1
|
- Compiled summer 2004
- Joan Jakubowski
|
2
|
- I. Physics Skills
- II. Mechanics
- III. Energy
- IV. Electricity and Magnetism
- V. Waves
- VI. Modern Physics
|
3
|
- A. Scientific Notation
- B. Graphing
- C. Significant Figures
- D. Units
- E. Prefixes
- F. Estimation
|
4
|
- Use for very large or very small numbers
- Write number with one digit to the left of the decimal followed by an
exponent (1.5 x 105)
- Examples: 2.1 x 103 represents 2100 and 3.6 x 10-4 represents
0.00036
|
5
|
- 1. Write 365,000,000 in scientific notation
- 2. Write 0.000087 in scientific notation
|
6
|
- 1.) 3.65 x 108
- 2.) 8.7 x 10-5
|
7
|
- Use graphs to make a “picture” of scientific data
- “independent variable”, the one you change in your experiment is graphed
on the “x” axis and listed first in a table
- “dependent variable”, the one changed by your experiment is graphed on
the “y” axis and listed second in a table
|
8
|
- Best fit “line” or “curve” is drawn once points are plotted. Does not
have to go through all points. Just gives you the “trend” of the points
- The “slope” of the line is given as the change in the “y” value divided
by the change in the “x” value
|
9
|
- 1. Direct Relationship means an increase/decrease in one variable causes
an increase/decrease in the other
- Example below
|
10
|
- 2. Inverse(indirect) relationship means that an increase in one variable
causes a decrease in the other variable and vice versa
- Examples
|
11
|
- 3. Constant proportion means that a change in one variable doesn’t
affect the other variable
- Example;
|
12
|
- 4. If either variable is squared(whether the relationship is direct or
indirect), the graph will curve more steeply.
|
13
|
- Uncertainty in measurements is expressed by using significant figures
- The more accurate or precise a measurement is, the more digits will be
significant
|
14
|
- 1. Zeros that appear before a nonzero digit are not significant
(examples: 0.002 has 1 significant figure and 0.13 has 2 significant
figures)
- 2. Zeros that appear between nonzero digits are significant (examples:
1002 has 4 significant figures and 0.405 has 3 significant figures)
|
15
|
- 3. zeros that appear after a nonzero digit are significant only if they
are followed by a decimal point (20. has 2 sig figs) or if they appear
to the right of the decimal point (35.0 has 3 sig figs)
|
16
|
- 1. How many significant figures does 0.050900 contain?
- 2. How many significant figures does 4800 contain?
|
17
|
- 1. 5 sig figs
- 2. 2 sig figs
|
18
|
- 1. Fundamental units are units that can’t be broken down
- 2. Derived units are made up of other units and then renamed
- 3. SI units are standardized units used by scientists worldwide
|
19
|
- Meter (m)– length, distance, displacement, height, radius, elongation or
compression of a spring, amplitude, wavelength
- Kilogram (kg)– mass
- Second (s)– time, period
- Ampere (A)– electric current
- Degree (o)– angle
|
20
|
- Meter per second (m/s)– speed, velocity
- Meter per second squared (m/s2)– acceleration
- Newton (N)– force
- Kilogram times meter per second (kg.m/s)– momentum
- Newton times second (N.s)-- impulse
|
21
|
- Joule (J)– work, all types of energy
- Watt (W)– power
- Coulomb (C)– electric charge
- Newton per Coulomb (N/C)– electric field strength (intensity)
- Volt (V)- potential difference (voltage)
- Electronvolt (eV)– energy (small amounts)
|
22
|
- Ohm (Ω)– resistance
- Ohm times meter (Ω.m)– resistivity
- Weber (Wb)– number of magnetic field (flux) lines
- Tesla (T)– magnetic field (flux) density
- Hertz (Hz)-- frequency
|
23
|
- Adding prefixes to base units makes them smaller or larger by powers of
ten
- The prefixes used in Regents Physics are tera, giga, mega, kilo, deci,
centi, milli, micro, nano and pico
|
24
|
- A terameter is 1012 meters, so… 4 Tm would be 4 000 000 000
000 meters
- A gigagram is 109 grams, so… 9 Gg would be 9 000 000 000
grams
- A megawatt is 106 watts, so… 100 MW would be 100 000 000
watts
- A kilometer is 103 meters, so… 45 km would be 45 000 meters
|
25
|
- A decigram is 10-1 gram, so… 15 dg would be 1.5 grams
- A centiwatt is 10-2 watt, so… 2 dW would be 0.02 Watt
- A millisecond is 10-3 second, so… 42 ms would be 0.042 second
|
26
|
- A microvolt is 10-6 volt, so… 8 µV would be 0.000 008 volt
- A nanojoule is 10-9 joule, so… 530 nJ would be 0.000 000 530
joule
- A picometer is 10-12 meter, so… 677 pm would be 0.000 000 000
677 meter
|
27
|
- 1.) 16 terameters would be how many meters?
- 2.) 2500 milligrams would be how many grams?
- 3.) 1596 volts would be how many gigavolts?
- 4.) 687 amperes would be how many nanoamperes?
|
28
|
- 1.) 16 000 000 000 000 meters
- 2.) 2.500 grams
- 3.) 1596 000 000 000 gigavolts
- 4.) 0.000 000 687 amperes
|
29
|
- You can estimate an answer to a problem by rounding the known
information
- You also should have an idea of how large common units are
|
30
|
- 2 cans of Progresso soup are just about the mass of 1 kilogram
- 1 medium apple weighs 1
newton
- The length of an average Physics student’s leg is 1 meter
|
31
|
- 1.) Which object weighs approximately one newton? Dime, paper clip, student, golf ball
- 2.) How high is an average doorknob from the floor? 101m, 100m, 101m,
10-2m
|
32
|
|
33
|
- A. Kinematics; vectors, velocity, acceleration
- B. Kinematics; freefall
- C. Statics
- D. Dynamics
- E. 2-dimensional motion
- F. Uniform Circular motion
- G. Mass, Weight, Gravity
- H. Friction
- I. Momentum and Impulse
|
34
|
- In physics, quantities can be vector or scalar
- VECTOR quantities have a magnitude (a number), a unit and a direction
- Example; 22m(south)
|
35
|
- SCALAR quantities only have a magnitude and a unit
- Example; 22m
|
36
|
- VECTOR quantities; displacement, velocity, acceleration, force, weight,
momentum, impulse, electric field strength
- SCALAR quantities; distance, mass, time, speed, work(energy), power
|
37
|
- Distance is the entire pathway an object travels
- Displacement is the “shortest” pathway from the beginning to the end
|
38
|
- 1.) A student walks 12m due north and then 5m due east. What is the
student’s resultant displacement? Distance?
