11 - Sph3u - Physics

Rounding

when at exactly .5 (not like .51) round to the even. so 2.5 > 2 & 3.5 > 4

Guess format (how to answer problems)

• Given: List the known information using appropriate symbols and numerical values, including units.
• Unknown: State the quantity or quantities you need to find.
• Equations: Identify the relevant equations or principles that relate the given and unknown quantities.
• Solve: Manipulate the equations algebraically and substitute the given values to solve for the unknown(s). Show all steps and calculations clearly.
• Sentence: Write a clear and concise sentence that answers the question posed in the problem, including the numerical value and correct units. Consider significant figures.

Example:

• Given: A car travels 150 kilometers in 2 hours.
• Unknown: Average speed of the car.
• Equations: speed = distance / time
• Solve: speed = 150 km / 2 h = 75 km/h
• Sentence: The average speed of the car is 75 kilometers per hour.

Significant Figures

• Nonzero Digits: Always significant.
  • Ex: 211.8 has four significant figures.
• Zeros Between Nonzeros: Always significant.
  • Ex: 20,007 has five significant figures.
• Leading Zeros: Never significant.
  • Ex: 0.0085 has two significant figures.
• Trailing Zeros (Whole Number with Decimal): Always significant.
  • Ex: 320. has three significant figures.
• • Trailing Zeros (Whole Number without Decimal): Never significant.
    • Ex: 32000 has two significant figures.
• Trailing Zeros (Decimal Number): Always significant.
  • Ex: 12.000 has five significant figures.
• Exact & Irrational Numbers: Infinite significant figures.
  • Ex: 1 meter = 100 centimeters, e, π
• Scientific Notation: Only consider the coefficient (A) for significant figures.
  • Ex:
    • 4.5 × 10³ has two significant figures.
    • 4.50 × 10³ has three significant figures.
    • 4.500 × 10³ has four significant figures.

Calculating with Sig Figs

Multiplying/Dividing

When multiplying or dividing the answer is limited to the lower number of sig digits of the terms being multiplied or divided
$77.8km/h\cdot 0.8967h\approx 69.8km$
$3500mol÷1.2L\approx 2900mol/L$

Adding/Subtracting

When Adding and subtracting values, the answer is always rounded to the same number of digits to the right of the decimal point as that of the least precise quantity.
$11.7cm+3.29cm+0.542cm=15.532=\text{15.5cm}$
$3.296+1.003-30.36+36.210=10.149=\text{10.15}$

Electricity

Electric Current

Electric current (I):

Measure of the rate of electron flow at a given point in a circuit, measured in amperes (A)

• I = Q/t, therefore 1A = 1C/s (… A = C/s)

A current of 1 ampere means $6.2\ast 10^{18}$ electrons are flowing through there very second

Direct Current (DC)

The movement of electrons in only one direction
$I=\frac{Q}{\Delta t}$

Alternating Current (AC)

The Movement of electrons in two directions

Ammeter

Electrical device that measures electric current (must connect in series)

Electric Potential Difference (Voltage)

Electric Potential Difference

The amount of electric potential energy associated with the charges

Electric Potential difference (V)

The change if electric potential energy associated with charges at two different points in a circuit
Q may be referred to as C
$V=\frac{\Delta E}{Q}$

Voltmeter

an electric device that measures electric potential difference (must be connected in parallel)

Resistance

Work

What is the resistance of a toaster with a voltage of 120v and a current of 6.5A
Given V & A
Unknown R
Equasion = $R=v/I$
Solve: $\frac{120V}{6.5A}=18\Omega$

What is the resistance of a car starter with 450A of current and 12V
Solve: $\frac{12}{450}\approx 2.7\ast 10^{-2}$

Electrical Resistance (R)

Ability of a material to oppose the flow of electric current; measured in ohms ($\Omega$)

