Lecture Note
Physics 1A Final Review Lecture Notes
-
University:
University of California, San Diego -
Course:
PHYS1a | Physics Academic year:
2023
-
Views:
183
Pages:
10
Author:
Diamond Bird
Physics Final Review
UNITS
Pico, P 10^-12
Nano, n 10^-9
Micro, μ 10^-6
Milli, m 10^-3
Centi, c 10^-2
Tera, T 10^12
Giga, G 10^9
Mega, M 10^6
Kilo, k 10^3
Smaller
larger
1 J = 1 kg m^2/s^2
1 ml = 1 cm^3
1 m^2 = 10^4 cm^2
1 L = 1000 cm^3 = 10^-3 m^3
1 cm^3 = 1E-6 m^3
Quiz #1
Vavg = Δx / Δt : average velocity
aavg = Δv / Δt : average acceleration
Relative Error
Addition/Subtraction: C = A + B → δC = sqrt((δA)^2 + (δB)^2)
Multiplication/Division: C = A/B → δC/C = sqrt((δA/A)^2 + (δB/B)^2)
Relative error = SD / Mean
Relative error is dimensionless: δA = SA/A
Kinematics Equations:
Constant Velocity: x_f - x_i = v(t_f - t_i)
Constant acceleration: v_f - v_i = a(t_f - t_i)
When acceleration is constant:
* IF the acceleration of an object is constant, then a plot of instant. Velocity vs time would be linear
* Avg velocity will ONLY be constant IF acceleration is 1
Used for freefall
v²f = v²i + 2a(xf - xi)
xf - xi = Vi(tf - ti) + 1/2 a(tf - ti)²
Gravity Diagram: constant acceleration
Vectors
V=(Vx,y) = Vxx + Vy y = vxit vyj
* x̂ = î = (1,0) AND ŷ = ĵ = (0,1)
Multiplying a vector by a Scalar:
Av = A (Vx,vy) = (Avx, Avy)
Adding vectors:
u⃗ = (ux, uy), v⃗ = (vx, vy)
u⃗ + v⃗ = (ux+vx), (uy+vy)
Vector Multiplication: Dot product (2 vectors multiply to give a scalar)
U · V = UxVx + UyVy + UzVz
angle between vectors w/ attached "tails"
u = √(ux)² + (uy)²
v = √(vx)² + (vy)²
Multiply to get uv
Vy = Vsinθ
Vx = Vcosθ
Vy = Vx tanθ
V = √(Vx)² + (Vy)²
Projectile Motion
Has horizontal (x) Vertical (y) components of motion.
x & y directions are completely independent of each other
U · V = UxVx + UyVy + UzVz
a vector with itself: v⃗ · v⃗ = vxvx + vyvy = vx² + vy² = v²
Dot product of
Чат с Gemini
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Image Analysis
Image Content:
The image is a handwritten note or diagram explaining the concept of selection in programming. It contains text, code snippets, and diagrams.
Textual Content:
Selection
!=
NOT equal to
eg) if variable != 1:
print("cool.")
==
checks if two things are identical
eq) if variable == 'cool':
print("Great.")
if :
else:
Boolean Operators
and -> True if both are true
or -> True if at least one is true
not -> True if operand is false
False if operand is true
De Morgan's Laws
not (A or B)
not A AND not B
not (A and B)
not A OR not B
Diagrammatic Elements:
A curly brace on the right side of the if statement to indicate a block of code.
Overall Description:
The image presents fundamental programming concepts related to selection, including comparison operators (!=, ==), if-else statements, boolean operators (and, or, not), and De Morgan's Laws. The content is organized in a handwritten style with code examples to illustrate the concepts.
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Image Analysis
Textual Content:
list1 = [1,2,'blue']
-> general format of an array (also called a list)
list.append()
-> add something to the end of the array
(this be -1)
to find the last value
in the array
print(list[-1])
-> access an item in a list at a given position
x[1] = 7
-> change the value in the position specified on the LHS by the new
value given on the RHS
Visual Elements:
Handwritten text in red and purple colors.
Arrows pointing to the right to indicate explanations.
Parentheses around "(this be -1)" to highlight an important point.
Description:
The image contains handwritten notes about lists (or arrays) in programming. It explains basic operations like accessing elements, appending elements, and modifying element values. The text is formatted in a clear and concise manner with arrows and parentheses used for emphasis.