- 2.) A student walks 50m due north and then walks 30m due south. What is
the student’s resultant displacement? Distance?
|
39
|
- 1.) 13m (NE) for displacement
- 17 m for distance
- 2.) 20m (N) for displacement 80 m for distance
|
40
|
- Speed is the distance an object moves in a unit of time
- Velocity is the displacement of an object in a unit of time
|
41
|
|
42
|
|
43
|
- 1.) A boy is coasting down a hill on a skateboard. At 1.0s he is
traveling at 4.0m/s and at 4.0s he is traveling at 10.0m/s. What
distance did he travel during that time period? (In all problems given
in Regents Physics, assume acceleration is constant)
|
44
|
- 1.) You must first find the boy’s average speed/velocity before you are
able to find the distance
|
45
|
|
46
|
- The time rate change of velocity is acceleration (how much you speed up
or slow down in a unit of time)
- We will only be dealing with constant (uniform) acceleration
|
47
|
|
48
|
|
49
|
- 1.) A car initially travels at 20m/s on a straight, horizontal road. The
driver applies the brakes, causing the car to slow down at a constant
rate of 2m/s2 until it comes to a stop. What was the car’s
stopping distance? (Use two different methods to solve the problem)
|
50
|
- First Method
- vi=20m/s
- vf=0m/s
- a=2m/s2
- Use vf2=vi2+2ad
to find “d”
- d=100m
|
51
|
|
52
|
- In a vacuum (empty space), objects fall freely at the same rate
- The rate at which objects fall is known as “g”, the acceleration due to
gravity
- On earth, the “g” is 9.81m/s2
|
53
|
- To solve freefall problems use the constant acceleration equations
- Assume a freely falling object has a vi=0m/s
- Assume a freely falling object has an a=9.81m/s2
|
54
|
- 1.) How far will an object near Earth’s surface fall in 5s?
- 2.) How long does it take for a rock to fall 60m? How fast will it be
going when it hits the ground
- 3.) In a vacuum, which will hit the ground first if dropped from 10m, a
ball or a feather?
|
55
|
|
56
|
|
57
|
- 3.) Both hit at the same time because “g” the acceleration due to
gravity is constant. It doesn’t depend on mass of object because it is a
ratio
|
58
|
- If you toss an object straight up, that is the opposite of freefall.
- So…
- vf is now 0m/s
- a is -9.8lm/s2
- Because the object is slowing down not speeding up
|
59
|
- 1.) How fast do you have to toss a ball straight up if you want it reach
a height of 20m?
- 2.) How long will the ball in problem #1 take to reach the 20m height?
|
60
|
|
61
|
|
62
|
- The study of the effect of forces on objects at rest
- Force is a push or pull
- The unit of force is the newton(N) (a derived vector quantity)
|
63
|
- When adding concurrent (acting on the same object at the same time)
forces follow three rules to find the resultant (the combined effect of
the forces)
- 1.) forces at 00, add them
- 2.) forces at 1800, subtract them
- 3.) forces at 900, use Pythagorean Theorem
|
64
|
- Forces at 00
- Forces at 1800
- Forces at 900
|
65
|
- 1.) Find the resultant of two 5.0N forces at 00? 1800?
And 900?
|
66
|
|
67
|
|
68
|
|
69
|
- The opposite of adding concurrent forces.
- Breaking a resultant force into its component forces
- Only need to know components(2 forces) at a 900 angle to each
other
|
70
|
- To find the component forces of the resultant force
- 1.) Draw x and y axes at the tail of the resultant force
- 2.) Draw lines from the head of the force to each of the axes
- 3.) From the tail of the resultant force to where the lines intersect
the axes, are the lengths of the component forces
|
71
|
- Black arrow=resultant force
- Orange lines=reference lines
- Green arrows=component forces
- y
- x
|
72
|
|
73
|
- Equilibrium occurs when the net force acting on an object is zero
- Zero net force means that when you take into account all the forces
acting on an object, they cancel each other out
|
74
|
- An object in equilibrium can either be at rest or can be moving with
constant (unchanging) velocity
- An “equilibrant” is a force equal and opposite to the resultant force
that keeps an object in equilibrium
|
75
|
- Black arrows=components
- Blue arrow=resultant
- Red arrow=equilibrant
- =
- +
|
76
|
- 1.) 10N, 8N and 6N forces act concurrently on an object that is in
equilibrium. What is the equilibrant of the 10N and 6N forces? Explain.
- 2.) A person pushes a lawnmower with a force of 300N at an angle of 600
to the ground. What are the vertical and horizontal components of the
300N force?
|
77
|
- 1.) The 8N force is the equilibrant (which is also equal to and opposite
the resultant) The 3 forces keep the object in equilibrium, so the third
force is always the equilibrant of the other two forces.
|
78
|
|
79
|
- The study of how forces affect the motion of an object
- Use Newton’s Three Laws of Motion to describe Dynamics
|
80
|
- Also called the law of inertia
- Inertia is the property of an object to resist change. Inertia is
directly proportional to the object’s mass
- “An object will remain in equilibrium (at rest or moving with constant
speed) unless acted upon by an unbalanced force”
|
81
|
- “When an unbalanced (net) force acts on an object, that object
accelerates in the direction of the force”
- How much an object accelerates depends on the force exerted on it and
the object’s mass (See equation)
|
82
|
- Fnet=the net force exerted on an object (the resultant of all forces on
an object) in newtons (N)
- m=mass in kilograms (kg)
- a=acceleration in m/s2
|
83
|
- Also called law of action-reaction
- “When an object exerts a force on another object, the second object
exerts a force equal and opposite to the first force”
- Masses of each of the objects don’t affect the size of the forces (will
affect the results of the forces)
|
84
|
- A drawing (can be to scale) that shows all concurrent forces acting on
an object
- Typical forces are the force of gravity, the normal force, the force of
friction, the force of acceleration, the force of tension, etc.
|
85
|
- Fg is the force of gravity or weight of an object (always
straight down)
- FN is the “normal” force (the force of a surface pushing up
against an object)
- Ff is the force of friction which is always opposite the
motion
- Fa is the force of acceleration caused by a push or pull
|
86
|
- If object is moving with constant speed to the right….
- Black arrow=Ff
- Green arrow=Fg
- Yellow arrow=FN
- Blue arrow=Fa
|
87
|
- When an object is at rest or moving with constant speed on a slope, some
things about the forces change and some don’t
- 1.) Fg is still straight down
- 2.) Ff is still opposite motion
- 3.) FN is no longer equal and opposite to Fg
- 4.) Ff is still opposite motion
- More……
|
88
|
- Fa=Ff=Ax=Acosθ
- FN=Ay=Asinθ
- Fg=mg (still straight down)
- On a horizontal surface, force of gravity and normal force are equal and
opposite
- On a slope, the normal force is equal and opposite to the “y” component
of the force of gravity
|
89
|
- Green arrow=FN
- Red arrow=Fg
- Black arrows=Fa and Ff
- Orange dashes=Ay
- Purple dashes=Ax
|
90
|
- 1.) Which has more inertia a 0.75kg pile of feathers or a 0.50kg pile of
lead marbles?
