General Physics Reviewer PDF

Module #1-18 (ctto: aeron.d & lamela) IMPORTANT MEASUREMENTS TO REMEMBER!!! Base Quantities: Derived Quantities: Length: Meter (m) Velocity: Meter per second (m/s) Mass: Kilogram (kg) Acceleration: Meter per second squared (m/s²) Time: Second (s)...

Module #1-18 (ctto: aeron.d & lamela) IMPORTANT MEASUREMENTS TO REMEMBER!!! Base Quantities: Derived Quantities: Length: Meter (m) Velocity: Meter per second (m/s) Mass: Kilogram (kg) Acceleration: Meter per second squared (m/s²) Time: Second (s) Force: Newton (N)  𝒌𝒈. 𝐦/𝐬² Work/Energy: Joule (J) > Regular Notation is the standard way of writing numbers. For example, seven hundred sixty can be written as 760. Scientific Notation > a way to express very large or small numbers using powers of ten. > Positive Exponent: the zeroes will be added to the right > Negative Exponent: the zeroes will be added to the left (decimal) EXAMPLE: Large Number: 280,000,000 ---> 2.8 x 10^8. Small Number: 0.0000056 ----> 5.6×10^-6 Rules for Significant Digits: 1. All nonzeros are significant. 2. All zeros between numbers are significant. 3. If decimals are present, Zeros to the left are not significant. Ex. 0.0078 - 2 sig. digits. 4. If decimals are present, Zeros to the right are significant. Ex. 0.07800 - 4 sig. digits. 5. If theres no decimal, Zeros to the right are not siginificant. Accuracy > closeness of value to the true value of the quantity being measured. > expressed using relative error: Relative error = |measured value−expected value| x 100 expected value x 100 EXAMPLE: The measured value of the apples is bout 5.5 kg, if the expected value for the mass of apples is 8 kg, what is the relative error? Relative error = |5.5 kg−8 kg| x 100 = 31.25% 8 kg Precision > refers to the degree of exactness with which a measurement is made and stated. > expressed as a relative or fractional uncertainty: Relative uncertainty = uncertainty x 100 measured quantity Example: The mass of the apples is 5.5 kg and it is found out that the measurement is 50% uncertain. What is the relative or fractional uncertainty obtained? Relative uncertainty = 0.5 x 100 = 9.09% 5.5 kg Scalar Quantities > Only has magnitude Example: Distance, Speed, Time, mass and work Vector Quantities > Has both magnitude and direction Example: Displacement, Velocity, Weight, Acceleration, Force, Impulse and Gravity Vector direction - can be due East, due West, due South or due North. some vectors do not lie exactly on the axis and are projected to a certain degree. Example: 30 degrees north of RTR building Vectors > can be represented by the use of an arrow with a head and a tail. > length of the arrow represents the magnitude of the vector > direction of the arrowhead represents the direction of the vector. > The tail is called the initial point or the origin of the vector. SOH CAH TOA & SHO CHA TAO Motion - the change in position of an object with respect to a reference point over time. Acceleration - the rate at which an object's velocity changes over time. Acceleration (a) = Δt Δv Average acceleration: the change in velocity over the time period during which that change occurs. Average Acceleration= Total Velocity Total elapsed time Constant acceleration - moving at the same rate of change of velocity. An example of constant acceleration is a ball being dropped from a height. As it falls, it accelerates downward at a constant rate of 9.8 m/s² due to gravity The measurement of acceleration is - 𝒎/𝒔² Uniformly Accelerated Motion (UAM) > is a motion with constant acceleration. Free fall is also known as uniform accelerated motion. FORMULAS for UAM: FORMULA y = ( initial + final velocity ) time Vi = (accel.)(time) - final velocity 2 Vf = (initial velocity) + (acceleration)(time) a = Vf²/2d y = (initial velocity) (time) + 1/2 (accel.)(time²) d = Vf²/2a Vf²= initial velocity + 2 (accel.) (distance) Free Fall > the motion of an object under the influence of gravity alone, with no air resistance. > Galileo said that all objects in free fall accelerate downward at approximately 9.8 m/s regardless of their mass. FORMULAS for Free Fall: y= vertical position Vi= initial velocity Vf=final velocity t=time g=acceleration due to gravity FORMULA y = ( initial + final velocity ) time  2 Vf = (initial velocity) - (gravity)(time) y = (initial velocity) (time) - 1/2 (gravity)(time²) Vf²= initial velocity - 2 (gravity) (vert. pos) PROJECTILE MOTION > a form of motion where an object moves in a bilaterally symmetrical, parabolic path. > occurs when force is applied at the beginning of trajectory. > only interferance with gravity. Components Initial Velocity > expressed as x & y components. Time of Flight > time from when the object is projected to the time it reaches the surface. Acceleration > 0 acceleration horizontally Velocity > horizontal: constant > vertical: varies linearly because acceleration is constant. Vy = u * sin θ - gt Trajectory - path followed by a projectile flying. Maximum Height  Vy = 0 Range > from the time it is launched to the time it comes back down to the same height at which it is launched. Formulas: > t^2 = 2d/g > t = 2Vig/ -g Circular Motion > circular path; direction of velocity always changes. > speed is tangent to the path and the force towards the center is constant. Tangential Speed (vT) > is the speed of an object in a circular motion. > depends on the distance from the object to the center.I > constant tangential speed is said to be uniform circular motion. Centripetal Acceleration (ac) > the acceleration directed toward the center of the circular path. Centripetal Acceleration = (tangential speed) ² radius of circular path Example: - A test car moves at a constant speed around a circular track. If the car is 48.2 m from the track’s center and has a centripetal acceleration of 8.05 m/s^2, what is the car’s tangential speed? Given: r = 48.2 m ac = 8.05 m/s^2 vt = ? derive the formula first then solve: ac =vt^2  vt = √acr  vt = √(8.05 m/s2)(48.2m) = 19. 