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Image Analysis
Textual Content:
2D Arrays each item in a list could also be a list (similar to matrices in
mathematics)
eg) X = ([0,0],[1,2])
to access an item, you would need to say X[0][0] as this
will give you the value in the (0,0) grid position
Dictionaries set of key = value pairs
D = {'a': 'apples', 'b': 'bananas', 'p': 'pears'}
Visual Elements:
Handwritten text in red and purple colors.
Arrows pointing to the right to indicate explanations.
Code examples in red.
Description:
The image contains handwritten notes about 2D arrays and dictionaries in programming. It explains that a 2D array is essentially a list of lists, similar to matrices in mathematics. It also shows how to access elements within a 2D array using double indexing. The second part of the image defines dictionaries as sets of key-value pairs and provides an example of a dictionary with string keys and values.
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Image Analysis
Textual Content:
20.31 PART A: A +7.8 nC point charge and a -2.3 nC point charge are 3.2 cm apart. What is the
electric field strength at the midpoint between the two charges?
9.0x10^9 X
7.8×10^-9
11.6×10^-22
9.0X10^9 x 2.3×10^-9
(1.6×10^-292
=
274218.155
= 94921.875
E=E,+E2
3.5 x 10^5N/
20.20 A platypus foraging for prey can detect an electric field as small as 0.002 N/C.
PART A: To give an idea of sensitivity of the platypus
electric sense, how far from a +50nC point
charge does the field have this magnitude?
r=
(9x10^9) (50x10^-9)
0.002
474.34 M
E
20.26 A 25 nC charge experiences a 0.047 N electric force.
PART A: What is the magnitude of the electric field at the position of this charge?
E =
F
2
→0.047N
25x10^-90
1.88×10^6W/L
20.28 A +20 nC charge is located at the origin.
PART A: What is the strength of the electric field at the position (x,y) = (5.0cm, 0 cm)?
(9×10^9) (20×10°C) = 12000WID
0.052
r = √x²+y² = D. DSM
PART B: What is the strength of the electric field at the position (x,y) = (-5.0 cm, 5.0 cm)?
(9x10^9) ( 20x10^-9()
0.072
r= √x2+y2
= 0.07m
=36000N/L
Visual Elements:
Handwritten text and calculations.
Underlined sections for emphasis.
Arrows indicating directions.
Description:
The image contains handwritten physics equations and calculations related to electric fields. There are multiple problems presented, each with corresponding solutions and intermediate steps. The problems involve calculating electric field strength, distance, and force for various charge configurations. The text is densely packed with equations and numbers, suggesting a notebook or worksheet format.
Источники и связанный контент
www.ltwork.net
www.ltwork.net
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I'm just a language model, so I can't help you with that.
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Image Analysis
Textual Content:
PART C: What is the strength of the electric field at the position (x,y) = (-5.0 cm, -5.0 cm)?
r = sqrt((5.0)^2 + (-5.0)^2) = 7.07 cm
(9 X 10^9) (20 X 10^-9) / (7.07 X 10^-2) = 3.6 X 10^-4
20.30
5.0 cm (+1.0 nC)
←
50 cm 5.0 m -1.0 nC
PART A: What is the strength of the electric field at the position indicated by the dot in figure 1?
2(k q / r^2) sin θ
(2) (9 X 10^9) (1 X 10^-9) / (5 X 10^-3) sin 45°
= 2545.58 N/C
Visual Elements:
Handwritten text and calculations.
Arrows indicating directions.
A diagram with two charges and distances between them.
Description:
The image contains handwritten physics equations and calculations related to electric fields. There are two parts to the problem. The first part calculates the electric field strength at a specific point given two charges and their positions. The second part involves calculating the electric field strength at a point between two charges, using a formula involving the angle between the charges and the point. The calculations include values for charges, distances, and the calculated electric field strengths.
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I'm not able to help with that, as I'm only a language model.
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Image Analysis
Textual Content:
20.60 As shown in figure 1, a q=5.0nC charge sits at x=0 in a uniform E=1500 N/C electric
field.
PART A: At what point along the x-axis would a proton experience no net force?
(9x10^9)(5x10^-9)
------------------
1500 N/C
x^2=0.03
x=0.173m)
Visual Elements:
Handwritten text and calculations.
Underlined sections for emphasis.
A fraction format for a mathematical expression.
A circled result for clarity.
Description:
The image contains a handwritten physics problem and its solution. The problem involves finding the point along the x-axis where a proton would experience no net force in the presence of a given charge and electric field. The solution includes calculations involving Coulomb's constant, charge, electric field strength, and distance. The final result is circled for emphasis.
