Physics 1st Quarter PDF

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St. Mary's School

Basilio, Andrei Martin M.

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physics measurement significant figures science

Summary

This document provides notes on measurement in physics, including significant figures and calculations. It covers expressing final answers and unit conversions. The concepts are relevant to secondary school physics.

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MEASUREMENT IN PHYSICS 1 ℎ𝑟 mins to hrs: 450.0 𝑚𝑖𝑛𝑠 × = 7.5 ℎ𝑟𝑠 SIGNIFICANT FIGURES 6...

MEASUREMENT IN PHYSICS 1 ℎ𝑟 mins to hrs: 450.0 𝑚𝑖𝑛𝑠 × = 7.5 ℎ𝑟𝑠 SIGNIFICANT FIGURES 60 𝑚𝑖𝑛𝑠 Final answer: 7.500 hrs  Significant Figures (SF or SigFig) are numbers in a 12 𝑖𝑛 measurement that are known with some degree ft to in: 18.0 𝑓𝑡 × = 216 𝑖𝑛 1 𝑓𝑡 confidence plus the last digit which is an estimation Final answer: 2.16 × 102 𝑖𝑛 or approximation. 24 ℎ𝑟𝑠 days to secs: 35.0 𝑑𝑎𝑦𝑠 × = 840 ℎ𝑟𝑠  The number of SF increases as the measuring material 1 𝑑𝑎𝑦 3600 𝑠𝑒𝑐𝑠 possess higher sensitivity. 840 ℎ𝑟𝑠 × = 3,024,000 𝑠𝑒𝑐𝑠 1 ℎ𝑟 - higher sensitivity provides more significant values Final answer: 3.02 × 106 𝑠𝑒𝑐𝑠 - the higher the sensitivity of the measuring material, gal to pt: 12.5 𝑔𝑎𝑙 × 3.785 𝐿 = 47.31254 𝐿 the higher the confidence of the measured value. 1 𝑔𝑎𝑙 1 𝑝𝑡 RULES IN DETERMINING SIGNIFICANT 47.31254 𝐿 × = 100.0264270613 𝑝𝑡 0.473 𝐿 VALUES Final answer: 1.00 × 102 𝑝𝑡  All non-zero digits are significant UNCERTAINTY IN MEASUREMENTS Ex: 745 (3 SF)  Measurement is the process of comparing an unknown  Trailing zeroes are significant if the number has a quantity to a standard unit. decimal point  Physical Quantities are characteristics or properties of Ex: 5200. (4 SF) an object that can be calculated or measured from other  Zeroes between non-zero digits are significant measurements. Ex: 70034 (5 SF) TWO TYPES OF PHYSICAL QUANTITIES  Leading zeroes are not significant  Fundamental Quantities are measured quantities Ex: 0.0034 (2 SF) and direct. EXPRESSING FINAL ANSWER WITH THE  Derived Quantities are calculated quantities and CORRECT NUMBER OF SIGNIFICANT FIGURES computed. Addition and Subtraction – follow the number of SF of  Accuracy refers to how close a measurement is to the the given with the least number of SF after the decimal true or accepted value. point  Precision refers to how close measurements of the 1. 0.0250 cm + 1.30 cm = 1.325 same item are to each other. Final answer: 1.33 cm (3 SF) TYPES OF ERROR 2. 2.201 cm – 1.20 cm = 1.001 Random Error Final answer: 1.00 cm (3 SF) - the uncertainty in measurement that are caused by Multiplication and Division – follow the number of SF of extraneous factors the given with the least number of SF - affects the precision of the measured values 1. 