Ohms law

$R=\frac{V}{I}therefore\Omega =\frac{V}{A}$

Factors that effect resistance

• type of material
• cross-sectional area
  • How long*wide the material is
• length
• temperature
  • hotter = more thermal energy

Resistors

A device that converts electrical energy to heat by reducing the flow of electric current

Kirchhoff’s law

Kirchhoff’s law	Series Circuit	Parallel
	$V_{series}=V_1+V_2+V_3+...$	$V_{parallel}=V_1=V_2=V_3=...$
	$I_{series}=I_1=I_2=I_3=...$	$I_{parallel}=I_1+I_2+I_3+...$
	$R_{series}=R_1+R_2+R_3+...$	$\frac{1}{R_{parallel}}=\frac{1}{R_1}=\frac{1}{R_2}=\frac{1}{R_3}+...$

Ohms Law

$I:Current$
$V:Voltage$
$R:Resistance$

Units

Quantity	Quanitiy symbol	Unit	Unit symbol
Mass	m	Gram	g
	I	Ampere	A
	V	Volts	V
	R	Ohms	$\Omega$

Tips and Tricks

LEDS

• Long leg = Positive; Short = Negative

Definitions

Q

Q represents electric charge, it is used in Coulomb’s law

J

Joules
A format of energy (E)
In electronics, it’s the same amount of energy, in electrical units. One joule is one watt of power, applied for one second (a watt-second); or a coulomb of electrical charge raised to a potential of one volt.

Magnetism

Flows from North to South\

Lenz’s Law

If a changing magnetic field induces a current in a coil, the electric current is in such a direction that its own magnetic field opposes the change that produced it

Conversions

$$\begin{array}{rl}& N=Numofcoils\\ & P:primary;S:Secondary\\ & \\ & \frac{V_P}{V_S}=\frac{N_P}{N_S}\\ & \\ & \frac{V_P}{V_S}=\frac{I_S}{I_P}\end{array}$$

Kinematics

Equations

$$\begin{array}{rl}& {\vec{v}}_f={\vec{v}}_i+\vec{a}\Delta t-\text{Find final velocity}\\ & \Delta \vec{d}={\vec{v}}_i\Delta t+\frac{1}{2}\vec{a}\Delta t^2-\text{Find displacement (given inital)}\\ & \Delta \vec{d}={\vec{v}}_f\Delta t-\frac{1}{2}\vec{a}\Delta t^2-\text{Find displacement (given final)}\\ & \Delta \vec{d}=\frac{1}{2}({\vec{v}}_i+{\vec{v}}_f)\Delta t-\text{Find displacement (without acceleration)}\\ & {\vec{v}}_f^2={\vec{v}}_i^2+2\vec{a}\Delta \vec{d}-\text{Find final velocity (without time)}\\ & \Delta t=\frac{2\vec{d}}{({\vec{v}}_f+{\vec{v}}_i)}-\text{Find time given final v \& a \& d}\end{array}$$

Scaler

Quantity that only has magnitude

Examples

Speed of 34km/h
4 hours

Vector

Quantity that has magnitude and direction

Examples

Velocity of 34km/h [E]
Displacement of 5km [S]

How to draw one

1. Draw a line
2. Draw a tip (arrow) pointing in the direction of choice

Adding/Sub

You quite literally add them up. if it’s negative, add the negative.

Displacement

the distance and direction of an object from a particular point in a straight line

e.g. The child was displaced 5km east

$$\Delta {\vec{d}}_T=\Delta {\vec{d}}_{\text{final}}-\Delta {\vec{d}}_{\text{initial}}$$

Tips

you must always have number, unit and direction
e.g. 5.2km [N]