- 2.) An unbalanced force of 10.0N acts on a 20.0kg mass for 5.0s. What is
the acceleration of the mass?
|
91
|
- 1.) The 0.75kg pile of feathers has more inertia because it has more
mass. Inertia is dependent on the “mass” of the object
|
92
|
|
93
|
- 3.) A 10N book rests on a
horizontal tabletop. What is the force of the tabletop on the book?
- 4.) How much force would it take to accelerate a 2.0kg object 5m/s2?
How much would that same force accelerate a 1.0kg object?
|
94
|
- 1.) The force of the tabletop on the book is also 10N (action/reaction)
|
95
|
|
96
|
- To describe an object moving 2-dimensionally, the motion must be
separated into a “horizontal” component and a “vertical” component
(neither has an effect on the other)
- Assume the motion occurs in a “perfect physics world”; a vacuum with no
friction
|
97
|
- 1.) Projectiles fired horizontally
- an example would be a baseball tossed straight horizontally away from
you
|
98
|
- Use the table below to solve these type of 2-D problems
|
99
|
- 2.) Projectiles fired at an angle
- an example would be a soccer ball lofted into the air and then falling
back onto the ground
|
100
|
- Use the table below to solve these type of 2-D problems
- Ax=Acosθ and Ay=Asinθ
|
101
|
- 1.) A girl tossed a ball horizontally with a speed of 10.0m/s from a
bridge 7.0m above a river. How long did the ball take to hit the river?
How far from the bottom of the bridge did the ball hit the river?
|
102
|
- 1.) In this problem you are asked to find time and horizontal distance
(see table on the next page)
|
103
|
|
104
|
|
105
|
- 2.) A soccer ball is kicked at an angle of 600 from the
ground with an angular velocity of 10.0m/s. How high does the soccer
ball go? How far away from where it was kicked does it land? How long
does its flight take?
|
106
|
- 2.) In this problem you are asked to find vertical distance, horizontal
distance and horizontal time. Finding vertical time is usually the best
way to start. (See table on next page)
|
107
|
|
108
|
- Find vertical “t” first using
- vf=vi+at
with….
- vf=0.0m/s
- vi=8.7m/s
- a=-9.81m/s2
- So…vertical “t”=0.89s and
horizontal “t” is twice that
- and equals 1.77s
|
109
|
- Find horizontal “d” using
|
110
|
- Find vertical “d” by using
|
111
|
- When an object moves with constant speed in a circular path
- The force (centripetal) will be constant towards the center
- Acceleration (centripetal) will only be a direction change towards the
center
- Velocity will be tangent to the circle in the direction of movement
|
112
|
- Fc=centripetal force, (N)
- v=constant velocity (m/s)
- ac=centripetal acceleration (m/s2)
- r=radius of the circular pathway (m)
- m=mass of the object in motion (kg)
|
113
|
|
114
|
|
115
|
- 1.) A 5kg cart travels in a circle of radius 2m with a constant velocity
of 4m/s. What is the centripetal force exerted on the cart that keeps it
on its circular pathway?
|
116
|
|
117
|
- Mass is the amount of matter in an object
- Weight is the force of gravity pulling down on an object
- Gravity is a force of attraction between objects
|
118
|
- Mass is measured in kilograms (kg)
- Mass doesn’t change with location (for example, if you travel to the moon your mass
doesn’t change)
|
119
|
- Weight is measured in newtons (N)
- Weight “does” change with location because it is dependent on the pull
of gravity
- Weight is equal to mass times the acceleration due to gravity
|
120
|
|
121
|
- g=acceleration due to gravity but it is also equal to “gravitational
field strength”
- The units of acceleration due to gravity are m/s2
- The units of gravitational field strength are N/kg
- Both quantities are found from the equation:
|
122
|
- 1.) If the distance between two masses is doubled, what happens to the
gravitational force between them?
- 2.) If the distance between two objects is halved and the mass of one of
the objects is doubled, what happens to the gravitational force between
them?
|
123
|
- 1.) Distance has an inverse squared relationship with the force of
gravity.
- So…since r is multiplied by 2 in the problem, square 2……so….22=4,
then take the inverse of that square which equals Ľ….so….the answer is
“Ľ the original Fg”
|
124
|
- 2.) Mass has a direct relationship with Fg and distance has an inverse squared
relationship with Fg.
- First…since m is doubled so is Fg and since r is halved,
square ˝ , which is Ľ and then take the inverse which is 4.
- Then…combine 2x4=8
- So…answer is “8 times Fg”
|
125
|
- 3.) Determine the force of gravity between a 2kg and a 3kg object that
are 5m apart.
- 4.) An object with a mass of 10kg has a weight of 4N on Planet X. What
is the acceleration due to gravity on Planet X? What is the
gravitational field strength on Planet X?
|
126
|
|
127
|
- The force that opposes motion measured in newtons (N)
- Always opposite direction of motion
- “Static Friction” is the force that opposes the “start of motion”
- “Kinetic Friction” is the force of friction between objects in contact
that are in motion
|
128
|
- The ratio of the force of friction to the normal force (no unit, since
newtons cancel out)
- Equation
- Ff= μFN
- μ=coefficient of friction
- Ff=force of friction
- FN=normal force
|
129
|
- The smaller the coefficient, the easier the surfaces slide over one
another
- The larger the coefficient, the harder it is to slide the surfaces over
one another
- Use the table in the reference tables
|
130
|
- 1.) A horizontal force is used to pull a 2.0kg cart at constant speed of
5.0m/s across a tabletop. The force of friction between the cart and the
tabletop is 10N. What is the coefficient of friction between the cart
and the tabletop? Is the friction force kinetic or static? Why?
|
131
|
- 1.)
- The friction force is kinetic because the cart is moving over the
tabletop
|
132
|
- Momentum is a vector quantity that is the product of mass and velocity
(unit is kg.m/s)
- Impulse is the product of the force applied to an object and time (unit
is N.s)
|
133
|
- p=momentum
- Δp=change in momentum= (usually) m(vf-vi)
- J=impulse
|
134
|
- p=mv
- J=Ft
- J=Δp
- pbefore=pafter
|
135
|
- 1.) A 5.0kg cart at rest experiences a 10N.s (E) impulse. What is the
cart’s velocity after the impulse?
- 2.) A 1.0kg cart at rest is hit
by a 0.2kg cart moving to the right at 10.0m/s. The collision causes the
1.0kg cart to move to the right at 3.0m/s. What is the velocity of the
0.2kg cart after the collision?
|
136
|
- 1.) Use J=Δp so……
- J=10N.s(E)=Δp=10kg.m/s(E)
- and since original p was 0kg.m/s and Δp=10kg.m/s(E),
- new p=10kg.m/s(E)
- then use….. p=mv so……..
- 10kg.m/s(E)=5.0kg x v so….
