unit conversion).
Always start by writing down what you are given, in this case, 1.48 g/mL, g in the numerator
and mL in the denominator. But wait—having read through the problem you realize that the
units of volume you need will be in cubic units—so, right from the start, because mL and cm3
are the same thing, change mL in the denominator to cm3. Then think—I need a set of
conversion factors that is going to convert grams to pounds and cubic cm to cubic inches. I
always begin by cancelling out the value in the numerator, so we know we want to get rid of
g and create lb. But, the conversion factor we are given converts lb to kg, so we will have to
convert g to kg first, and then convert kg to lb.. Because we want to cancel g put this in the
denominator of the conversion factor; because we want to create kg put this in the
numerator. We know that 1 kg equal 1000 g so put these numbers in the conversion factor.
Then, in the next factor we want to get rid of kg and create lb so put kg in the denominator
and lb in the numerator. We are given that 1 kg equals 2.205 lb, so put these numbers in.
Now that we have created lb, go back and get rid of cm3, and create in3. Because we want to
cancel cm3, in the conversion factor put cm3 in the numerator, and then put in3 in the
denominator. Then put the values given for the conversion factor, remembering that
because the units are cubed, the values must also be cubed. Finally, perform the calculation.
There are 3 sig figs in the answer. Answer 5.35x10-2 lb/in3
1.48 g
cm3
14.
2.205lb 2.543 cm3 5.35x10-2 lb
x
x
x
=
1 in3
in3
1000 g
1 kg
1 kg
Convert 0.0898 ft3 to L. (2.54 cm = 1 in., 1 L = 1 dm3) (show a unit conversion)
Always start by writing down what you are given, in this case, .0898ft3. Then think—I need a
conversion factor that gets rid of ft3 and creates L so I put ft3 in the denominator of
conversion factor so it cancels. But, I can’t put L in the numerator because I haven’t been told
how many L is in a ft3. But I have been told how many cm are in 1 in, so I can start by
converting ft3 to in3, and then convert that to cubic cm. So, create cubic inches in the
numerator. Then, you know that there are 12 inches in 1 ft—add these numbers in
remembering you have to cube the values because the units are cubed.
Then, convert cubic inches to cubic cm—put cubic inches in the denominator of the next
conversion factor so it cancels and cubic cm in the numerator to create it. Add in the given
numbers for each—2.54 cm and 1 in—again, remembering to cube these values. Now notice,
we are not given how many cubic cm are in 1 L, we are given that 1 dm3 equals 1 L. So, we
have to convert cubic cm to cubic dm. So, in the next conversion factor put cubic cm in the
denominator to cancel it and cubic dm in the numerator. There are 10 cm in 1 dm so put
these numbers in, remembering to cube them. Finally, convert cubic dm to L by cancelling
dm and creating L in a conversion factor. Complete the calculations. There are 3 sig figs in
the answer. Answer 2.54 L
.0898 ft 3 x
123 in3
1 ft 3
x
2.543 cm3
1 in3
x
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1 dm3
x
1L
103 cm3 1 dm3
= 2.54L
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5
15.
It is estimated that uranium is relatively common in the earth's crust, occurring in
amounts of 4 g / metric ton. A metric ton is 1000 kg. At this concentration, what mass of
uranium is present in 1.2 mg of the earth's crust? (show a unit conversion).
Always start by writing down what you are given. In this case figuring out what you are
given may be difficult. What you are really given to start with is that there are 1.2 mg of
crust. Then think—I You want a conversion factor that gets rid of mg of crust, and creates
grams of Uranium. So, put mg crust in the denominator to make it cancel. But, you cannot
put grams of Uranium in the numerator because you have not been told how many grams of
uranium are 1 mg of crust. But you have been told how many grams of Uranium are in 1
metric ton. Therefore, you should put metric tons of crust in the numerator of the conversion
factor. Now you need to reason out how many mg are in 1 metric ton so you can complete the
conversion factor. You have been given that 1 metric ton is 1000 kg. You also know that 1 kg
is 1000 g. Each of these is a factor of 103. Therefore, 1 metric ton is equal to 106 g. But, we
want mg. You know that there are 1000 mg in 1 g, so this is another factor of 103. 1 metric
ton is equivalent to 109 mg. So now we can put these numbers into our conversion factor.