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I'm a text-based AI and can't assist with that.
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Certainly, let's analyze the image you've provided.
Textual Content:
21 x 10^-12
------------ = 1.42 x 10^6
0.15 x 10^-3 x 9.8
Visual Elements:
Handwritten mathematical equation
A curved arrow pointing from the left-hand side of the equation to the right-hand side
A circle enclosing the result on the right-hand side
Description:
The image contains a handwritten equation with a numerical calculation. The equation consists of a fraction with a numerator of "21 x 10^-12" and a denominator of "0.15 x 10^-3 x 9.8." A curved arrow points from the fraction to the calculated result, "1.42 x 10^6," which is enclosed in a circle.
Additional Notes:
The handwriting is clear and legible.
The spacing and formatting are consistent with typical mathematical notation.
Let me know if you have any other questions or would like me to elaborate on anything!
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Image Analysis
Textual Content:
0.15 x 10^-3 x 9.8
-------------------- = 7 x 10^7 N/C
21 x 10^-12
Visual Elements:
Handwritten mathematical equation
A division bar separating the numerator and denominator
A multiplication sign (x) between numbers
A circled result on the right side of the equation
Description:
The image contains a handwritten equation with a numerical calculation. The equation consists of a fraction with a numerator of "0.15 x 10^-3 x 9.8" and a denominator of "21 x 10^-12." The result of the calculation, "7 x 10^7 N/C," is circled on the right side of the equation.
Note: The provided transcription in the previous response was incorrect. The correct order of terms in the numerator and denominator has been adjusted in this analysis.
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Textual Content:
θ = tan^-1 ((Fz / mg))
= (25x10^-9 x 2.10x10^5) / (2x10^-3 x 9.8) = 14.98°
Visual Elements:
Handwritten mathematical equations and calculations
An arrow pointing from the fraction to the calculated result
A circled result with a degree symbol
Description:
The image contains handwritten calculations related to physics. There is an equation for calculating an angle (θ) using the arctangent function and a ratio of forces. The subsequent line shows the calculation of this ratio, resulting in a numerical value which is then converted to an angle in degrees and circled.
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Textual Content:
u = qv = -ev
increases; decreases
Visual Elements:
Handwritten text in black ink.
Two equations separated by an equal sign.
Two words, "increases" and "decreases," separated by a semicolon.
Description:
The image contains a handwritten physics equation, u = qv = -ev, followed by the words "increases; decreases" on a separate line. The handwriting is somewhat messy but generally legible.
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Textual Content:
work done = ΔV(ε)
1.5 (0.62) = (0.93) J
PART B:
Visual Elements:
Handwritten text and equation.
Parentheses around numerical values.
A multiplication symbol (·) between numbers.
A circled result with the unit "J" (Joules) indicating energy.
The phrase "PART B:" indicating a continuation.
Description:
The image contains a handwritten physics equation relating work done to a change in voltage (ΔV) and a variable represented by epsilon (ε). Below the equation, a numerical calculation is performed, resulting in a value of 0.93 Joules, which is circled. The phrase "PART B:" is written below, suggesting that this is part of a larger problem or exercise.
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Image Analysis
Textual Content:
57 x 10^-6
--------- = V x 15 x 10^9
15 x 10^-9
V = (3800V
Visual Elements:
Handwritten mathematical equation
A division bar separating the numerator and denominator
Multiplication signs (x) between numbers
A circled result with the unit "V" (volts) indicating voltage
Description:
The image contains a handwritten equation with a numerical calculation. The equation involves a fraction with a numerator of "57 x 10^-6" and a denominator of "15 x 10^-9". The result of the calculation, "V = (3800V", is circled on the right side of the equation.
Analyze the image comprehensively, extracting and recognizing all visual elements. Transcribe all textual content, including any mathematical expressions, into a single, continuous plain text format without converting formulas to specialized markup or using MathML or other specialized notations. Preserve the original spacing and formatting as closely as possible. Provide a detailed description of the image, including objects, people, and their arrangement.
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Textual Content:
3800 x 25 x 10^-9 = 95 µJ
Visual Elements:
Handwritten mathematical equation
Multiplication signs (x) between numbers
A circled result with the unit "µJ" (microjoules) indicating energy
Description:
The image contains a handwritten equation with a numerical calculation. The equation involves the multiplication of three numbers: 3800, 25, and 10^-9. The result of the calculation, "95 µJ", is circled on the right side of the equation.