- v=2m/s(E)
|
137
|
- 2.) Use pbefore=pafter
- Pbefore=0kg.m/s + 2kg.m/s(right)
- =2kg.m/s(right)
- Pafter=2kg.m/s(right)=3kg.m/s +
- P(0.2kg cart) so….p of
0.2kg cart must be -1kg.m/s or 1kg.m/s(left)
- more…..
|
138
|
- So if p after collision of 0.2kg cart is 1kg.m/s(left) and
- p=mv
- 1kg.m/s(left)=0.2kg x v
- And v=5m/s(left)
|
139
|
- A. Work and Power
- B. Potential and Kinetic Energy
- C. Conservation of Energy
- D. Energy of a Spring
|
140
|
- Work is using energy to move an object a distance
- Work is scalar
- The unit of work is the Joule (J)
- Work and energy are manifestations of the same thing, that is why they
have the same unit of Joules
|
141
|
- Power is the rate at which work is done so there is a “time” factor in
power but not in work
- Power and time are inversely proportional; the less time it takes to do
work the more power is developed
- The unit of power is the watt (W)
- Power is scalar
|
142
|
- W=work in Joules (J)
- F=force in newtons (N)
- d=distance in meters (m)
- ΔET=change in total energy in Joules (J)
- P=power in watts (W)
- t=time in seconds (s)
|
143
|
- W=Fd=ΔET
- ***When work is done vertically, “F” can be the weight of the object Fg=mg
|
144
|
- 1.) A 2.5kg object is moved 2.0m in 2.0s after receiving a horizontal
push of 10.0N. How much work is done on the object? How much power is
developed? How much would the object’s total energy change?
- 2.) A horizontal 40.0N force causes a box to move at a constant rate of
5.0m/s. How much power is developed? How much work is done against
friction in 4.0s?
|
145
|
- 1.) to find work use W=Fd
- So…W=10.0N x 2.0m=20.0J
- To find power use P=W/t
- So…P=20.0J/2.0s=10.0W
- To find total energy change it’s the same as work done so……
- ΔET=W=20.0J
|
146
|
- 2.) To find power use
- So… P=40.0N x 5.0m/s=200W
- To find work use P=W/t so…200W=W/4.0s
- So….W=800J
|
147
|
- 3.) A 2.0kg object is raised vertically 0.25m. What is the work done
raising it?
- 4.) A lift hoists a 5000N object vertically, 5.0 meters in the air. How
much work was done lifting it?
|
148
|
- 3.) to find work use W=Fd with F equal to the weight of the object
- So…..W=mg x d
- So...W=2.0kgx9.81m/s2x0.25m
- So…W=4.9J
|
149
|
- 4.) to find work use W=Fd
- Even though it is vertical motion, you don’t have to multiply by “g”
because weight is already given in newtons
- So…W=Fd=5000N x 5.0m
- And W=25000J
|
150
|
- Gravitational Potential Energy is energy of position above the earth
- Elastic Potential Energy is energy due to compression or elongation of a
spring
- Kinetic Energy is energy due to motion
- The unit for all types of energy is the same as for work the Joule (J).
All energy is scalar
|
151
|
- ΔPE=change in gravitational potential energy in Joules (J)
- m=mass in kilograms (kg)
- g=acceleration due to gravity in (m/s2)
- Δh=change in height in meters (m)
- Equation ΔPE=mgΔh
- ***Gravitational PE only changes if there is a change in vertical
position
|
152
|
- 1.) How much potential energy is gained by a 5.2kg object lifted from
the floor to the top of a 0.80m high table?
- 2.) How much work is done in the example above?
|
153
|
- 1.) Use ΔPE=mgΔh to find potential energy gained so ΔPE=5.2kgx9.81m/s2x0.80m
- So…ΔPE=40.81J
- 2.) W=ΔET so..W is also 40.81J
|
154
|
- KE=kinetic energy in Joules (J)
- m=mass in kilograms (kg)
- v=velocity or speed in (m/s)
|
155
|
- 1.) If the speed of a car is doubled, what happens to its kinetic
energy?
- 2.) A 6.0kg cart possesses 75J of kinetic energy. How fast is it going?
|
156
|
- 1.) Using KE=1/2mv2 if v is doubled, because v if squared KE
will be quadrupled.
- 2.) Use KE=1/2mv2 so…..
- 75J=1/2 x 6.0kg x v
- And…..v=5m/s
|
157
|
- In a closed system the total amount of energy is conserved
- Total energy includes potential energy, kinetic energy and internal
energy
- Energy within a system can be transferred among different types of
energy but it can’t be destroyed
|
158
|
- In a perfect physics world since there is no friction there will be no
change in internal energy so you don’t have to take that into account
- In a perfect physics world energy will transfer between PE and KE
|
159
|
- In the real world there is friction so the internal energy of an object
will be affected by the friction (such as air resistance)
|
160
|
- ET=total energy of a system
- PE=potential energy
- KE=kinetic energy
- Q=internal energy
- ***all units are Joules (J)
|
161
|
- In a real world situation, ET=KE+PE+Q because friction exists
and may cause an increase in the internal energy of an object
- In a “perfect physics world” ET=KE+PE with KE+PE equal to the total
“mechanical energy of the system object
|
162
|
- position #1
- position #2
- position #3
more..
|
163
|
|
164
|
- Position #1
-
Position #2
-
Position #3
|
165
|
- Position #1 the ball/bob has not
starting falling yet so the total energy is all in gravitational
potential energy
- Position #2 the ball/bob is halfway down, so total energy is split
evenly between PE and KE
- Position #3 the ball/bob is at the end of its fall so total energy is
all in KE
|
166
|
- 1.) A 2.0kg block starts at rest and then slides along a frictionless
track. What is the speed of the block at point B?
-
A
-
7.0m
- B
|
167
|
- Since there is no friction, Q does not need to be included
- So…use ET=PE+KE
- At position B, the total energy is entirely KE
- Since you cannot find KE directly, instead find PE at the beginning of
the slide and that will be equal to KE at the end of the slide more…..
|
168
|
- PE (at position A) =mgΔh=2.0kgx9.81m/s2x7.0m =137.3J
- KE (at position A) =0J because there is no speed
- So ET (at position A)=137.3J
- At position B there is no height so the PE is 0J
- More….
|
169
|
- At position B the total energy still has to be 137.3J because energy is
conserved and because there is no friction no energy was “lost” along
the slide
- So….ET(position B)=137.3J=0J+KE
- So…KE also equals 137.3J at position B
- More…
|
170
|
- Use KE=1/2mv2
- So…KE=137.3J=1/2x2.0kgxv2
- So v (at position B)=11.7m/s
|
171
|
- position #1
- position #2
- position #3
- 1.) From what height must you drop the 0.5kg ball so that the it will be
traveling at 25m/s at position #3(bottom of the fall)?
- 2.)How fast will it be traveling
at position #2 (halfway down)?
