Physics 1st Quarter PDF

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)