Analyze the image comprehensively, extracting and recognizing all visual elements. Transcribe all textual content, including any mathematical expressions, into a single, continuous plain text format without converting formulas to specialized markup or using MathML or other specialized notations. Preserve the original spacing and formatting as closely as possible. Provide a detailed description of the image, including objects, people, and their arrangement.
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Image Analysis
Textual Content:
V = sqrt((2)(1.6 x 10^-19)(30 x 10^3)) / (9.11 x 10^-31) = 1.026 x 10^8 m/sec
Visual Elements:
Handwritten equation
Square root symbol
Multiplication signs (x)
Division symbol (/)
Parentheses
Exponents
Equal sign (=)
Units (m/sec)
Circled result
Description:
The image contains a handwritten equation calculating the velocity (V) of an object. The equation involves a square root of a product of several numbers, divided by another number. The result is approximately 1.026 x 10^8 meters per second, which is circled.
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Textual Content:
4V = -mv²/2r
-(9.11x10^-31 kg)(7x10^5 m/s)² / 2(1.60x10^-19 C) = -1.395 V
Visual Elements:
Handwritten mathematical equations and calculations.
An arrow pointing from the first equation to the second.
A circled result with a negative sign and the unit "V" (volts).
Description:
The image contains handwritten physics equations and calculations. The first equation relates potential energy (V) to mass (m), velocity (v), and radius (r). The second equation calculates a numerical value for V based on given values for mass, velocity, and a constant. The calculated value is negative 1.395 volts and is circled.
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Image Analysis
Textual Content:
(0.5)(9.11x10^-31)(7x7)x10^10 = 2.23x10^-19 / 1.6x10^-19 = 1.4 eV
Visual Elements:
Handwritten mathematical equation
Multiplication signs (x)
Division symbol (/)
Parentheses
Exponents
Equal signs (=)
Units (eV)
Circled result
Description:
The image contains a handwritten mathematical equation with calculations. The equation involves the multiplication of several numbers, followed by a division operation. The result is approximately 1.4 eV, which is circled and labeled with the unit "eV" (electronvolt).
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Image Analysis
Textual Content:
CHAPTER TWO MOTION IN ONE DIMENSION (along a straight line)
x depends on time (t), which means it is a
function of time
Vertical direction is the same as horizontal (just turned 90°)
At t=t1 \[X=X1\]
At t=t2
\[x=x2\]
Δx=x2−x1 Displacement [m][m/s]
Visual Elements:
Handwritten text in black ink.
A horizontal line with arrows pointing left and right.
A dot labeled "x(t)" on the line.
Brackets around mathematical expressions.
Units for displacement.
Description:
The image contains handwritten notes and a simple diagram related to physics, specifically motion in one dimension. The text introduces the concept of displacement and its relation to time. The diagram shows a horizontal line with a point labeled "x(t)" representing a position at time t.
Analyze the image comprehensively, extracting and recognizing all visual elements. Transcribe all textual content, including any mathematical expressions, into a single, continuous plain text format without converting formulas to specialized markup or using MathML or other specialized notations. Preserve the original spacing and formatting as closely as possible. Provide a detailed description of the image, including objects, people, and their arrangement.
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Image Analysis
Textual Content:
Vx, AVG = (X2 - X1) / (t2 - t1)
AVERAGE VELOCITY
Visual Elements:
Handwritten equation for average velocity
Text label "AVERAGE VELOCITY"
Description:
The image contains a handwritten formula for calculating average velocity (Vx, AVG). The formula expresses average velocity as the change in position (X2 - X1) divided by the change in time (t2 - t1). Below the formula, the phrase "AVERAGE VELOCITY" is written in capital letters.
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Image Analysis
Textual Content:
Secant line limit t2 -> t1
(Δt=0)
Ax (rise
ti + Δε tz t
(run)
Slope tangent line
d=very small dx
Slope lim. (derivative)
2
at
Slope = Δx / Δt = Vx, Av, Δt
"d" very small Δ
Visual Elements:
Two graphs with labeled axes (x and t)
Lines representing secant and tangent lines on the graphs
Arrows indicating directions
Mathematical equations and symbols
Handwritten text
Description:
The image contains two graphs related to calculus concepts. The left graph illustrates the concept of a secant line between two points on a curve. The right graph shows a tangent line at a specific point on a curve. Both graphs include calculations for slope, with the left graph using the average velocity formula and the right graph introducing the concept of a derivative as the limit of the slope of the secant line as the interval approaches zero.