Now we can create a conversion factor that cancels metric tons and creates grams of
Uranium. The conversion factor you have been given states that there are 4 g U for every 1
metric ton of crust—so, put metric tons in the denominator so it cancels, and 4 g U in the
numerator so we keep this. Then complete the calculations. . There is 1 sig fig in the answer.
Answer 5 ng
1.2 mg crust x
1 metric ton crust
10 mg crust
9
x
4 gU
1 metric ton crust
= 5 ng
16.
A 20.0 mL sample of glycerol has a mass of 25.4 grams. What is the density of
glycerol in ounces/quart?
(1.00 ounce = 28.4 grams, and 1.00 liter = 1.06 quarts)
Always start by writing down what it is that you are given. In this case, determining this
could be confusing. You are given that there is a total mass of 25.4 grams and you are told
that it is contained in a volume of 20.0 mL. This becomes your given value with the mass in
the numerator and volume in the denominator—values are already in the form of density!
Then think—I need a set of conversion factors that is going to convert grams to ounces and
mL to quarts. I always begin by cancelling out the value in the numerator, so we know we
want to get rid of g and create ounces. Begin by placing g in the denominator so that it
cancels and oz in the numerator. We have been told that there are 28.4 g in 1.00 ounces so
place these numbers in the conversion factor. To create the other conversion we know we
want to get rid of mL so we put mL in the numerator of the conversion factor to cancel—but,
we are not told how many mL are in a quart in the conversion we are given, so we will first
have to change mL to L. So, put L in the denominator to create it. We know there are 1000
mL in 1L so add these numbers in. Then we can convert L to quarts with a final conversion
factor. We want to cancel L so put L in the numerator. We want to create quarts so we put
quarts in the denominator. We have been told that 1.06 qt is equal to 1.00 L. Add these
numbers in. Complete the calculations. There are 3 sig figs in the answer. Answer 42.2
oz/qt
25.4 g
1.00 oz 1000 mL 1.00 L 42.2oz
x
x
=
1.06 qt
qt
20.0 mL 28.4 g
1L
x
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17.
During a physics experiment, an electron is accelerated to 87 percent of the speed
of light. What is the speed of the electron in miles per hour? (speed of light = 3.00  108
m/s, 1 km = 0.6214 mi)
Always start by writing down what it is that you are given. In this case, determining this
could be confusing. You are given that an electron is accelerated to 87% of the speed of light,
which is the same thing as .87 x the speed of light, or, with the actual value, .87 x (3.00 x 108
m/s) This becomes your given value with meters in the numerator and seconds in the
denominator. Then think—I need to create a set of conversion factors that convert meters
into miles and seconds into hours. I always begin by cancelling out the value in the
numerator, so we know we want to get rid of m and create mi. Begin by placing m in the
denominator so that it cancels. However, we cannot put mi in the numerator because we
have not been told how many meters are in a mile—but, we have been told how many km are
in a mile, so we can first convert from m to km. So, create km by putting it in the numerator.
I know there are 1000 m in a km so I add these numbers to the conversion factor. Then
convert km to mi—put km in the denominator of the conversion factor so it cancels. Next put
mi in the numerator to create it. As we are given that 1 km equals .6214 mi, add these
numbers in.
Now that meters has been converted to miles, next convert seconds to hours. You should
already know a conversion factor for this—1 hr is 3600 seconds. Place seconds in the
numerator of the conversion factor to cancel this. Then place hr in the denominator to create
it. Finally, place the corresponding numbers. Complete the calculations. There are 2 sig figs
in the final answer. Answer 5.8 x 108 mi/hr
.87x(3.00x108 m) 1km .6214mi 3600s 5.8x108 mi
x
x
x
=
s
1000m
1km
1hr
hr
18.