7 m/s r Tangential Acceleration (aT) > The acceleration of an object moving in a circular motion due to a change in speed. Centripetal Force (Fc) > The net force directed toward the center of the circle. > also known as the “center-seeking force”.  Fc = m * ac  Fc = m * (vT) ² r Laws of Motion (Newton my beloved) First Law of Inertia - An object at rest stays at rest and an object in motion stays in motion with the same velocity unless acted upon by an external force. Second Law of Acceleration * “The acceleration produced by a net force on an object is directly proportional to the magnitude of the net force, is in the same direction as the net force, and is inversely proportional to the mass of the object” > Force is directly proportional to mass. > Force is directly proportional to acceleration. > Mass is inversely proportional to acceleration. Third Law of Interaction > For every action, there is an equal/opposite reaction. Free Body Diagrams are diagrams used to show the relative magnitude and direction of all forces acting upon an object in a given situation. Example problems: A force of 500 N (West) was applied by a nurse on a box of PPE’s with a mass of 100 kg. What is the acceleration of the box if a frictional force of 250 N acts in the opposite direction? Given: Fa = 500 N (East) m = 100 kg Ff = 450 N a =? Solution: W = mg = 100 kg (9.8 m/s2) = 980 kg m/s2 = 980 N N = W = 980 N Fnet = 500 N – 450 N = 50 N (West) (since normal force and weight cancels each other) a = Fnet/m a = 50 N (West) 100kg a= 0.5 m/s2 (West) Extra Info on Third Law of Interaction: One force is called the action force. The other is called the reaction force. Static friction (fs) - acts on objects when they are resting on a surface. It includes all cases in which the frictional force is sufficient to prevent relative motion between surfaces. Sliding friction or Kinetic friction (fk) - defined as a force that acts between moving surfaces. Anything that rubs, slides or slips has some kinetic friction. Sample Problems: - A smooth wooden block is placed on a smooth wooden tabletop. You find that you must exert a force of 20.0 N to keep 50.0-N block moving at a constant velocity. a. What is the coefficient of sliding friction for the block and the table? b. If a 30.0-N brick is placed on the block, what force will be required to keep the block and brick moving at constant velocity? Solution: a. N = W. Thus Fa = Ff = μW or μ = Fa/W = 20.0 N/50.0 N = 0.40 b. Weight of the brick W’ = 30.0 N Ff = μN Fa’ = Ff = μ(W + W’) Ff = μN = (0.40) (50.0 N + 30.0 N) Fa’ = ? = (0.40) (80.0 N) N = W + W’ = 32.0 N Fa’ = Ff Module 13: Work (J) > the amount of force applied to an object (in the same directions as the motion) over a displacement. Formula: Work = Force x Displacement  W = F x d If the force is at an angle to the displacement: Work = (Force x cosine Ø) (Displacement)  W = (f)(d) * cos Ø Power (J/s) > a quantity that measures the rate at which work is done or energy is transformed. Power = work/time or P= W/t since, W (work)= F(force) x d(displacement), then, P =(Force )(displacement)/time or P= (F)(D)/t Energy Mechanical Energy > energy due to the position of something or the movement of something. Potential Energy > stored energy. Kinetic Energy > the energy an object possesses due to its motion. Gravitational Potential Energy > The energy associated with an object due to the object’s position relative to the gravitational source. Elastic Potential Energy > energy stored in a compressed or stretched spring or object. K = force constant delta x = change in size. Work-Energy Theorem > states that whenever work is done, energy changes. Formula: Wnet = (1/2) (m)(v)^2 final − (1/2) (m)(v)^2 initial The sum of objects kinetic energy and potential energy is the object’s mechanical energy. Formula: ME = KE + PE MODULE 16: Center of Mass (COM) > the point at which the total mass can be considered to be concentrated. > The center of gravity usually coincides with the center of mass, which is the average position of all particles of the object. Geometric Centers (Centroid) > corresponds with the center of gravity in the case when the body is homogeneous. Formula in finding center of mass: > xCM, yCM and zCM = coordinates of the center of mass of the system > x1+, y1+ and z1+ = coordinates of each elements making up the system > m1, m2 and m3 = represent the mass of each element making up the system Momentum > a quantity that describes the difficulty in changing the state of motion of a moving object. It is a vector quantity meaning it has magnitude and direction. Impulse - the product of the force and the time it takes for the force to be applied and has unit (kg.m/s) - change in momentum. Formula: Ft = m(vf – vi) Impulse-Momentum Theorem formula: I=∆p Example: How much impulse does a 10 N force produce if it is applied in 0.05s? Given: Solution: F = 10 N I = Ft t = 0.05s = (10 N)(0.05s) Find: I I = 0.5 kg.m/s Module 18: The Law of Conservation of Momentum - states that the total momentum of a closed system remains constant if no external forces act on it. The coefficient of restitution(e) is the negative ratio of the relative velocity of two colliding bodies after a collision to the relative velocity before the collision. In equation, e = (vA2 – vB2) (vA1 – vB1) where: vA2 and vB2 = velocities of bodies A and B after collision vA1 and vB1 = velocities of bodies A and B before collision Types of collision: - elastic collision the bodies separate after colliding with each other. The coefficient of restitution is equal to 1. When the total momentum is conserved but the total kinetic energy is not conserved, then the collision is inelastic. The coefficient of restitution for inelastic collision is between 0 to 1. If the interacting bodies stick together and move as one after a collision, the collision is said to be perfectly inelastic. The coefficient of restitution for inelastic is 0. (di pa ayos pero tinatamad naq - aeron)