Uniform vs Non-Uniform

Uniform - Speed is equal and in a straight line

Non-Uniform - Speed is non-equal or not in a straight line

Speed & Velocity

Average speed

($V_{av}$) the total distance divided by total time taken

$$v_{av}=\frac{\Delta d}{\Delta t}-\text{Units: m/s}$$

Average Velocity

${\vec{v}}_{av}$ the total displacement divided by the total time for that displacement
$a=\frac{v_f^2-v_i^2}{2\Delta d}$

$${\vec{v}}_{av}=\frac{\Delta \vec{d}}{\Delta t}-\text{Units: m/s}$$

Acceleration

Acceleration (${\vec{a}}_{av}$) - rate of change of velocity divided by time

$${\vec{a}}_{av}=\frac{\Delta \vec{v}}{\Delta t}-{\text{Units: m/s}}^2$$

Positive: The direction of acceleration is the same as current velocity
Negative: It’s counteraction the current direction of motion (deceleration)

Gravity

$g=9.8m/s^2$

Calculating complex displacements

$$\begin{array}{rl}& 90°apart\\ & \Delta {\vec{d}}_T=\sqrt{\Delta d_1^2+\Delta d_2^2}|\tan \theta =\frac{\Delta d_2}{\Delta d_1}\end{array}$$

Forces

Inertia

The ability of an object to resist change in motion
Once something is going, it doesn’t want to stop (and when it stopped it doesn’t want to start)

Mass vs weight

• Mass
  • The amount of matter in an object
• Weight
  • The amount gravitational attraction that is acting on an object

Common forces:

$$\begin{array}{rl}& Gravity({\vec{F}}_g)\\ & Normal({\vec{F}}_N)-\text{perp against a surfance}\\ & Tension({\vec{F}}_T)\\ & Applied({\vec{F}}_a)\\ & Friction({\vec{F}}_f)\end{array}$$

Net forces

The sum of all of the forces acting on an object.
5N of force to left, 10N to right. Net force: 5N right

Newtons (N)

$N=kg\ast m/s^2$

Force of gravity

Measured in Newtons (N)
${\vec{f}}_g=m\vec{g}$ (mass (Kg) * $\vec{g}$)

Gravitational field strength

$$\begin{array}{rl}& g=\frac{Gm}{r^2}\\ & G=\text{Universal Gratational Constant}\\ & (6.67x10^{-11}{m}^3/kg\cdot s^2)\end{array}$$

Fundamental forces

1. Strong Nuclear
   1. Forces holding an atom together
2. Weak Nuclear
   1. Between molecules (eg. h2o to h2o) or DNA’s helix shape
3. Electromagnetic
   1. Caused by electric charges
4. Gravitational

Newton’s Laws

Second Law

$${\vec{F}}_{net}=m\vec{a}$$

Energy & Work

• Energy: The ability to do work
• Work: The energy transferred to an object by a applied force over a measured distance

Formulas

W → Work (Joules)
F → Force applied (N)
$\Delta d$ → displacement (m)

$$\begin{array}{rl}& W=\vec{F}\Delta \vec{d}\\ & W=F\Delta d(\cos \Theta )\\ & (\Theta \text{being angle between force applied and the displacement)}\\ & \\ & 1J=1N\cdot m\\ & =kg\ast \frac{m^2}{s^2}\\ & \\ & E_k=\frac{1}{2}m{\vec{v}}^2\\ & \\ & W=1\frac{J}{S}\end{array}$$

Work

Case 1: Positive work

Force applied and displacement are in the same direction

• E.g. lifting a bag

Case 2: Negative work

If force applied and displacement are in opposing directions

• E.g. a pully or a rocket running

Case 3: Work against gravity

If the force applied and displacement are both vertical and no acceleration occurs the work against gravity is positive

• E.g. an elevator

Case 4: Zero work

If force and displacement are perpendicular

Work Energy Princible

The net amount of mechanical work equals the object’s change in kinetic energy

$$\begin{array}{rl}& W=\Delta E_k\\ & W_{net}=E_{K_i}-E_{k_i}\\ & W_{net}=\frac{m{v}_f^2}{2}-\frac{m{v}_i^2}{2}\end{array}$$