In the spring of 2008, petrol cost £1.029 per litre in London. On the same day, the
exchange rate was $1 = £0.497. What was the price of London petrol in dollars ($) per
gallon? (1 gal = 3.7854 L)
Always start by writing down what it is that you are given, in this case, £1.029 per L, with
£1.029 in the numerator and 1L in the denominator. Then think-I need a set of conversion
factors that converts £ to $ and L to gal. I always begin by cancelling out the value in the
numerator, so we know we want to get rid of £ and create $. Begin by placing £ in the
denominator so that it cancels and $ in the numerator. We have been told that there are .497
£ per 1 dollar so place these numbers in the conversion factor. Then, to convert L to gallons
we want to get rid of L so we will place this in the numerator of the conversion factor so it
cancels. Then, to create gallons we will place this in the denominator. We are told that 1
gallon equal 3.7854 L so add these numbers in. Complete the calculations. There are 3 sig
figs in the final answer. Answer $7.84/gal
£ 1.029
1L
x
3.7854 L $7.84
=
1 gal
gal
£ .497
$1
x
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19.
Manganese makes up 1.3  10–4 percent by mass of the elements found in a
normal healthy body. How many grams of manganese would be found in the body of a
person weighing 261 lb? (2.205 lb = 1 kg)
Always start by writing down what it is that you are given. In this case, determining this
could be confusing. You are given that 1.3x10-4 % of a 261 lb body is manganese. Because a
percent is a fraction, this is the same as saying (1.3 x 10-6 ) x 261 lb is manganese, and this
becomes your given value. Then think—starting with this number of pounds, I need a
conversion factor that gets rid of lb and gives me grams. To get rid of pounds, place it in the
denominator so it cancels. But, we cannot yet put grams in the numerator to create this
because we have not been told how many grams are in 1 pound. We have been told, however,
that there are 2.205 lb in 1 kg, so we will start by putting kg in the numerator. Then we can
convert to grams. Place kg in the denominator of the next conversion factor so it cancels.
Then place g in the numerator. We know that there are 1000 g in 1 kg so add these numbers
into this conversion factor. Answer .15 g
(1.3 x 10-6 )(261 lb) x
1kg
1000g
x
= .15g
2.205lb 1kg
2.0
A scientist obtains the number 0.045006700 on a calculator. If this number
actually has four (4) significant figures, how should it be written?
Remember that if a number has a decimal place, leading 0’s are not significant, so the
significant digits start with the number four in the hundredths place. Only that and
following three places should be used, and the 0 before the 6 should be rounded up because it
represents an estimated value:
Answer 0.04501
21.
Express the number 0.000191 in scientific notation.
Remember that with scientific notation the number you multiply by 10x should be a number
between 1.0000… and 9.9999….. This number should only include the significant digits in the
value, with the decimal point after the first digit—so in this case you would start with 1.91
times 10x.
You should visually be able to “see” what the exponent will be for the first several decimal
places without having to count. For example, at least be able to see immediately that:
.1
.01
.001
.0001
tenths
hundredths
thousandths
ten thousandths
10-1
10-2
10-3
10-4
The number of the exponent will be the place of the first non-zero digit—in this case the first
non-zero digit is in the fourth (10,000ths) decimal place so 10-4. Within just a few seconds
(meaning 2 or 3 seconds) you should know that the answer is 1.91 x 10-4. Answer 1.91 x 10-4
22.
Express 165,000 in exponential notation.
Remember that in a number with no decimal point all trailing zeros are not significant
(unless you have witnessed the actual measurement to know that a trailing 0 is actually an
estimate). So, 165,000 only has three significant figures and the value you are multiplying
must be 1.65.