- *Assume no friction
|
172
|
- 1.) At position #3, total energy will be all in KE because there is no
height and no friction
- So…use ET=KE=1/2mv2
- KE=1/2 x 0.5kg x (25m/s)2
- So…KE=156.25J=PE (at position#1)
- So…ΔPE=156.25J=mgΔh
- And Δh=31.86m
|
173
|
- 2.) Since position #2 is half way down total energy will be half in PE
and half in KE
- So…KE at position #2 will be half that at position #3
- So…KE at position #2 is 78.125J
- Then use KE (at #2)=78.125J =1/2 x 0.5kg x v2
- v=17.68m/s at position #2
|
174
|
- Energy stored in a spring is called “elastic potential energy”
- Energy is stored in a spring when the spring is stretched or compressed
- The work done to compress or stretch a spring becomes its elastic
potential energy
|
175
|
- Fs=force applied to stretch or compress the spring in newtons
(N)
- k=spring constant in (N/m) ***specific for each type of spring
- x=the change in length in the spring from the equilibrium position in
meters (m)
|
176
|
|
177
|
|
178
|
- 1.) What is the potential energy stored in a spring that stretches 0.25m
from equilibrium when a 2kg mass is hung from it?
- 2.) 100J of energy are stored when a spring is compressed 0.1m from
equilibrium. What force was needed to compress the spring?
|
179
|
- 1.) Using PEs=1/2kx2 you know “x” but not “k”
- You can find “k” using Fs=kx
- With Fs equal to the weight of the hanging mass
- So… Fs=Fg=mg=2kgx9.81m/s2
- Fs=19.62N=kx=k x 0.25m
- k=78.48N/m
- More…
|
180
|
- Now use PEs=1/2kx2
- PEs=1/2 x 78.48N/m x (0.25m)2
- So PEs=2.45J
|
181
|
- 2.) To find the force will use Fs=kx, but since you only know
“x” you must find “k” also
- Use PEs=1/2kx2 to find “k”
- PEs=100J=1/2k(0.1m)2
- k=20 000N/m
- use Fs=kx=20 000N/m x 0.1m
- Fs=2000N
|
182
|
- 1.) Thermal Energy is heat energy which is the KE possessed by the
particles making up an object
- 2.) Internal Energy is the total PE and KE of the particles making up an
object
- 3.) Nuclear Energy is the energy released by nuclear fission or fusion
- 4.) Electromagnetic Energy is the energy associated with electric and
magnetic fields
|
183
|
- A. Electrostatics/Electric Fields
- B. Current Electricity
- C. Series Circuits
- D. Parallel Circuits
- E. Electric Power and Energy
- F. Magnetism and Electromagnetism
|
184
|
- Atomic Structure—the atom consists of proton(s) and neutron(s) in the
nucleus and electrons outside the nucleus.
- The proton and neutron have similar mass (listed in reference table)
- The electron has very little mass (also in reference table)
|
185
|
- A proton has a positive charge that is equal in magnitude but opposite
in sign to the electron’s negative charge
- A neutron has no charge so it is neutral
|
186
|
- The unit of charge is the coulomb (C)
- Each proton or each electron has an “elementary charge” (e) of 1.60x10-19C
- The magnitude of the charge on both an electron and a proton are the
same, only the signs are different
|
187
|
- An object has a neutral charge or no net charge if there are equal
numbers of protons and neutrons
- An object will have a net “negative” charge if there are more electrons
than protons
- An object will have a net “positive” charge if there are less electrons
than protons
|
188
|
- Transfer of charge occurs only through movement of electrons
- If an object loses electrons, it will have a net positive charge
- If an object gains electrons, it will have a net negative charge
|
189
|
- When the 2 spheres touch
- Next….
|
190
|
- And then are pulled apart. This is what happens
|
191
|
- On the previous page, only charge is transferred
- Total charge stays the same
- So charge is always conserved
- There is “CONSERVATION OF CHARGE” just like with energy and momentum
|
192
|
- 1.) Sphere #1 touches sphere #2 and then is separated. Then sphere #2
touches #3 and is separated. What are the final charges on each sphere?
|
193
|
- 1.) #1 has a charge of +2e
- #2 has a charge of +3e
- #3 has a charge of +3e
|
194
|
- Fe=electrostatic force in newtons (N) can be attractive or
repulsive
- k=electrostatic constant 8.99x109N.m2/C2
- q=charge in coulombs (C)
- r=distance of separation in meters (m)
- E=electric field strength in (N/C)
|
195
|
|
196
|
- An electric field is an area around a charged particle in which electric
force can be detected
- Electric fields are detected and mapped using “positive” test charges
- Field lines are the imaginary lines along which a positive test charge
would move (arrows show the direction)
|
197
|
- Negative charge Positive charge
- Parallel Plates
- +
- -
|
198
|
- Negative and positive charges
- Field lines
|
199
|
- Two positive charges
- Field lines
|
200
|
- Two negative charges
- Field lines
|
201
|
- 1.) What is the magnitude of the electric field strength when an
electron experiences a 5.0N force?
- 2.) Is a charge of 4.8x10-19C? possible? A charge of
- 5.0x10-19C?
- 3.) What is the electrostatic force between two 5.0C charges that are
1.0x10-4m apart?
|
202
|
|
203
|
- 2.) Only whole number multiples of the elementary charge are possible.
To find out if charge is possible, divide by 1.60x10-19C.
- 4.8x10-19C is possible because when it is divided you get 3,
which is a whole number.
- 5.0x10-19C is not possible because when it is divided you get
3.125 which is not a whole number.
|
204
|
|
205
|
- Current is the rate at which charge passes through a closed pathway (a
circuit)
- The unit of current is ampere (A)
|
206
|
- Must have a “potential difference” supplied by a battery or power source
- Must have a “pathway” supplied by wires
- Can have a resistor(s)
- Can have meters (ammeters, voltmeters, ohmmeters)
|
207
|
- Current is the flow of charge past a point in a circuit in a unit of
time. Measured in amperes (A)
- Potential Difference is the work done to move a charge between two
points. Measured in volts (V)
- Resistance is the opposition to the flow of current. Measured in ohms (Ω)
|
208
|
- I=current in amperes (A)
- Δq=charge in coulombs (C)
- t=time in seconds (s)
- V=potential difference in volts (V)
- W=work in Joules (J)
- R=resistance in ohms (Ω)
- ρ=resistivity in (Ω.m)
- L=length of wire in meters (m)
- A=cross-sectional area of wire in square meters (m2)
|
209
|
|
210
|
- A material is a “good” conductor if electric current flows easily
through it
- Metals are good conductors because they have many electrons available
and they are not tightly bound to the atom
- If a material is an extremely poor conductor, it is called an “insulator
|
211
|
- Resistivity is an property inherent to a material (and its temperature)
that is directly proportional to the resistance of the material
- Use the “short, thick, cold” rule to remember what affects resistance
- Short, thick, cold wires have the lowest resistance (best conduction)
|
212
|
- 1.) How many electrons pass a point in a wire in 2.0s if the wire
carries a current of 2.5A?