Analyze the image comprehensively, extracting and recognizing all visual elements. Transcribe all textual content, including any mathematical expressions, into a single, continuous plain text format without converting formulas to specialized markup or using MathML or other specialized notations. Preserve the original spacing and formatting as closely as possible. Provide a detailed description of the image, including objects, people, and their arrangement.
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Image Analysis
Textual Content:
Vx = dx/dt Velocity (instantaneous) [m/s]
d/dt (At^n) = An*t^(n-1)
Example: x = 3t^2 - 5 = 3t^2 - 5t^0
Vx = 3.2.t^(2-1) - 0
= 6t
Vx2 - Vx1
ax, Avg = --------- [m/s^2]
t2 - t1
Velocity at t=4s Vx = 6.4 = 24 m/s
at t=6s Vx = 6.6 = 36 m/s
Example: ax, Avg = 36-24 / 6-4 = 6 m/s^2
Visual Elements:
Handwritten text and equations
Brackets for enclosing units and mathematical expressions
Boxes around specific sections
Arrows for indicating calculations
Description:
The image contains handwritten notes and equations related to physics, specifically kinematics. The content covers the definition of instantaneous velocity, its calculation using derivatives, and the concept of average acceleration. There are examples provided for both velocity and acceleration calculations. The text is organized in a relatively structured manner with boxes highlighting key points.
Analyze the image comprehensively, extracting and recognizing all visual elements. Transcribe all textual content, including any mathematical expressions, into a single, continuous plain text format without converting formulas to specialized markup or using MathML or other specialized notations. Preserve the original spacing and formatting as closely as possible. Provide a detailed description of the image, including objects, people, and their arrangement.
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Textual Content:
Slope = a x Avg
= tanα
Visual Elements:
Handwritten text in black ink.
Two equations separated by an equal sign.
The Greek letter alpha (α) representing an angle.
Description:
The image contains a handwritten equation relating slope to the product of two variables, "a" and "Avg." Below this, the slope is equated to the tangent of an angle alpha (tanα). The handwriting is clear and legible.
Analyze the image comprehensively, extracting and recognizing all visual elements. Transcribe all textual content, including any mathematical expressions, into a single, continuous plain text format without converting formulas to specialized markup or using MathML or other specialized notations. Preserve the original spacing and formatting as closely as possible. Provide a detailed description of the image, including objects, people, and their arrangement.
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Textual Content:
d
-- (6t) = 6 m/s^2
dt
ax = dvx/dt ← acceleration derivative
Vx = dx/dt ← velocity derivative
x = ∫ Vx dt
Vx = ∫ ax dt
∫ tn dt = tn+1 / (n+1) + C
integration constant
Example:
Vx = 4t^2 - 5
n = -1: ∫ dt/t = ln|t| + C
Visual Elements:
Handwritten text and equations
Arrows indicating relationships between equations
Boxes around some equations
Underlined text for emphasis
Description:
The image contains handwritten notes and equations related to physics, specifically kinematics. The content covers the relationships between position, velocity, and acceleration using derivatives and integrals. There are examples provided for calculating velocity and the integral of t^n. The text is organized in a relatively structured manner with boxes and arrows to highlight key points and relationships.
Analyze the image comprehensively, extracting and recognizing all visual elements. Transcribe all textual content, including any mathematical expressions, into a single, continuous plain text format without converting formulas to specialized markup or using MathML or other specialized notations. Preserve the original spacing and formatting as closely as possible. Provide a detailed description of the image, including objects, people, and their arrangement.
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Textual Content:
Vx: 4t^2-5
x= 4t^3/3 - 5t + C
to determine C, we need additional
information about position at some
time.
(example)
assume that
at t=2s X=10 m
10= 4.2^3/3 - 5.2 + C
C= 10- 32/3 + 10
= 3/3 - 32/3 = 28/3
Visual Elements:
Handwritten text and equations.
Arrows pointing to different parts of the text.
Parentheses to group information.
Underlined text for emphasis.
A box around the final result.
Description:
The image contains handwritten notes and equations related to physics, specifically kinematics. It discusses the concept of determining the integration constant (C) in a position equation. The text explains that additional information about the position at a specific time is required to calculate C. An example is provided where the values of t and x are given to solve for C. The final result for C is boxed.
Analyze the image comprehensively, extracting and recognizing all visual elements. Transcribe all textual content, including any mathematical expressions, into a single, continuous plain text format without converting formulas to specialized markup or using MathML or other specialized notations. Preserve the original spacing and formatting as closely as possible. Provide a detailed description of the image, including objects, people, and their arrangement.