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8
As with decimal numbers less than 1, for numbers greater than 1 you should be able to “see”
what the exponent will be without having to count digits, for almost any number. This is
helped by knowing the following:
10
tens
100
hundreds
1000
thousands
10000
ten thousands 104
100,000
hundred thousands
1,000,000
millions
1,000,000,000 billions
109
1,000,000,000,000 trillions
101
102
103
105
106
1012
For this problem, you should immediately know that because the number is in hundred
thousands, the exponent will be 105 and you should know the answer again within 2 or 3
seconds.
Answer 1.65 x 105
23.
Express the number 0.0470 in scientific notation.
24.
Express the number 6.04 × 10–3 in common decimal form.
25.
Express the number 2.38 × 104 in common decimal form.
Given the discussion of the 26 and 27, you can immediately see that there are three
significant figures beginning with the 4 in the hundredths place so the number you are
multiplying is 4.70. You can also see this number begins in the hundredths place so the
exponent value will be -2. Answer 4.70 x 10-2
Remember that the for a number expressed in scientific notation the number of digits of the
multiplied number is also the number of significant figures—to express measurements in
unambiguous numbers of significant figures is the purpose of scientific notation. Therefore,
the given number has three significant figures—6.04. Because you can visualize the number
of decimal places associated with a negative exponent you know that for 10-3 would be a
number that started in the thousandths place. Answer .00604
Because you can visualize the number of places associated with a positive exponent you know
that 104 would be a number that started in the ten thousandths place. 2.38 x 10,000 would
be:
Answer 23,800
26.
We generally report a measurement by recording all of the certain digits plus ____
uncertain digit(s).
a)
no
b)
one
c)
two
d)
three
e)
four
You are aware of our rules for reporting measurements—we report all certain digits, and
the first estimated or uncertain digit. Answer B
27.
The beakers shown below have different precisions as shown.
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Suppose you pour the water from these three beakers into one container. What would be
the volume in the container reported to the correct number of significant figures?
9
In the first drawing least count marks are 1 mL so we can estimate to .1 mL—I would
estimate 26.4 mL. In the second drawing the least count marks are 10 mL so we can estimate
to 1 mL—I would estimate 27 mL. In the third drawing the least count marks are .1 mL so we
can estimate to the nearest .01 mL—I would estimate 26.42 mL. Because we are adding
values together the number of sig figs is constrained by the least number of decimal places,
which will be 0 decimal places. So, adding the three values together we get: Answer
26.4mL + 27mL + 26.42mL = 79.82 mL = 80 mL
28.
You are asked to determine the perimeter of the cover of your textbook. You
measure the length as 33.16 cm and the width as 24.83 cm. How many significant figures
should you report for the perimeter?
Because we are adding values together the number of sig figs is constrained by the least
number of decimal places, which will be 2 decimal places—remembering that for a
perimeter we need to add in each length and side value twice, we get
33.16cm+ 33.16cm+ 24.83cm+ 24.83cm = 115.98cm . So, our final answer must have five
significant figures.
Remember that this corresponds to the idea that we can increase the precision of our
answers by increasing the values of the overall measurements. The uncertainty of the
estimated digit in the 0.01 decimal place of the final answer affects this value less than it
does any of the individual measurements which each have only four significant figures.
Answer 5 sig figs
29.
Consider the numbers 23.68 and 4.12. The sum of these numbers has ____
significant figures, and the product of these numbers has ____ significant figures.
For the sum, the number of sig figs is constrained by the 2 decimal places—the
answer must also have 2 decimals places, so for this sum, the final answer will have four
sigfigs. Multiplications are constrained by the value of the smallest number of significant
figures. In this case, 4.12 has only 3 sig figs so the answer must have 3 sig figs. Answer 4,3
30.
Using the rules of significant figures, calculate the following:
6.167  81
5.10
Remember for mixed calculations, for sig figs we perform each operation as we come to it.