87475.20 cm × 0.0030 cm = 262.4256 cm2 - uncontrollable Final answer: 2.6 × 102 cm2 (2 SF) Systematic Error 2. 87475.20 cm2 ÷ 0.0030 cm = 29,158,400 cm - caused by not properly calibrated measuring devices and Final answer: 2.9 × 107 cm (2 SF) they are related to inaccuracy of measurement - affects the accuracy SCIENTIFIC NOTATION - controllable  Scientific Notation is a neat and quick way of writing very large or very small numbers How to resolve conflicts brought by both errors?  First coefficient is always single digit  Conducting experiment with several trials using  Moving the decimal point to the LEFT = exponent is different measuring tools will help you resolve POSITIVE conflicts brought by both errors. Ex: 154,055. 94 → 1.54 × 105 SCALAR AND VECTORS  Moving the decimal point to the RIGHT = exponent  Scalar Quantities are quantities that has given is NEGATIVE magnitude only. Ex: 0.000623 → 6.23 × 10−4 Ex: speed, distance, time CONVERSION OF UNITS  Vector Quantities are quantities that have both magnitude and direction given. Ex: displacement, velocity, acceleration SCALAR SUM  MATHEMATICAL o Add the given quantities as ordinary positive numbers.  GRAPHICAL o A line segment is used to represent quantities. FACTOR-LABEL METHOD 𝑑𝑒𝑠𝑖𝑟𝑒𝑑 𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑚𝑒𝑛𝑡 VECTOR SUM 𝑔𝑖𝑣𝑒𝑛 𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑚𝑒𝑛𝑡 ×  MATHEMATICAL 𝑔𝑖𝑣𝑒𝑛 𝑢𝑛𝑖𝑡 Examples: o Add the given quantities as integers, either 1 𝑖𝑛 positive or negative depending on the cm to in: 25.0 𝑐𝑚 × = 9.842519685 𝑖𝑛 direction. 2.54 𝑐𝑚 Final answer: 9.84 in o North and East are POSITIVE while South and West are NEGATIVE. GENERAL PHYSICS 1 QUARTER 1 Basilio, Andrei Martin M. (12-STEM 02)  GRAPHICAL VECTOR SUM (GRAPHICAL) o A ray line is used to represent quantities. Examples: SCALAR SUM (MATHEMATICAL) SCALAR SUM (GRAPHICAL) VECTOR SUM (MATHEMATICAL) REWRITING GIVEN VECTORS IN COMPONENT FORM Given problem: The horizontal movement along the runway is 2000.o meters eastward, then it takes off at an angle of 10 from the runway and travelled along this to a distance of 970,000.0 meters. VERTICAL COMPONENT: The vertical component of the motion along the runway is ZERO because it lies directly along the horizontal which means that the x- component is 2000.0 meters east. DIAGONAL MOTION: The diagonal motion is at 10 from the runway. GENERAL PHYSICS 1 QUARTER 1 Basilio, Andrei Martin M. (12-STEM 02) MAGNITUDE OF THE X AND Y COMPONENTS: TOTAL X AND Y COMPONENTS:  Constant velocity  Constant slope  No acceleration VELOCITY vs TIME GRAPH MAGNITUDE OF THE RESULANT VECTOR:  Constant velocity DIRECTION OF THE RESULTANT VECTOR:  No acceleration Final Answer: 971,969.7 m, 9 58’ 46.3” North of East KINEMATICS (MOTION IN A STRAIGHT LINE) DESCRIPTORS OF MOTION  Distance (d) – m, km  Velocity constantly changes over time  Displacement (𝑑⃗) – m, N ; km, W  Constant acceleration  Speed (V) – m/s  Velocity (𝑉⃗⃗ ) – m/s, N ; km/h, W DISPLACEMENT AS AREA UNDER VELOCITY  Acceleration (𝑎⃗) – m/s2 vs TIME CURVES DISPLACEMENT vs TIME GRAPH 1. G: l = 6.0 s, w = 40.0 m/s, E U: 𝑑⃗ = ?  