Transcript Welcome to Physics 100 !!!!

Welcome to Physics 100 !!!!
Dr. Gregory G. Wood
Fall 2005
A bit about me….
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Just joined CSUCI
Married to Dr. Tabitha Swan-Wood
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Expecting a baby girl Jan. 15, 2005
Quiz
1)
2)
3)
4)
5)
What is your name?
What is your major?
What is the email address you would like
correspondence about this course sent to?
Why are you taking physics? (Try to put something
more than “because I have to…”)
Tell me a bit about yourself.
What is Physics?
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Broadly defined: A scientific method
used to explain physical phenomena in
the universe using the tools of
mathematics.
Examples of Physics…
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Classical Mechanics:
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Motion of the planets
1-D Atomic Chain Transverse Modes-Java Applet
Examples of Physics…
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Light & Waves:
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Rainbows & Prisms
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Ultra-Sounds
Examples of Physics…
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Light & Waves:
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Atomic Scale Imaging
Examples of Physics…
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Quantum Mechanics & Solid State:
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Transistors and Solid Electronics
Physics Permeates Your Life
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TV
Radio
Computers
Automobiles
Plasma Screens
Medical Instruments: MRI, Ultrasound,
X-rays,…
Chapter 1
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Skim Chap. 1 and make sure you
understand it.
Important prefixes:
Power
Prefix
Abbreviation
103
kilo
k
10-2
centi
c
10-3
milli
m
10-6
micro
m
Chapter 1
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Dimensional Analysis:
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Units must be equal on both sides of an
equation
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When adding or subtracting units must be equal
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
[units] = [units]
[units] + [units]
Examples
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Yes: 5 m/s = 3 m/s + 2 m/s
No: 5 m/s = 3 m/s2 + 2 m/s
Chapter 1
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Scientific Notation
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3,240 = 3.24 x 103
Converting Units
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To convert 23 seconds to units of hours:
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23 sec x 1 min x 1 hour = 0.00639 hr = 6.39 x 10-3 hr
60 sec
60 min
Chapter 1
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How to Approach Physics Problems:
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Carefully read the problem
Visualize the problem
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If complicated, try to separate the events
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Sketch a diagram of what’s happening
Set-up the appropriate physics equations
Solve the equations
Check your answer: units and magnitude
Think about your answer
Chapter 1
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A note on grading and partial credit
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Give a solid attempt at every problem
Sketches will be worth something
Chapter 1
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Distance vs. Displacement
Chapter 2
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Position & Displacement…
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Create an axis
0m
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position
6m
At t=0sec Marm is at 0m and at t=3sec Marm is
at 6m
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Total Displacement (6m-0m) = 6m
Total Time (3sec-0sec) = 3sec
Chapter 2
0m
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Total Displacement (6m-0m) = 6m
Total Time (3sec-0sec) = 3sec
Average Speed
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6m
At t=0sec Marm is at 0m and at t=3sec Marm is at 6m
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position
Average Speed = Total Displacement
Total Time
=
6m =
3 sec
3m
sec
Average Velocity
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Equals average speed plus a direction
3m
To the right
sec
Chapter 2
Slope of tangent line = instantaneous velocity
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Position vs. Time
Time [sec]
Position [m]
0.8
1
2
2.2
4
3
6
3.5
Average Velocity:
between t=0 sec and t=6 sec
Instantaneous Velocity:
at t=6 sec
Chapter 2
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Instantaneous Velocity
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Lim Dx
Dt0 Dt
Dx
Dt
Dt
Dx
Chapter 2
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Average Acceleration
aave
v f  vi
Dv


Dt
t f  ti
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Instantaneous Acceleration
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Units:
Dv
a  Dt0
lim
Dt
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Dv
Dt

m/s = m
s
s2
Chapter 2
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Negative Acceleration (Deceleration)
a
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vf < v i
because
aave
v f  vi
Dv


Dt
t f  ti
+
v
Chapter 2
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Motion under Constant Acceleration
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Useful Equations:
v  v0  at
1
( v0  v )
2
1
x  x0  ( v0  v )t
2
1
x  x0  v0t  at 2
2
v 2  v02  2a ( x  x0 )
vav 
( 2  7)
( 2  9)
( 2  10)
( 2  11)
( 2  12)
Constant Accleration
dv
Dv
a
 lim Dt 0
dt
Dt
v f  v0
a
t
v f  v0  at
(2.7)
Aspen the dog starts at 2.0 m/s and
accelerates at 3.3 m/s2 for 1.1 s before
reaching top speed. What is her top speed?
Vf=2.0 m/s + 3.3 m/s2 x 1.1s = (2.0+3.63)m/s
= 5.6 m/s (two sig figs).
v vs. t
at
A1=½ t * at
A2=v0t
t
v vs. t for Constant Acceleration
Distance = Area under v vs. t curve
(because d = v*t)
at
d = A1 + A2
A1=½ t * at
d = v0t + ½ t * at
x – x0 = v0t + ½ at2
x = x0 + v0t + ½ at2
(2-11)
A2=v0t
t
How far?
Distance is the area under the
Velocity vs. time graph thus:
1
d  v0t  att
2
Aspen the dog starts at 2.0 m/s and
accelerates at 3.3 m/s2 for 1.1 s before
reaching top speed. How far does
she travel?
D = area rectangle + area triangle
Area triangle = ½ base x height
1 2
x  x0  v0t  at
2
(2.11)
Position vs. Time Plots
Constant Velocity
Constant Acceleration
1
x  x0  v0t  at 2
2
a = b + ct + dt2
Upward Parabola
Constant Deceleration
a = b + ct – dt2
Larger slope equals larger velocity
Downward Parabola
Without time information…
v f  v0  at
(2.7)
1 2
x  x0  v0t  at
2
(2.11)
Solve (2.7) for time and substitute into (2.11) and you will find:
v 2f  v02  2a( x  x0 )
(2.12)
Aspen goes from 2.0 m/s to 5.6 m/s over a distance of 4.2 m,
find acceleration.
Chapter 2
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Example 1 (prob. #12)
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It was a dark and stormy night, when
suddenly you saw a flash of lightening.
Three-and-a-half seconds later you heard
the thunder. Given that the speed of
sound in air is about 340 m/s, how far
away was the lightening bolt?
Chapter 2
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Example 2 (prob. #40)
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When you see a traffic light turn red you
apply the brakes until you come to a sop.
If your initial speed was 12 m/s, and you
were heading due west, what was your
average velocity during braking?
Assume constant velocity.
Chapter 2
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Example 3 (prob. #100)
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You drop a ski glove from a height h onto
fresh snow, and it sinks to a depth d
before coming to rest. (a) In terms of g
and h, what is the speed of the glove
when it reaches the snow? (b) What are
the magnitude and direction of the
glove’s acceleration as it moves through
the snow, assuming it to be constant?
Give your answer in terms of g, h, and d.
Lab One Activity: Measure a
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4-setups: measure time and distance
Assume constant acceleration
Each group uses different angle of
incline
Use a variety of distances – average
all a values
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