The first operation is 6.167 + 81. As this is an addition, we are limited by the number of
decimal places—the smallest number of decimal places in this calculation is 0. If we do the
addition we get 87.167. We use this number for the rest of our calculation to limit rounding
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error but in the final calculation, this number has only two sig figs. It is useful to indicate
this in some in your work so you don’t forget—underlining or circling the number of sigfigs
works best. We then divide 87.167 by 5.10 to get 17.09157. As the numerator is constrained
to 2 sig figs, and this last operation is a division, the final answer must also have two
significant figures. Thus, the final answer is 17. Answer 17
6.167 + 81 87 .167
=
= 17.09157 = 17
5.10
5.10
31.
Using the rules of significant figures, calculate the following: 4.0021-0.247
As this is a subtraction problem we are limited to the smallest number of decimal places,
which is three. Perform the calculation using the numbers given and limit the result to three
decimal places. The answer ends up with four significant figures but this is okay even though
the subtracted value only has three sig figs—again, subtractions are only limited by decimal
places, not least number of sig figs in the values. Answer 3.755
4.0021- .247 = 3.755
32.
How many significant figures are there in the number 0.04560700?
Remember that if a decimal number has leading zero’s these ARE NOT significant. For a
decimal number we begin counting at the first non-zero number. This and all following
numbers, including 0’s, are significant. Beginning with the first non-zero digit, there are
seven sig figs. Answer 7
33.
How many significant figures are there in the number 0.0006313?
34.
How many significant figures are there in the number 3.1400?
Remember that if a decimal number has leading zero’s these ARE NOT significant. For a
decimal number we begin counting at the first non-zero number. This and all following
numbers, including 0’s, are significant. Beginning with the first non-zero digit, there are 4
sig figs. Answer 4
Remember that if a decimal number has trailing zero’s these ARE significant. For a decimal
number we begin counting at the first non-zero number after the decimal point. This and all
following numbers, including 0’s, are significant. Beginning with the first non-zero digit,
there are 5 sig figs. Answer 5
35.
How many significant figures should be reported for the difference between
18.6150 mL and 18.57 mL?
As this is a subtraction problem we are limited to the smallest number of decimal places,
which is two. Perform the calculation using the numbers given and limit the result to two
decimal places. Be very careful here—the answer ends up with only one significant figure—
while there are two decimal places one of these is a leading 0. But this is okay even though
both given values have more sig figs—again, subtractions are only limited by decimal places,
not least number of sig figs in the values. Answer 1 significant figure.
18.6150mL -18.57mL = .045mL = .05mL 1significant figure
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3.478 g  1.164 g
– 0.349 g/mL?
2.00 mL
Remember for mixed calculations, for sig figs we perform each operation as we come to it.
The first operation is 3.478 x 1.164. As this is a multiplication, we are limited by the smallest
number of sig figs in the values—in this case, both have 4 sig figs so the answer of this
operation is limited to 4 sig figs. If we do the multiplication we get 4.048392. We use this
number for the rest of our calculation to limit rounding error but in the final calculation, this
number has only 4 sig figs. It is useful to indicate this in some in your work so you don’t
forget—underlining or circling the number of sigfigs works best. We then divide 4.048392g
by 2.00mL to get 2.024196. As this operation is a division, it will be constrained to the
smallest number of sig figs in either value, in this case, 3 (from 2.00 mL). Notice that this
gives our answer 2 decimal places. We then subtract .349 from 2.024196, remembering that
this last value really only has 2 decimal places, so our final answer is also limited to 2
decimal places. The subtraction gives 1.675196, which, limited to 2 decimal places rounds to
1.68 g/mL.
36.
What is the best answer to report for
3.478g x 1.164 g
4.048 3 92g **
- .349g / mL =
- .349g / mL =
2.00mL
2.00 mL
2.2 0 4196 - .349 = 1.675196 = 1.68
I will give the first person who points out the error in the above problem a candy bar.
37.
What is the best answer to report for (515  0.0025) + 24.57?
Remember for mixed calculations, for sig figs we perform each operation as we come to it.