No motion takes place F: 𝑑⃗ = 𝑙 × 𝑤 S: 𝑑⃗ = (6.0 s) (40.0 m/s) 𝑑⃗ = 240 m, East ⃗ A: 𝑑 = 2.4 x 102 m, East GENERAL PHYSICS 1 QUARTER 1 Basilio, Andrei Martin M. (12-STEM 02) 𝑉⃗⃗𝑓 = 19.0 m/s + 10.0 m/s 𝑉⃗⃗𝑓 = 29.0 m/s, East ⃗⃗𝑓 = 29.0 m/s, East A: 𝑉 RELATIVE VELOCITIES AND FRAME OF REFERENCE 2. G: b = 4.0 s, h1 = 20.0 m/s, E, h2 = 50.0 m/s, E U: 𝑑⃗ = ? 1 F: 𝑑⃗ = 𝑏(ℎ1 + ℎ2 ) 2 1 RELATIVE VELOCITIES AND THE MEDIUM AT S: 𝑑⃗ = (4.0 s) (20 m/s + 50 m/s) WHICH OBJECTS MOVE 2 𝑑⃗ = 140 m, East A: 𝑑⃗ =1.4 x 102 m, East CONSTANT ACCELERATION 3. G: b = 6.0 s, h = 50 m/s, E AVERAGE VELOCITY U: 𝑑⃗ = ? 1 F: 𝑑⃗= 𝑏ℎ 2 1 S: 𝑑⃗ = (6.0 s) (50.0 m/s) 2 𝑑⃗ = 150 m, East A: 𝑑⃗ = 1.5 x 102 m, East TOTAL DISPLACEMENT VELOCITY AS AREA UNDER ACCELERATION vs TIME CURVES CONSTANT VELOCITY vs CONSTANT ACCELERATION  If horizontal, both can happen but in different span of time.  If vertical, only acceleration can be constant due to gravitational pull. G: for rectangle: l = 5.0 m/s, E, w = 3.0 s for triangle 1: b = 2.0 s, h = 5.0 m/s, E FREE FALL for triangle 2: b = 1.0s, h = -2.0 m/s  A free fall occurs if the motion is solely affected by ⃗⃗𝑖 = 10.0 m/s, E 𝑉 gravity. U: ∆ 𝑉⃗⃗ = ?, 𝑉 ⃗⃗𝑓 = ?  The initial velocity is always zero. F: ∆ 𝑉⃗⃗ = [(𝑙 × 𝑤) + (1 𝑏ℎ) + (1 𝑏ℎ)] , 2 2 ⃗⃗𝑓 = ∆ → + → 𝑉 𝑉 𝑉𝑖 1 ⃗⃗ = {[(5.0 m/s) (3.0 s)] + [ (2.0 s) (5.0 m/s)] + S: ∆ 𝑉 2 1 [ (1.0 s) (-2.0 m/s)]} 2 ∆𝑉⃗⃗ = 19.0 m/s, East GENERAL PHYSICS 1 QUARTER 1 Basilio, Andrei Martin M. (12-STEM 02) CASE 2: PROJECTILE RELEASED DIAGONALLY FORMULA TO BE USED: CIRCULAR MOTION OBJECT THROWN UPWARD DEFINITION OF TERMS:  The final velocity at the peak is always zero.  Tangent is away from the circle along a path tangent to any point on it.  Radial is directed towards the center of the circle along the radius.  Uniform means always remaining the same in all cases.  Non-Uniform means not uniform and is varying. CENTRIPETAL FORCE  Centripetal Force is a center-seeking force.  Centripetal Acceleration keep holds the object from skidding out of the path. PROJECTILE MOTION DEFINITION OF TERMS: TANGENTIAL VELOCITY  Peak is the location at which the maximum y-  Tangential Velocity is perpendicular to the centripetal displacement measures. force.  Trajectory is the path taken by the projectile.  Once the string is cut, the object will move tangential  Time of Flight is the time taken by the projectile from to the circumference of the circle’s path. initial location to final location.  An object moving in a uniform circular motion cannot  Range is the maximum x-displacement. attain a constant velocity because of the change in  Independent means free from influence and cause no position. effect.  