- 2.) A 10 ohm resistor has 20C of charge passing through it in 5s. What
it the potential difference across the resistor?
|
213
|
|
214
|
|
215
|
- 3.) What is the resistance of a 5.0m long aluminum wire with a
cross-sectional area of 2.0x10-6m2 at 200C?
|
216
|
|
217
|
- Series circuit is a circuit with a single pathway for the current
- Current stays the same throughout the circuit
- Resistors share the potential difference from the battery (not
necessarily equally)
- The sum of the resistances of all resistors is equal to the equivalent
resistance of the entire circuit
|
218
|
|
219
|
- Ammeter measures the current in a circuit
- Voltmeter measures the potential difference across a resistor or battery
- Ohmmeter measures the resistance of a resistor
|
220
|
- Cell
- Battery
- Switch
- Voltmeter
- ammeter
|
221
|
- Resistor
- Variable Resistor
- Lamp
|
222
|
|
223
|
- 1.) Two resistors with resistances of 4Ω and 6Ω are connected
in series to a 25V battery. Determine the potential drop through each
resistor, the current through each resistor and the equivalent
resistance.
|
224
|
- 1.) You can use R=V/I as soon as you have 2 pieces of information at
each resistor. You can also use the series circuit laws.
- First, use Req=R1+R2 to find Req
- 4Ω+6Ω=10Ω=Req (use this as the R at the
battery)
- More…
|
225
|
- Then use R=V/I at the battery to find I at the battery
- So…10Ω=25V/I and I=2.5A and
all currents are equal in a series circuit, so I1and I2=2.5A
- More….
|
226
|
- Now that you have 2 pieces of information at each resistor, you can use
R=V/I to find potential differences
- At the 4Ω resistor, 4Ω=V/2.5A
- so V1=10V
- At the 6Ω resistor, 6Ω=V/2.5A
- So…V2=15V
|
227
|
- Parallel circuits have more than one pathway that the current can pass
through
- Current is shared (not necessarily equally) among the pathways
- Voltage is the same in each pathway as across the battery
- Equivalent resistance is the sum of the reciprocal resistances of the
pathways (Req is always less than R of any individual
pathway)
|
228
|
|
229
|
|
230
|
- 1.) Three resistors of 2Ω, 4Ω and 4Ω are connected in
parallel to an applied potential difference of 20V. Determine equivalent
resistance for the circuit, the potential difference across each
resistor and the current through each resistor.
|
231
|
- 1.)
- So equivalent resistance=1Ω
- More…..
|
232
|
- Use V=V1=V2=V3
- So since V=20V, V1=V2=V3=20V
- More……
|
233
|
- Using R=V/I, substitute the potential differences and resistances for
each resistor to find the current for each resistor
- I=20A
- I1=10A
- I2=5A
- I3=5A
|
234
|
- Electric power is the product of potential difference and current
measured in watts just like mechanical power
- Electric energy is the product of power and time measured in joules just
like other forms of energy
|
235
|
|
236
|
- 1.) How much time does it take a 60W light bulb to dissipate 100J of
energy?
- 2.) What is the power of an electric mixer while operating at 120V, if
it has a resistance of 10Ω?
- 3.) A washing machine operates at 220V for 10 minutes, consuming 3.0x106J
of energy. How much current does it draw during this time?
|
237
|
- 1.) Use W=Pt
- So…100J=60Wxt
- So…t=0.6s
- 2.) Use
- So…P=1440W
|
238
|
- 3.) Use W=VIt but first need to
change the 10 minutes to 600 seconds.
- So…3.0x106J=220V(I)600s
- So…I=22.7A
|
239
|
- Magnetism is a force of attraction or repulsion occurring when spinning
electrons align
- A magnet has two ends called “poles”
- The “N” pole is “north seeking”
- The “S” pole is “south seeking”
- Like “poles” repel
- Unlike “poles” attract
|
240
|
- If you break a magnet, the pieces still have N and S poles
- The earth has an “S” pole at the North Pole
- The earth has a “N” pole at the South Pole
- The “N” pole of a compass will point towards earth’s “S” pole which is
the geographic North pole
|
241
|
- There is a “field” around each magnet
- Imaginary magnetic field or flux lines show where a magnetic field is
- You can use a compass to map a magnetic field
|
242
|
- The weber (Wb) is the unit for measuring the number of field lines
- The tesla (T) is the unit for magnetic field or flux density
- 1T=1Wb/m2
|
243
|
|
244
|
|
245
|
|
246
|
- Moving a conductor through a magnetic field will induce a potential
difference which may cause a current to flow (conductor must “break” the
field lines for this to occur)
- A wire with a current flowing through it creates a magnetic field
- So…magnetism creates electricity and electricity creates magnetism
|
247
|
- Must move wire into and out of page to induce potential current by
breaking field lines
|
248
|
- A.) Wave Characteristics
- B.) Periodic Wave Phenomena
- C.) Light
- D.) Reflection and Refraction
- E.) Electromagnetic Spectrum
|
249
|
- A wave is a vibration in a “medium” or in a “field”
- Sound waves must travel through a medium (material)
- Light waves may travel in either a medium or in an electromagnetic field
- Waves transfer energy only, not matter
|
250
|
- A pulse is a single disturbance or vibration
- A periodic wave is a series of regular vibrations
|
251
|
- 1.) Longitudinal Waves are waves that vibrate parallel to the direction
of energy transfer. Sound and earthquake p waves are examples.
|
252
|
- 2.) Transverse waves are waves that vibrate perpendicularly to the
direction of energy transfer. Light waves and other electromagnetic
waves are examples.
|
253
|
- Frequency is how many wave cycles per second
- The symbol for frequency is f
- The unit for frequency is hertz (Hz)
- 1Hz=1/s
- Frequency is pitch in sound
- Frequency is color in visible light
|
254
|
- High frequency wave
- Low frequency wave
|
255
|
- Period is the time required for one wave cycle
- The symbol for period is T
- The unit for period is the second (s)
- The equation for period is T=1/f
- Period and frequency are inversely proportional
|
256
|
- Short period wave
- Long period wave
|
257
|
- The amplitude of a wave is the amount of displacement from the
equilibrium line for the wave (how far crest or trough is from the
equilibrium line)
- Symbol for amplitude is A
- Unit for amplitude is meter (m)
- Amplitude is loudness in sound
- Amplitude is brightness in light
|
258
|
- High amplitude wave
- Low amplitude wave
|
259
|
- Wavelength is the distance between points of a complete wave cycle
- Symbol for wavelength is λ
- Unit for wavelength is meter (m)
- A wave with a high frequency will have short wavelengths and short
period
|
260
|
- Short wavelength λ
- λ
- Long wavelength
|
261
|
- Points that are at the same type of position on a wave cycle (including
same direction from equilibrium and moving in same direction) are “in
phase”
|
262
|
- Points that are in phase are whole wavelengths apart (i.e., 1λ
apart, 2λ apart, 3λ apart, etc.)