The first operation is 515 x 0.0025. As this is a multiplication, we are limited by the smallest
number of sig figs in the values—in this case, 2 sig figs (for .0025) so the answer of this
operation is limited to 2 sig figs. If we do the multiplication we get 1.2875. We use this
number for the rest of our calculation to limit rounding error but in the final calculation, this
number has only 2 sig figs. It is useful to indicate this in some in your work so you don’t
forget—underlining or circling the number of sigfigs works best. We then add 1.2875 to
24.57 to get 25.8575. As this is an addition problem sig figs are limited to the smallest
number of decimal places. Because the previous number is limited to 2 sigfigs, which is the .1
place, the first number is only supposed to have 1 decimal place, so this limits the final
answer to one decimal place, or, 25.9.
(515x.0025) + 24.57 = 1. 2 875 + 24.57 = 25.8575 = 25.9
38.
The degree of agreement among several measurements of the same quantity is
called __________. It reflects the reproducibility of a given type of measurement.
a)
accuracy
b)
error
c)
precision
d)
significance
e)
certainty
A) Incorrect. Accuracy is the determination of how close one or more measurements are to
the correct answer. B) Incorrect. Error is a determination of how far away from the
expected value a measurement is. C) Correct. For a group of measurements, precision is the
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term used to describe how close together the values of the measurements are. If they are
close together, they are precise and the measurement is reproducible. This is not to be
confused with the term precision as it is used for a single measurement. In that context it
refers to the degree of uncertainty for the measurement. D) Incorrect. The term significance
refers to the number of certain and uncertain digits present in a single measurement. E)
Incorrect. Certainty refers to the degree to which determine a measurement—all digits in a
measurement except the last are certain, while the last digit is uncertain. Answer C
39.
As part of the calibration of a new laboratory balance, a 1.000-g mass is weighed
with the following results:
Trial
1
2
3
Mass
1.201 ± 0.001
1.202 ± 0.001
1.200 ± 0.001
The balance is:
a)
Both accurate and precise.
b)
Accurate but imprecise.
c)
Precise but inaccurate.
d)
Both inaccurate and imprecise.
e)
Accuracy and precision are impossible to determine with the available
information.
All three measurements are very close together and are therefore precise as a group of
measurements. However, as a group, they are nowhere near the expected value and so are
not accurate. So, the measurements are precise but inaccurate. Answer C
Use the following to answer questions 19-20:
Consider the following three archery targets:
I.
40.
a)
b)
c)
d)
e)
II.
III.
Which of the following figure(s) represent a result having high precision?
Figure I only
Figure II only
Figure III only
Figure I and Figure II
Figure II and Figure III
Measurements that are close together have a high precision, regardless of whether or not
they are accurate. Therefore, the results in both diagrams II and III are precise. Answer E
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41.
a)
b)
c)
d)
e)
13
Which of the following statements concerning these figures is correct?
Figure I represents systematic error and Figure II represents random error.
Figure I represents random error and Figure II represents systematic error.
Figure I and Figure II represent random error.
Figure I and Figure II represent systematic error.
Figure III represents no errors.
In a group of measurements systematic error is error related to equipment problems, and so
measured values differ from expected results by about the same amount and in the same
direction, as shown in II. Random error is error related to operator error and random
variation observed in all processes, so values differ from the expected results in all directions
and by varying amounts, as shown in I. In figure III, although all values are close to the
expected value and so are accurate and precise, there will always still be small amounts of
error as shown in figure III. As figure I represents random error, figure II represents
systematic error, and figure III still represents small amounts of error, B is correct. Answer B
42.
Which of the following is the least probable concerning five measurements taken
in the lab?
a)
The measurements are accurate and precise.
b)
The measurements are accurate but not precise.
c)
The measurements are precise but not accurate.
d)
The measurements are neither accurate nor precise.
e)
All of these are equally probable.