Motion along x-axis is always constant. LAWS OF MOTION AND THEIR APPLICATIONS  Motion along y-axis uniformly accelerates due to DEFINITION OF TERMS: gravitational pull.  Mass is the amount of matter present in a body.  Initial velocity along y is zero.  Equilibrium is the state of rest where there is no change. CASES IN PROJECTILE  Contact Force is a force that acts at the point of CASE 1: PROJECTILE RELEASED contact between two objects. HORIZONTALLY  Tension is the tightness of a rope or string when you try to stretch it.  Weight is the force with which a body is pulled towards the center of the Earth due to gravity.  Laws of Motion are the three statements describing the relations between the forces acting on a body and the motion of the body. FORMULA TO BE USED: THREE LAWS OF MOTION Proponent: Sir Isaac Newton 1ST LAW: LAW OF INERTIA  A body in motion remains in motion or a body at rest remains at rest, unless acted upon by a force. 2ND LAW: LAW OF ACCELERATION  Force equals mass times acceleration: 𝐹 = 𝑚 × 𝑎  Force is directly proportional to mass and acceleration. GENERAL PHYSICS 1 QUARTER 1 Basilio, Andrei Martin M. (12-STEM 02) 3RD LAW: LAW OF INTERACTION c. What is the amount of normal force exerted by the  For every action, there is an equal and opposite ramp? reaction. ⃗⃗ = cos 25 𝑊 𝑁 ⃗⃗⃗⃗ + 𝑚𝑎𝑥 ⃗⃗ = (0.90630777) (−9.604 𝑁) 𝑁 INERTIAL REFERENCE FRAME ⃗⃗ = 8,704 𝑁 𝑁  Zero acceleration ⃗⃗ = 8,704 N, upward Final answer: 𝑁  Velocity measurements always depends on the inertial reference frame the observer adopts. d. If instead of staying stationary, the car accelerates 0.15 m/s2 due to an application of a force through the cable CONTACT AND NON- CONTACT FORCES along the ramp, how much tension must be in the cable?  Contact Forces are forces that require direct physical ⃗⃗ = sin 25 𝑊 𝑇 ⃗⃗⃗⃗ + 𝑚𝑎𝑥 contact between two objects. Ex: frictional, spring, and muscular/applied forces ⃗⃗ = (0.4226182617) (-9,604 N) + (980 kg) (0.15 𝑇  Non-Contact Forces are forces that act on an object m/s2) without the need for physical contact. ⃗⃗ = 4,206 𝑁 𝑇 Ex: magnetic, gravitational, and electrostatic forces Final answer: 𝑇 ⃗⃗ = 4.2 x 102 N, to the right TYPES OF FRICTIONAL FORCES Sample problem: A 980-kg car is held in place by a light cable parallel on a very smooth (frictionless) ramp. The ramp itself rises at 25.0 above the horizontal. a. What is the weight of the car? ⃗⃗⃗⃗ = 𝑚 𝑔⃗ 𝑊 ⃗⃗⃗⃗ = (980.0 kg) (-9.80 m/s2) 𝑊 ⃗⃗⃗⃗ = -9,604.0 kg m/s2 𝑊 Final answer: 𝑊 ⃗⃗⃗⃗ = 9.60 x 102 N, downward b. How much tension is being carried by the cable? ⃗⃗ = sin 25 𝑊 𝑇 ⃗⃗⃗⃗ + 𝑚𝑎𝑥 ⃗⃗ = (0.4226182617) (−9.604 𝑁) 𝑇 ⃗⃗ = 4,059 𝑁 𝑇 Final answer: 𝑇 ⃗⃗ = 4,509 N, to the right GENERAL PHYSICS 1 QUARTER 1 Basilio, Andrei Martin M. (12-STEM 02)

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