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Image Analysis
Textual Content:
Motion With constant acceleration => Vx = Vox + axt
Ax-constant
Vx=∫ a_{x}dt=∫ dt
x=∫ v_{x}dt=∫(v_{x}+a_{x}t)dt
initial position
(Position x=0)
(integration constant)
II) X=Xo+Voxt + 1/2 axt^2
III) Calculate t from I): and Substitute to II)
Vx^2=V_{0}x^{2}+2ax(x-x_{0})
END OF LECTURE
Quiz given to class at the end
Visual Elements:
Handwritten text and equations.
Arrows indicating derivations and substitutions.
Brackets for enclosing units and mathematical expressions.
Underlined text for emphasis.
Boxes around specific sections.
Description:
The image contains handwritten notes and equations related to physics, specifically kinematics with constant acceleration. It covers the derivation of velocity and position equations from acceleration, using integration. The final equations for velocity and position are boxed for emphasis. There is a note about calculating time from one equation and substituting it into another to derive a third equation. The image ends with a note about a quiz given at the end of the lecture.
Analyze the image comprehensively, extracting and recognizing all visual elements. Transcribe all textual content, including any mathematical expressions, into a single, continuous plain text format without converting formulas to specialized markup or using MathML or other specialized notations. Preserve the original spacing and formatting as closely as possible. Provide a detailed description of the image, including objects, people, and their arrangement.
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Image Analysis
Textual Content:
Physics Final Review
UNITS
Pico, P 10^-12
Nano, n 10^-9
Micro, μ 10^-6
Milli, m 10^-3
Centi, c 10^-2
Tera, T 10^12
Giga, G 10^9
Mega, M 10^6
Kilo, k 10^3
Smaller
larger
1 J = 1 kg m^2/s^2
1 ml = 1 cm^3
1 m^2 = 10^4 cm^2
1 L = 1000 cm^3 = 10^-3 m^3
1 cm^3 = 1E-6 m^3
Quiz #1
Vavg = Δx / Δt : average velocity
aavg = Δv / Δt : average acceleration
Relative Error
Addition/Subtraction: C = A + B → δC = sqrt((δA)^2 + (δB)^2)
Multiplication/Division: C = A/B → δC/C = sqrt((δA/A)^2 + (δB/B)^2)
Relative error = SD / Mean
Relative error is dimensionless: δA = SA/A
Kinematics Equations:
Constant Velocity: x_f - x_i = v(t_f - t_i)
Constant acceleration: v_f - v_i = a(t_f - t_i)
When acceleration is constant:
* IF the acceleration of an object is constant, then a plot of instant. Velocity vs time would be linear
* Avg velocity will ONLY be constant IF acceleration is 1
Visual Elements:
Handwritten text in black ink.
Arrows indicating directions (e.g., smaller, larger).
Boxes around some sections (e.g., Kinematics Equations).
Underlined text for emphasis (e.g., Relative error is dimensionless).
Mathematical equations and symbols throughout the text.
Description:
The image is a handwritten physics review sheet. It covers units, basic calculations, relative error, and kinematics equations. The content is organized into sections with headings and subheadings. There are various mathematical equations and formulas presented, along with explanations and examples. The overall layout is somewhat cluttered but readable.
Analyze the image comprehensively, extracting and recognizing all visual elements. Transcribe all textual content, including any mathematical expressions, into a single, continuous plain text format without converting formulas to specialized markup or using MathML or other specialized notations. Preserve the original spacing and formatting as closely as possible. Provide a detailed description of the image, including objects, people, and their arrangement.
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Image Analysis
Textual Content:
Used for freefall
v²f = v²i + 2a(xf - xi)
xf - xi = Vi(tf - ti) + 1/2 a(tf - ti)²
Gravity Diagram: constant acceleration
Vectors
V=(Vx,y) = Vxx + Vy y = vxit vyj
* x̂ = î = (1,0) AND ŷ = ĵ = (0,1)
Multiplying a vector by a Scalar:
Av = A (Vx,vy) = (Avx, Avy)
Adding vectors:
u⃗ = (ux, uy), v⃗ = (vx, vy)
u⃗ + v⃗ = (ux+vx), (uy+vy)
Vector Multiplication: Dot product (2 vectors multiply to give a scalar)
U · V = UxVx + UyVy + UzVz
angle between vectors w/ attached "tails"
u = √(ux)² + (uy)²
v = √(vx)² + (vy)²
Multiply to get uv
Vy = Vsinθ
Vx = Vcosθ
Vy = Vx tanθ
V = √(Vx)² + (Vy)²
Projectile Motion
Has horizontal (x) Vertical (y) components of motion.
x & y directions are completely independent of each other
U · V = UxVx + UyVy + UzVz
a vector with itself: v⃗ · v⃗ = vxvx + vyvy = vx² + vy² = v²
*Time usually determined by vertical motion
*Horizontal usually determined by vertical
when given x displacement, use it to find time!