- Points that are in phase are multiples of 360 degrees apart (i.e., 360
degrees apart, 720 degrees apart, 1080 degrees apart, etc.)
|
263
|
- A B C
- D E F
- Points A, B and C are in phase
- Points D, E and F are in phase
- Points A and B are 360 degrees apart and 1λ apart
- Points D and F are 720 degrees apart and 2λ apart
|
264
|
- A C
-
D
- B
- Points A and B are 180 degrees apart and ˝ λ apart (not in phase)
- Points C and D are 90 degrees apart and Ľ λ apart (not in phase)
|
265
|
- Speed of a wave is equal to the product of wavelength and frequency
- Symbol for speed is v
- Unit for speed is m/s
- Equation for speed is v=fλ
- Can also use basic speed equation v=d/t if needed
|
266
|
- The speed of a wave depends on its “type” and the medium it is traveling
through
- The speed of light (and all electromagnetic waves) in a vacuum and in
air is 3.00x108m/s
- The speed of sound in air at STP (standard temperature and pressure) is
331m/s
|
267
|
- 1.) A wave has a speed of 7.5m/s and a period of 0.5s. What is the
wavelength of the wave?
- 2.) If the period of a wave is doubled, what happens to its frequency?
- 3.) Points on a wave are 0.5λ apart, 7200 apart, 4λ
apart. Which points are in phase?
|
268
|
- 1.) First use T=1/f to find f, then use v=fλ to find λ
- So…0.5s=1/f and f=2.0Hz
- Then…7.5m/s=2.0Hz(λ) and λ=3.75m
- 2.) T and f are inversely related so if T is doubled f is halved.
|
269
|
- 3.) Points are in phase only if they are whole wavelengths apart or
multiples of 360 degrees apart. So..the points that are 720 degrees
apart are in phase and the points that are 4λ apart are in phase.
The points that are 0.5λ apart are not in phase.
|
270
|
- 5.0m
- 4.) What is the wavelength of the wave shown above?
|
271
|
- 4.) The wave is 5.0m long and contains 2.5 cycles. You want to know the
length of 1 cycle.
- So…5.0m/2.5 cycles=2.0m/cycle
- So…λ=2.0m
|
272
|
- When waves interact with one another many different “phenomena” result
- Those phenomena are the doppler effect, interference, standing waves,
resonance and diffraction
|
273
|
- Some phenomena are easier to explain using circular waves
- A is the λ from
- crest to crest
- is the
- wavefront (all
- points on the
- circle are in A A
- phase)
|
274
|
- If the source of waves is moving relative to an observer, the observed
frequency of the wave will change (actual f produced by the source
doesn’t change and movement must be fast for observer to notice)
|
275
|
- If source is moving towards the observer, f will increase (higher pitch
if sound wave, shift towards blue if light wave)
- If source is moving away from the observer, f will decrease (lower pitch
if sound wave, shift towards red if light wave)
|
276
|
|
277
|
- In the diagram on the previous page, the wave source is moving towards
Joe
- In the diagram on the previous page, the wave source is moving away from
Shmoe
|
278
|
- The effect on Joe is that the f of the wave will be higher for him
(shorter λ and T, higher pitch or bluer light)
- The effect on Shmoe is that the f of the wave will be lower for him
(longer λ and T, lower pitch or redder light)
|
279
|
- When waves travel in the same region and in the same plane at the same
time (superposition) they can interfere with each other
- In “constructive” interference, waves build on one another to increase
amplitude
- In “destructive” interference, waves destroy on another to decrease
amplitude
|
280
|
- Maximum constructive interference occurs when waves are in phase
- Maximum destructive interence occurs when waves are 180 degrees out of
phase
|
281
|
|
282
|
|
283
|
- Standing waves occur when waves having same A and same f travel in
opposite directions. Vibrates so that it looks like waves are stationary
|
284
|
- A body with an ability to vibrate has a natural frequency at which it
will vibrate
- If you put an object that is vibrating with that same natural frequency
next to that body, the body will also start vibrating (don’t need to
touch)
- This is called “resonance”
|
285
|
- Examples of resonance:
- Find the “Tacoma Narrows Bridge” video on the internet
- A singer being able to break a glass with her voice
|
286
|
- When waves bend behind a barrier instead of going straight through that
is called diffraction
- Wave
- fronts
|
287
|
- Light is the part of the electromagnetic spectrum that is visible to
humans
- The speed of light symbol is c
- The speed of light is constant, 3.00x108m/s in air or a
vacuum
- Instead of using v=fλ, when it’s light you can substitute c=fλ
|
288
|
- No object can travel faster than the speed of light (one of Einstein’s
ideas)
- The speed of light in a material is always less than c
|
289
|
- 1.) Determine the wavelength in a vacuum of a light wave having a
frequency of 6.4x1014Hz.
- 2.) What is the frequency of a light wave with a wavelength of 5.6x10-7m?
|
290
|
- 1.) Use c=fλ
- 3.00x108m/s=6.4x1014Hz(λ)
- So… λ=4.7x10-7m
- 2.) Use c=fλ
- 3.00x108m/s=f(5.6x10-7m)
- So…f=5.4x1014Hz
|
291
|
- Reflection occurs when a ray of light hits a boundary and bounces back
into the same medium
- Refraction occurs when a ray of light enters a new medium and changes
direction because of a change in speed
|
292
|
- The law of reflection
- θi θr
- Dotted line is the “normal”
- (reference line)
|
293
|
- In a mirror the image of an object is flipped laterally (your right hand
is your left hand in a mirror image)
- To view an object in a plane mirror, you need a minimum of ˝ the height
of the object for the height of the mirror
|
294
|
- When light enters a different medium, the change in direction depends on
the density of the new medium
- If new medium is denser, the light will slow down and bend towards the
normal
- If new medium is less dense, the light will speed up and bend away from
the normal
|
295
|
|
296
|
- Symbols
- n=index of refraction (no units)
- v=speed of the wave in m/s
- λ=wavelength in m
- c=speed of light in a vacuum (3.00x108m/s)
- θ1=angle of incidence
- θ2=angle of refraction
|
297
|
- θ1
- air
- glass θ2
- Light travels from air into glass which is more dense so it slows down
and bends towards the normal
|
298
|
- θ1
- air
- glass θ2
- Light travels from glass into air which is less dense so it speeds up
and bends away from the normal
|
299
|
- Absolute index of refraction is the ratio of the speed of light in a
vacuum to the speed of light in a specific medium
- The symbol is n
- The higher the “n” number the denser the medium
- The lower the “n” number the less dense the medium
|
300
|
- Higher “n” means slower v of light in the medium
- Lower “n” means higher v of light in the medium
- Use reference tables to find “n” numbers
|
301
|
- 1.) What is the speed of light in diamond?