Thermal Energy

$$\begin{array}{rl}& Q=mc\Delta T\\ & Q_{released}+Q_{absorbed}=0\end{array}$$

Water Capacity

$4.18\cdot 10^{3}J/Kg/C°$

Homework

2024-07-05

1

Since volts add up in series
$120-60-12=\mathrm{48...}{V}_2=48v$

2

Since Amps add up in a parallel circuit…
$2.0A-1.1A-0.3A=0.6A$

3

Since amps split and volts are the same, we can find that all V’s are $2.4V$
$V_T=V_1=2.4V$
$0.71A-0.28A-0.17A=0.26A...I_3=0.26A$

4

This one’s weird but it’s still parallel and it adds up;
$9.5A-2.2A-3.1A=I_2=4.2A$

5

This one is series so we need to find $V_2$
$36V-12V-13V=V_2=11V$

6

Okay so we need to find a couple of Amp messurements
since the amps split where the wires split we can assume $I_2\&I_3$ are both $1.1A$
$I_1$ has never split so it must be the same at $I_I$ so if we add it up
$1.1A+2.3A=3.4A=I_1$
$3.4A-1.1A=2.3A=I_3$

7

A)
Since we have all values.. we can just add them up
$2.2+\mathrm{3.3.}+1.1=6.6v=V_t$
B)
This is also series so we know the voltage adds up, if the total if 9v we can just subtract the values we know, so..
$9V-4.1V-2.1V=2.8V=\Delta V_2$

2024-07-08

Simple Series And Parallel Circuits.pdf - Google Drive

1

	V	I	R
T	6	0.2	30
1	1	0.2	5
2	5	0.2	25

2

	V	I	R
T	50	7.5	6.7
1	50	4.2	12
2	50	3.3	15

3

Type: Series

	V	I	R
T	6	5	1.2
1	3	5	0.6
2	1	5	0.2
3	2	5	0.4

4

Type: Parallel

	V	I	R
T	24	8.4	2.85
1	24	1.2	20
2	24	2.4	10
3	24	4.8	5

5

Type: Series

	V	I	R
T	3	0.086	35
1	1.29	0.086	15
2	1.72	0.086	20

6

Type: Series

	V	I	R
T	6	3	2
1	3	3	1
2	1.5	3	0.5
3	1.5	3	0.5

7

Type: Parallel

	V	I	R
T	120	2.2	54.5
1	120	1	120
2	120	1.2	100

8

Type: Parallel

	V	I	R
T	12	0.96	12.5
1	12	0.6	20
2	12	0.24	50
3	12	0.12	100

Extra hard question (Combo circuit) 🥰

Combination Circuit Example.pdf - Google Drive

	V	I	R
T	24	2	12
1	16	2	8
2 (parallel)	8	2	4

2024-07-09

1

Use Conversions

$$\begin{gathered} \text{Cross multiply} \\ \frac{240}{(\frac{550}{110})}=48V \end{gathered}$$

2

Use Conversions #2

$$240/12\cdot 0.15=3A$$

2024-07-11

1

use #2 from Equations
Given: initial (0 speed) time (15s) acceleration ($2.0m/s^2$)
Unknown: displacement
equation: #2
solve: $1/2(2.0)15^2$
Sentence: The care was displaced $2.2\ast 10^2$m [N] away

2

Given: $v_1$, $v_2$ and distance
Unknown: acceleration
equation: Acceleration formula
solve: $31.25/(2(50))=0.3125=0.31m/s^2$

3

Todo: Figure out how to do this

Given: distance, acceleration
Unknown: time(s)
Equation: #2

$$\begin{array}{rl}\text{Cart A}& \\ 1000& =20t+0.5(0)\Delta t^2-\text{why does removing ˆ2 make more sense}\\ & =1000=20t\\ & =\frac{1000}{20}=t\\ & =50=t\\ & \\ \text{Cart B}& \\ 100& =0\Delta t+0.5(0.333m/s^2)\Delta t^2\\ & =0.5(0.333m/s^2)\Delta t^2\\ & =\frac{\sqrt{0.333_{m/s}\Delta t^2}}{2}\end{array}$$

Solve:
Sentence:

2024-07-19

Friction

A 75 kg hockey player glides across the ice on his skates with steel blades. The coefficient of kinetic friction is 0.01. What is the magnitude of the force of friction acting on the skater.