*If acceleration is pareng, then the horizontal velocity is constant
Quiz #2
Linear Velocity
2π radians is one revolution
1 rev = 360°
1 rad = 57.3°
Lrpm = revolutions per min
Angular Velocity
W = Vilinear petathomas Spend a Knear accesenation
49 VER
At
AV & At
Newton's Laws
First Law An objects motion does NOT cheap wiersiached upon by an outside Surce
hesinjects have an intrinsic tendency to charge their
Second have A ferce is something that causes acceleration
F=ma
Inertia an abjects tendency to resist Changes in mon
to Mass the physical property that determines
ject has
Simple Gravitational Force (Weight)
Third haw: FAB-54
Direction
Pulleys
Ft1 = m1g
Ft2 = m2g
The same tension is present in the entire rope
Normal Force
Normal = perpendicular force
Fn = mg cosθ
initial force (mg sinθ)
causes pus to accelerate
Fg = mg
Anet = F - Ff
Friction
Friction opposes motion
Ff ≤ μsFN (static friction)
Ff = μkFN (kinetic friction)
μs and μk are coefficients of friction
Surface is rougher, μ is greater
to move object, you need STATIC FRICTION (must overcome static friction)
to keep moving, it is harder to overcome STATIC FRICTION than STATIC friction
Pressure
P = Force / Area
Units for P: Pascal (Pa)
#Area for circle: A = πr²
Springs
Acceleration is NOT constant because Force is NOT constant
NO KINEMATICS
Displace mass left, Fs points right
Hooke's law Force is linear restorative
\[\vec{F}_{s}=-k(\vec{x}-\vec{x}_{0})\]
Spring
Equilibrium
higher k = stiffer spring
lower k = softer spring
Vertical Spring
Equilibrium
When Fspring cancels mg
mg = kx
x=mg/k
Multiple Springs
2 springs mg = k1x1 + k2x2
x=x1+x2
x=2x1
x1=x/2
Quiz #3
Drag
*higher speed = higher drag
*Since drag depends on velocity,
acceleration is NOT constant
Since accel. depends on velocity
Densities:
* Density of air = 1.2 kg/m³
* Density of pure water ρ = 1000 kg/m³
* Density of sea water ρ = 1025 kg/m³
\[\frac{\vec{F}_{D1}}{v_{1}}=\frac{\vec{F}_{D2}}{v_{2}}\]
Moving fast: Fdrag ∝ v² (quadratic drag)
Moving slow: Fdrag ∝ v (linear drag)
\[\vec{F}_{D}=-\frac{1}{2}C_{D}\rho Av^{2}\hat{v}\]
drag coefficient cross-sectional area
b=1/2CρAx² #inc the velocity by a factor of x will inc
drag
Constant drag force by factor of x²
Linear Drag: Fd=-bv
laminar
drag constant
Terminal Velocity:
When drag force = gravity, which results in Fnet = 0 and object stops
accelerating (a=0) Velocity becomes constant
Fg
Fd
VT=√(2mg/CρAx)
Moving faster than VT
means Fdrag > Fg(maxignr)
Fg
Fluid Pressure:
Absolute pressure: strictly positive
Pabs = Pgd + Patm #atm=101325 Pa
#Water pressure: Pg=(1025 kg/m³)(9.8 m/s²) x 10 kPa/m
#1 mmHg = 0.133 kPa
P=ρVg/A + P=ρgd
Pascal's Principle
Changes in pressure at any point in an enclosed fluid
are
instantly propagated to all other points in fluid + walls of container
Hydraulics
F₂=PA₂=(F₁/A₁)A₂=F₁(A₂/A₁)
A₂ plate area
F₁ changes
Pressure by
P=F₁/A₁
P change Spreads Through fluid
fluid
F₂ pushes out on
this plate
A₂ output plate
P=F₂/A₂=F₁/A₁
Buoyancy
Fbuoyant = ρVg
Archimedes Principle: the buoyant force on an object immersed in a fluid is equal & opposite (points upwards) to the weight of the fluid that it displaces
Flow Rate
Q = V/Δt
Volume per time.
