Transcript adv_BasicMathReviewx

Basic Math
Significant Figures and
Scientific Notation
Bell work problem

Usain Bolt ran 100 m in 9.69 seconds. A
cheetah can run 700 yards in 20 seconds.
Who's faster?
Note 1 yard = 0.9144 meters.
Bell work problem

Usain Bolt ran 100 m in 9.69 seconds. A
cheetah can run 700 yards in 20 seconds.
Who's faster?
Note 1 yard = 0.9144 meters.
(100)m
Usain Bolt 
 10.32 m/s
(9.69)s
0.9144 m
CheetahDis tance  700 yards 
 640.08 m
1 yard
(640.08)m
CheetahSpe ed 
 32 m/s
(20)s
Significant Figures
How long is the snake ?
1m
1m
0.1m = 10 cm
1m
0.01m = 1 cm
Significant Figures
How long is the snake ?
0.3 m
1m
0.27 m
1m
0.1m = 10 cm
0.270 m
1m
0.01m = 1 cm
Basic Rules for Significant Digits

Rule 1- which digits are significant: The
digits in a measurement that are
considered significant are
–
–
all of those digits that represent marked
calibrations on the measuring device plus one
additional
digit to represent the estimated digit (tenths of the
smallest calibration).
Adding and Subtracting

The sum or
difference of
measurements may
have no more
decimal places than
the least number of
places in any
measurement.
Adding and Subtracting

The sum or
difference of
measurements may
have no more
decimal places than
the least number of
places in any
measurement.
Ans: 29.8 m
Multiply and Divide

When multiplying or dividing, the
number of significant figures retained
may not exceed the least number of
digits in either the of the factors.
0.304 cm x 73.84168 cm =
Multiply and Divide

When multiplying or dividing, the
number of significant figures retained
may not exceed the least number of
digits in either the of the factors.
0.304 cm x 73.84168 cm =
22.4 cm2
Note: only display three (3) significant figures because 0.304 only has three
significant figures.
Big Numbers
What is the number:
602,214,150,000,000,000,000,000
 Avogadro’s number.
 How do we usually see this?
6.0221415x1023
or
6.0221415e+23
 Why do we display it in this form?

Physics Big and Little Numbers
Gravitational Constant (very small)
6.67 × 10-11 C = 6.67e-11
 Coulomb’s Constant (very big):
8.99 × 109 = 8.99e+9
 Charge on an electron (very small)
1.60217646 × 10-19 = 1.602e-19

Entering Scientific Notation on
Calculator
Practice with
gravitational
constant:
6.67e-11
 Type “6.67” 6.67
 Type “2nd-,” 6.67E
 Type Negative
(NOT minus) 11 6.67E-11

Practice Problems

Force between two large bodies:
Nm
(5.97E 24kg)  (60,627kg)
(6.67E  11
)
2
2
kg
(3.18E8m)
2
Practice Problems

Force between two large bodies:
Nm
(5.97E 24kg)  (60,627kg)
(6.67E  11
)
2
2
kg
(3.18E8m)
2
239 N
Practice Problems

Force between two charges:
Nm (1.6e  19C)( 3.2e  19C)
8.99e  9 2 *
2
C
(1.0e  10m)
2
Practice Problems

Force between two charges:
Nm (1.6e  19C)( 3.2e  19C)
8.99e  9 2 *
2
C
(1.0e  10m)
2
4.6e  8N
Review: Conversions to/from
Scientific Notation

Convert 3,600,000 to scientific notation
–

Convert 0.00435 to scientific notation
–

4.35e-3 or 4.35 x 10-3
Convert 8.99E9 (or 8.99 x 109)to standard
notation
–

3.6E6 or 3.6 x 106
8,990,000,000
Convert 6.67E-5 (or 6.67 x 10-5) to standard
notation
–
0.0000667
Unit Conversions
Importance of Units and
how to convert between
Units
Why do we use units?

Someone wants to sell you a car for 100.
–

Is this a good deal?
You cannot determine if this is a good deal
unless you know the units.
–
–
–
–
$1k units ($100,000 for a car? – too high)
$1 units ($100 – probably pretty good)
$10 units ($1,000 – depends on car)
$0.01 (i.e. 1¢) – ($1.00 for car – great deal!)
Units (cont’d)

There are seven (7) base units in the SI
system. All other units are derived from
these.
–
How many can you name?
Units
Units (cont’d)
acceleration
force
frequency
volume
velocity
velocity
Units Prefixes
Converting units



What if someone wanted to sell you their car for
5,000 £ (British pounds). Is this a good deal?
We need to convert units
We can convert units because





it is always okay to multiply something by 1.
Any number divided by itself (or its equivalent) is always
equal to 1
As long as numerator and denominator are equivalent, the
number is equal to 1.
$1.5782 is equal to 1 £. How much is 5,000 £ ?
$1 is equal to 0.6336 £. Does this change the conversion?
Converting Units
 $1.5782 
5,000£  
  $7,891
 1£ 
OR
 $1 
5,000£  
  $7,891
 0.6336£ 
More on Converting Units


Always okay to multiply by 1
We are NEVER changing the original
quantity (just the way we look at it)
–

Which is more $1 or 100¢ ?
Ensure the undesirable units ‘cancel out’
–
–
One instance should be in numerator (top)
Other instance should be in denominator (bottom)
Practice:

The distance to San Antonio is 197 miles. How
many km is this? (1 km = 0.62 miles)
–