Since the surface is horizontal $F_{normal}=weight$
$W=75kg\cdot 9.8m/s^2$
$W=735.75N$
Force of friction $F_{friction}=\mu k\cdot N$
$F_{friction}=0.01\cdot 735.75$
$F_{friction}=7.3575$

Forces 1

$m=0.5kgT=2s{V}_i=3m/sa=1.2m/s^2$
a)

$$\begin{array}{rl}& {\vec{F}}_{net}=m\vec{a}\\ & ={\vec{F}}_a-{\vec{F}}_f=m\vec{a}\\ & =\overrightarrow{F_a}=m\vec{a}+{\vec{F}}_f\\ & =\vec{F}a=(0.50kg)(1.2m/s^2)+0N\\ & =0.6N\\ & =0.60N\end{array}$$

Therefor the string exerts a force of 0.60 Newtons on the string.
b) displacement

$$\begin{array}{rl}& \Delta \vec{d}={\vec{v}}_i\Delta t+0.5a\Delta t\\ & =(3m/s)(2)+0.5(1.2m/s^2)(2^2)\\ & =6m+2.4m\\ & =8.4m[right]\end{array}$$

Therefore the displacement of the cart is 8.4m [right]
c) Mechanical work

$$\begin{array}{rl}& W=\vec{F}\Delta \vec{d}\\ & =(0.6N)(8.4m)\\ & =5.04J\end{array}$$

… the mechanical work done is 5.0J

Yayyy

Energy & Work

$$\begin{array}{rl}& W=F\Delta d(\cos \Theta )\\ & =(125N)(12.0)(\cos 40°)\\ & =1149.066665J\\ & =1150J\\ & \therefore \text{The person has exerted 1150J of mechanical work}\end{array}$$

Kinetic energy

1

$$\begin{array}{rl}& \text{m = 70kg; v = 12m/s; d = 100m}\\ & E_k=\frac{1}{2}m{\vec{v}}^2\\ & =\frac{1}{2}(70Kg)(12m/s{)}^2\\ & =5.0\ast 10^{3}J\\ & \therefore \text{the kinetic energy of the athlete is 5000J}\end{array}$$

2

(note: convert grams to kg)

$$\begin{array}{rl}& E_k=0.5m{v}^2\\ & =30J=0.5\frac{0.15v^2}{2}\\ & v=\sqrt{\frac{2(30J)}{0.15kg}}\\ & =\sqrt{600}\\ & =20\\ & \therefore \text{The speed of the bird is 20m/s }\end{array}$$

2024-07-24

Temp to mass

Given: Aluminium $9.2\ast 10^3$, energy absorbed
Unknown: Mass
Equation: $Q=mcT$
Solve:

$$\begin{array}{rl}& 1.0\ast 10^{4}J=m(9.2\cdot 10^{2}J/kg\cdot t)(5.0°C)\\ & =m(4600)\\ & \\ & m=\frac{4600}{10000}kg\\ & =2.1739130435\\ & =2.2kg\end{array}$$

Sentence: therefore the mass of the block is 2.2kg

Thermal Energy

$$\begin{array}{rl}& T_{fe}=80.0°C{V}_{H2O}=100mL;T_f=22°C{T}_{h2o}=20.0°C{c}_{H2O}=4.18\ast 10^3\\ & \end{array}$$

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FORMAT

Given:
Unknown:
Equation:
Solve:
Sentence:

WORk

###### Title
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