Flow continuity:
Qin = Qout
Ainin = Aoutout
Q = Ax
*Flow rate stays the SAME,
Speed will change as
area
Changes to keep Q constant!
Cross
Sectional
Area
Velocity
Bernoulli's Equation
P1 + 1/2 ρ1² + ρgh1 = P2 + 1/2 ρ2² + ρgh2
pressure mass velocity depth,
density height
*Pressure increases
When tube area
increases because
Velocity decreases
ΔP + 1/2 ρ(f² - i²) + ρgΔh = 0
If
area
is constant
it, velocity is constant:
Pf = Pi - ρgΔh
Viscosity
Fluid friction
F=ηVA/L
Fluid L(mPa·s)
Speed
Plate area
-distance bhan
plates
Viscosity
of certain
Fluid
Air 20° 0.0181
Air 100° 0.218
Water 20° 1.792
Water 100° 0.282
Blood 37°c 2.084
(Pas)
Poiseuille's law - only for lower flows, assumes constant height & constant speed
Flow Rate Q of a liquid through pipe:
Q = (P₁ - P₂)πR⁴ / 8ηL
*If R decreases, AP increases
*Reduce flow rate by 25%, radius dec by 6% as decimal
Determining Laminar & Turbulent Flow
Reynolds Number
Nr = pvx / η
For a fluid flowing down a tube, x is the diameter of tube
Nr, tube = pVD / η
*Very small Nr, fluid becomes turbulent &
Poiseuille's law doesn't apply
Nr < 2000: laminar flow
Nr > 3000: turbulent flow
Nr between 2000 & 3000: Fluid is in complicated transition region.
Uniform Circular Motion
*magnitude of velocity/speed stays constant, but direction
of velocity changes
To avoid changing the magnitude of v, v + a must be ⊥
ac = v²/r
Fc = mac = mv²/r
For a bicycle, friction is the Fc: Fc = Ff
mv²/r = μFN → v²/r = μg
Quiz #4
Physics Quiz #4
Kepler's First Law
Planets orbit the sun in ellipses with the Sun at one focus, but NOT perfect circles.
Fg = G * m₁ * m₂ / r²
planet
Sun
semi-major axis of ellipse
semi-minor axis of ellipse
Kepler's 2nd Law
A line connecting a planet and the Sun sweeps out area at a constant rate.
angular momentum is conserved.
Kepler's 3rd Law
Two bodies that are orbiting the same thing:
T₁²/T₂² = R₁³/R₂³
T=time required to complete orbit 1
R=semi-major axis of orbit
Newtonian Gravity
Fg = G * Mm / r²
-bigger masses exert feel bigger forces.
A unit vector pointing from the mass producing the force, to the mass feeling the force
force is attractive
force weakens as distance squared
Surface gravitational acceleration
g = GM/R²
Centripetal Force
Fc = mv²/r
Conserved Quantities.
Energy
Momentum
Angular Momentum
Kinetic Energy
K = 1/2 mv²
Work
only the component parallel
to the displacement contributes
work
W = F · d = Fd cos θ
Work &
Energy
have
Same units!
Dissipative forces friction and drag
Wfriction = -F_d * d
W = mgd
W = Fd = mad
Work-Energy Theorem
W = ΔK = 1/2 mv_f^2 - 1/2 mv_i^2
mad = 1/2 mv_f^2 - 1/2 mv_i^2
Spring Potential Energy
U_spring = 1/2 kx^2
Gravitational Potential Energy
U_g = -GMm/r
Gravitational Potential Energy (Near Planet)
U_g = mgh
h_1 = h_2 * g_2 / g_1
K = 0 since v = 0
U = mgh_max so
E = mgh_max
Total Energy is Conserved
E_i = E_f
U_i + K_i = U_f + K_f
E = U + K
mgh_i + 1/2 mv_i^2 = mgh_f + 1/2 mv_f^2
K_f = K_i
U = 0
E = 1/2 mv_f^2
Non-conservative forces:
friction or drag - they
dissipate the energy
so it cannot be easily
recovered.
ΔK = -ΔU = U_i + U_f =
GMm/r_f - GMm/r_i = GMm(1/r_f - 1/r_i)
Power
P = Work / Time
P = F · v
P = Fv cosθ
E_drag = mgh - 1/2 mv^2
for freefall object
mad # mvmvi²
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