If measurements are accurate, as they would all cluster around the expected value, they must
also be precise. Therefore, while measurements can be precise and not accurate, as in
systematic error, or they can be both not precise and inaccurate, or they can precise and
accurate, they cannot be accurate but not precise. Answer B.
43.
You measure water in two containers: a 10-mL graduated cylinder with marks at
every mL, and a 1-mL pipet marked at every 0.1 mL. If you have some water in each of the
containers and add them together, to what decimal place could you report the total
volume of water?
a)
0.01 mL
b)
0.1 mL
c)
1 mL
d)
10 mL
e)
none of these
For the 10 mL cylinder with marks at every mL, you could estimate to the nearest 0.1 mL. For
the 1 mL pipet, you could estimate to the nearest .01 mL. When we add measurements, we
limit the number of significant figures to the smallest number of decimal places—in this case,
to the nearest 0.1 mL. Answer B
44.
a)
b)
c)
d)
The agreement of a particular value with the true value is called
accuracy
error
precision
significance
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e)
certainty
45.
a)
b)
c)
d)
e)
The amount of uncertainty in a measured quantity is determined by:
both the skill of the observer and the limitations of the measuring instrument
neither the skill of the observer nor the limitations of the measuring instrument
the limitations of the measuring instrument only
the skill of the observer only
none of these
14
We determine how accurate a measurement is by trying to figure out how close it is to an
expected value. Answer A
Not only is there error in the ability of a human to determine values on a scale of
measurement, there is also variability in the manufacture of such scales. Error related to
both must be considered. Answer A
Use the following to answer questions 73-74:
The density of a liquid is determined by successively weighing 25, 50, 75, 100, and 125 mL of the liquid in a
250-mL beaker.
46.
If volume of liquid is plotted along the horizontal axis, and total mass of beaker
plus liquid is plotted on the vertical axis:
a)
The x, or horizontal, intercept is the negative value of the weight of the beaker.
b)
The y, or vertical, intercept is the weight of the empty beaker.
c)
The slope of the line is 1.0.
d)
The line will pass through the origin.
e)
The slope of the line is independent of the identity of the liquid.
It might help to make a quick sketch of the plot of this investigation
The plot should be linear because we would expect the same increment
of mass for every 25 mL of liquid added to the beaker. Because the slope
(rise over run) would be equivalent to mass over volume, the slope should reflect the
density—but it is unlikely that this would be a value of one, unless the liquid is water
(eliminate (c)) . So, the slope is dependent on the identity of the liquid (eliminate (e)). Where
the line intercepts the x axis would indicate both a 0 mass and negative volume (eliminate
(a). It should make sense that if there are 0 mL of liquid in the beaker that the
line created by the plot will intercept the y axis at the mass of the beaker, so the line cannot
pass through the origin (eliminate (d), confirm (b)) . Answer B
47.
Considering the plot of total mass (y-axis) versus volume (x-axis), which of the
following is true?
a)The plot should be rather linear because the slope measures the density of a liquid.
b)The plot should be curved upward because the slope measures the density of a liquid.
c)The plot should be curved upward because the mass of the liquid is higher in successive
trials.
d)The plot should be linear because the mass of the beaker stays constant.
e)
None of the above.
(a) True. The measurement of mass over volume is the density of a substance. As we are
incrementing the volume by the same amount with each measurement (increasing 25 mL)
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we would predict that the mass would also increase by the same amount with each
increment, meaning that the value of rise over run (the slope), which would be equivalent to
the mass over volume (density) would remain constant—this means that the plot should be
linear. (b) False. Because the line should be linear because the slope measures the density,
(see (a)) this would be inconsistent with statement (b). (c) False. While the mass of the
liquid is higher in successive trials, it is increased by the same amount with each trial, and so,
the relationship is linear. (d) False. While the mass of the beaker stays constant, the mass of
the liquid in the beaker increases, so while the mass of the beaker is constant, this is not why
the plot is linear. (e) False. Answer A
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