- 2.) What is the ratio of the speed of light in corn oil to the speed of
light in water?
|
302
|
- 3.) A ray of monochromatic light having a frequency of 5.09x1014Hz
is traveling through water. The ray is incident on corn oil at an angle
of 600 to the normal. What is the angle of refraction in corn
oil?
|
303
|
|
304
|
|
305
|
|
306
|
- Electromagnetic waves include gamma rays, x rays, ultraviolet rays,
infrared rays, microwaves, t.v. and radio waves, long waves
- Electromagnetic waves are produced by accelerating charges (produce
alternating electric and magnetic fields)
|
307
|
- High energy electromagnetic waves have high f and small λ
- Low energy electromagnetic waves have low f and long λ
- Energy in a wave can be increased by increasing f, decreasing λ and
T and increasing the duration of the wave
|
308
|
- In the visible light spectrum, the violet end is high f, short λ
and high energy
- In the visible light spectrum, the red end is low f, long λ and low
energy
|
309
|
- A. Quantum Theory
- B. Atomic Models
- C. Nucleus
- D. Standard Model
- E. Mass-Energy
|
310
|
- E-m energy is absorbed and emitted in specific amounts called “quanta”
according to the quantum theory
- A photon is the basic unit of
- e-m energy (particle of light)
- In wave theory E-m energy is emitted continuously
- Energy of a photon is directly related to its frequency
|
311
|
- Light acts like a wave during diffraction, interference and the doppler
effect
- Light acts like a particle (quantum idea) during the photoelectric
effect and atomic spectra
|
312
|
- Equation
- Ephoton=energy of a photon in joules (J)
- h=Planck’s constant
- (6.63x10-34J.s)
- f=frequency in hertz (Hz)
- λ=wavelength in meters (m)
- c=speed of light in vacuum 3.00x108m/s
|
313
|
- Photoelectric Effect—Einstein used the particle idea of light (quantum
theory) to explain why light incident on a photosensitive metal would
only cause electrons to be emitted if its frequency was high enough
|
314
|
- When light acts like a particle, it has momentum just like a particle
- Compton Effect—when an e-m photon hits an electron, the photon transfers
some of its energy and momentum to the electron
- Particles can also act like waves by having a wavelength (usually too
small to detect)
|
315
|
- Thomson Model of the atom introduced the idea of an atom containing both
positive and negative charges (balancing each other because atom is
neutral)
- negative charge
- positive charge
|
316
|
- Rutherford Model of the atom introduced “nucleus” idea with small
positive core at center surrounded by orbiting negative electrons
- positive charge
- negative charge
|
317
|
- Bohr Model of the atom kept Thomson and Rutherford ideas but added
“energy levels” to explain why electrons don’t crash into the nucleus
- Positive charge
- Negative charge
|
318
|
- Bohr Model Ideas
- Electrons can only gain or lose energy in specific amounts (quanta)
- Electrons can occupy only certain orbits that have fixed radii
- Orbits closer to the nucleus have lower energy, those farther away,
higher energy
|
319
|
- Bohr Model Ideas (cont.)
- Electrons can jump up to higher orbits by “absorbing” quanta of energy
- Electrons will emit quanta of energy when they fall down to lower
orbits
- The orbit closest to the nucleus is called “ground state”
|
320
|
- Bohr Model Ideas (cont.)
- Any orbit (level) above ground state is called an “excited state”
- An atom is “ionized” when an electron has been removed
- “The ionization potential” is the amount of energy needed to remove an
electron from the atom
- Use energy level diagrams in the reference tables
|
321
|
- Cloud Model was introduced because the Bohr model couldn’t explain atoms
with many electrons
- Cloud model has same basic setup, nucleus and electrons, but there are
no fixed orbits
- Electrons are spread out in electron clouds instead
|
322
|
- Each element has a specific spectrum (lines of specific frequencies of
e-m energy that are either emitted or absorbed)
|
323
|
- Emission Spectra (Bright-line) occur when electrons fall to lower energy
levels and “emit” quanta of energy
- Absorption Spectra (Dark line) occur when electrons absorb energy in
order to jump up energy levels leaving behind dark lines
|
324
|
- The nucleus is the core of the atom made up of protons and neutrons
- The nucleus contains almost all of the mass of the atom
- The nucleus is positively charged because of the protons
|
325
|
- Each proton has a charge of +1e which is equal to +1.60x10-19C
- Each neutron has a charge of 0e so it is neutral
- The mass of a proton or a neutron is 1.67x10-27kg
|
326
|
- The nuclear force (strong force) is the strongest force known, that is
why nuclear reactions (fission and fusion) create such large amounts of
energy
|
327
|
- The particles of an atom; the electrons, neutrons and protons, are not
the whole story
- An electron is a fundamental particle, but a neutron and proton are not
- A fundamental particle can’t be broken down into smaller particles (at
least we think so)
|
328
|
- There are 3 categories of fundamental particles; hadrons, leptons and
force carrier particles
- Hadrons can interact with all 4 fundamental forces of nature. Protons
and neutrons are hadrons
- Leptons can interact with all the forces except the strong nuclear
force. Electrons are leptons
|
329
|
|
330
|
- A Baryon is a particle that can be changed into a proton or a neutron
- A Meson is a particle of intermediate mass
|
331
|
|
332
|
- A quark is a particle that makes up baryons and mesons
- Quarks have partial charges, +1/3e, -1/3e, +2/3e or -2/3e
- See reference tables for lists of quarks
- Each particle has an antiparticle having all same characteristics except
opposite charge and magnetic moment
|
333
|
|
334
|
|
335
|
- 1.) Strong nuclear force is by far the strongest, very close ranged
(nuclear distances)
- 2.) Electromagnetic force is the next strongest and closed ranged
- 3.) Weak nuclear force is the third strongest, very close ranged
- 4.) Gravitational force is weakest and very long ranged
|
336
|
- According to Albert Einstein, mass and energy are different
manifestations of the same thing
E=mc2
- E=energy in joules (J)
- m=mass in kilograms (kg)
- c=3.00x108m/s
|
337
|
- The mass-energy relationship can also be seen in the following
relationship
- 1u=931Mev
- u=universal mass unit
- Mev=megaelectronvolt (106eV)
- ****eV is an energy unit for very small amounts of energy
- 1eV also equals 1.60x10-19J
|
338
|
- Just like energy has to be conserved, so does mass energy
- Mass can be converted to energy and energy can be converted to mass
- But mass cannot be created or destroyed
|
339
|
- 1.) How much energy would be produced if a 1.0kg mass was completely
converted to energy?
- 2.) Determine the energy of a photon with a frequency of 3.5x1014Hz.
|
340
|
- 3.) The rest mass of a neutron is 1.0087u. Determine the energy
equivalent in megaelectronvolts.
- 4.) What type of e-m energy will be either absorbed or emitted when an
electron falls from the n=3 to the n=1 level in a hydrogen atom?
|
341
|
- 1.) Use E=mc2
- E=1.0kg(3.0x108m/s)2
- E=9x1016J
- 2.) Use Ephoton=hf
- Ephoton=6.63x10-34J.s(3.5x1015Hz) Ephoton=2.3x10-18J
|
342
|
- 3.) Use 1u=931MeV
- So…1.0087u x 931MeV/1u=
- 939MeV for the energy equivalent
|
343
|
- 4.) First subtract energy levels so…13.6eV-1.51eV=12.09eV
- Then change to joules using 1eV=1.60x10-19J
- So…12.09eV=1.93x10-18J
- Then use E=hf
- 1.93x10-18J=6.63x10-34J.s(f)
- So….f=2.92x1015Hz which is ultraviolet light
- (find that in reference table)
|
344
|
|