A sack of cement is 50 kg. How heavy is this in
lbs (1 kg = 2.2 lbs)
–

318 km
110 lbs
The speed limit is 65 mph. How fast is this in
km/hr (1 mph = 1.60934 km/hr)
–
105 km/hr
More Complicated practice

How many seconds in a day?
–

86,400
How many seconds in 3:45:15?
–
13,515
More Complicated practice

How many seconds in a day?
–

86,400
How many seconds in 3:45:15?
–
13,515
More Complicated practice
(cont’d)

A dropped object achieves a speed of
39.8 m/s. How fast is that in mph? (there
are 1.60934 km in 1 mile).
More Complicated practice
(cont’d)
A dropped object achieves a speed of
39.8 m/s. How fast is that in mph? (there
are 1.60934 km in 1 mile).
39.8 𝑚 1 𝑘𝑚
1 𝑚𝑖𝑙𝑒𝑠
= 0.0247 𝑚𝑖𝑙𝑒𝑠
𝑠
𝑠
1000 𝑚 1.60934 𝑘𝑚

0.0247 𝑚𝑖𝑙𝑒𝑠 60 𝑠
𝑠
1 𝑚𝑖𝑛
60 𝑚𝑖𝑛
= 89 𝑚𝑖𝑙𝑒/ℎ𝑟
1 ℎ𝑟
Practice

Vesna Vulovic survived the longest fall on
record without a parachute when her plane
exploded and she fell 6miles, 551 yards.
What is the distance in meters? (there are
1.60934 km in 1 mile, and 1inch = 2.54 cm).
Practice

Bicyclists in the Tour de France reach
speeds of 34.0 miles per hour (mi/h) on flat
sections of the road. What is this speed in (a)
kilometers per hour (km/h) and (b) meters
per second (m/s)?
Functions
Using functions and
using basic algebra to
manipulate functions
Why we model
Bell work problem

Calculate the change in distance (Δd) for an
object that falls for 3 seconds (t) if it has an
initial velocity (vi) of 4.9 m/s, and an
acceleration (a) of 9.8 m/s2. Use the
following formula:
1
2
d  vi  t  a  t
2
Bell work answer

Calculate the change in distance (Δd) for an object
that falls for 3 seconds with (t) if it has an initial
velocity (vi) of 4.9 m/s, and an acceleration (a) of 9.8
m/s2. Use the following formula:
1 2
d  vi  t  a  t
2
2
2
d  (4.9m / s)(3s)  (1 / 2)(9.8m / s )(3s)
d  14.7m  44.1m  58.8m
Solving Physics Problems
1. Read the problem carefully.
2. Identify the quantity to be found.
3. Identify the quantities that are given in the problem.
4. Identify the equation that contains these quantities.
5. Solve the equation for the unknown.
6. Substitute the value given in the problem, along with
their proper units, into the equation and solve it.
7. Check to see if the answer will be correct in the
correct units. (dimensional analysis)
8. Check your answer to see if it reasonable.
Plug and Chug



Simply plugging in the numbers represented
by the variables
And calculating the desired result
Must put the equation in the “proper” form
before you can do this
–
–
Rearrange the formula so that the desired
variable is all by itself on the left:
i.e. Y= {some messy equation}
Practice – Using Formula Charts

Use appropriate formula to calculate
“average velocity” if an object moves a
distance of 923 miles in 14 hours. (you can
leave the answer in mph)
Practice – Using Formula Charts

Use appropriate formula to calculate
“average velocity” if an object moves a
distance of 923 miles in 14 hours. (you can
leave the answer in mph)
vavg
d 923 miles


 66 mph
t 14 hours
Practice – Using Formula Charts

Use the appropriate formula to calculate net
force that is applied if a mass of 3,900 kg (a
Hummer) is accelerated 2.43 m/s2 .
–

9501 kg·m/s2
Use the appropriate formula to calculate the
acceleration required to move an object with an
initial velocity (vi) of 0 m/s, a final velocity (vf) of
26.8 m/s, over a displacement (Δd) of 147.4 m.
–
2.43 m/s2
Rearranging simple functions


Sometimes the desired variable is in the
wrong place, so we must “solve” for the
desired variable (get it all by itself and on
top)
On simple formulas we can Cross multiply
–
This works because we can always multiply each
side of function by same number
Applying to
Velocity/Time/Distance



Use formula charts
Show that there is “one” function
Create two (2) additional ones
Variable Triangles

If you are planning on
a scientific/math
oriented career
(including business
medical fields,
computers) – need to
master basic algebra
skills (including
formula
manipulations)
Δd
v
Δt
Variable Triangles - practice

Can create for any
‘proportional’
equation
Variable Triangles - practice

Awkward and does
not really work for
added variables or
complex equations
Practice

Solve the following equation for displacement
(Δd). Then solve for time (Δt).
vavg
d  vavg  t
d

t
d
t 
vavg
Rearranging simple functions

Moving a value
–
–

always can add to one side, if we add the same to
the other side
Always can subtract from one side if we subtract
same from the other side
Solve for x:
y  mx  b
Using “rearranged” functions

Using the averaged velocity function, solve
the following:
–
The average velocity of the speed of light is 3e8
m/s. It takes 8 minutes (8*60 seconds) for light to
reach the earth from the sun. How far is the sun?

–
144,000,000 km
If Michael Phelps swims 1.667 m/s for 2 minutes
(120 seconds), how far has he swam?

200 m
Advanced Practice

Solve for vi =
1
2
d  vi  t  a  t
2
Advanced Practice

Solved for vi =
d  a  t
vi 
t
Or:
1
2

d 1
vi 
 2 a t
t
2