logging in or signing up 2011_ TAKS _Physics _ Review_1-63 erad206 Download Post to : URL : Related Presentations : Share Add to Flag Embed Email Send to Blogs and Networks Add to Channel Uploaded from authorPOINT lite Insert YouTube videos in PowerPont slides with aS Desktop Copy embed code: Embed: Flash iPad Dynamic Copy Does not support media & animations Automatically changes to Flash or non-Flash embed WordPress Embed Customize Embed URL: Copy Thumbnail: Copy The presentation is successfully added In Your Favorites. Views: 225 Category: Entertainment License: All Rights Reserved Like it (0) Dislike it (0) Added: April 29, 2011 This Presentation is Public Favorites: 0 Presentation Description No description available. Comments Posting comment... Premium member Presentation Transcript Slide 1: Review TAKS PhysicsMotion: Motion Motion can be described as change in position of a body . Everything in the universe has motion. Movement occurs in different amounts and in different directions.Position/Time Graphs: Position/Time Graphs Position-versus-Time graphs are used to show the motion of an object. Time is ALWAYS the x axis Position is ALWAYS the y axis Horizontal lines in a position/time graph mean Straight slope lines in a position/time graph mean Curved slope lines in a position/time graph mean The object isn’t moving Movement at a constant speed AccelerationVelocity: Velocity Average velocity is the change of distance of an object over time . Another name for velocity is speed ! (Velocity has a direction, speed does not.) The unit of speed is meters/second (m/s) . velocity = distance / time v = d / tA student practicing for a track meet ran 250 meters in 30 seconds. a. What was her average speed? : A student practicing for a track meet ran 250 meters in 30 seconds. a. What was her average speed? V= d/t V=250m/30s V= 8.33m/s b. If on the following day she ran 300m in 30s,by how much did her speed increase? V= 300m/30s = 10m/s 10m/s-8.33m/s= 1.67m/sSlide 6: 1. How many meters can Swimmer 1 cover in 30 sec? 2. How far will Swimmer 2 go in 30 sec? 3. Predict how far Swimmer 1 can go in 60 sec. 4. Predict how far Swimmer 2 can go in 60 sec. 5. Which swimmer has the greater speed? 6. Calculate the speed of Swimmer 1. 7. Calculate the speed of Swimmer 2. 60 m 30 m 120 m 60 m Swimmer 1 2 m/s 1 m/sDistance/Time Graphs: Distance/Time Graphs Distance-versus-Time graphs are used to show the speed or velocity of an object. Time is ALWAYS the x axis Distance is ALWAYS the y axisSlide 8: If the car has a high velocity (or speed), the graph has a steep slope. If the car has a low velocity (or speed), the graph has a shallow slope Circle the graph showing the object with the higher speed. High Speed (90 mph) Low Speed (20 mph) Distance/Time GraphsSlide 9: A student collects these data as a marble rolls along the edge of a meter stick placed flat on a concrete floor. The average speed of the rolling marble is – A. 0.62 cm/s B. 1.6 cm/s C. 16.0 cm/s D. 62.0 cm/s Position (cm) Time (s) 0 0 10 6.2 25 15.6 40 25.1 55 34.4 70 43.8 85 53.1 100 62.5Slide 10: What is the average velocity of the car between the 10s and the 20s stages of the journey (in m/s)? . Sports Car Example 0 5 10 15 20 25 30 Time (s) V = d/t V = 50m/10s = 5m/s a. What is the total distance covered during the entire trip? b. What is the average speed for the whole trip? (m) 100+100= 200m 200 m/30s = 6.67 m/s (Since the car returns to its start, its average velocity for the trip is 0 m/s.)Slide 11: A particle in a magnetic field moves as follows: 0-2 s v= 2-7 s v= 7-8 s v= Find the velocity for each part of the motion. 4 m/ 2 s = 2 m/s 0 m / 5 s = 0 m/s -4 m/ 1 s = -4 m/sSlide 12: 1. Speed= 1600m/s Time = 1.5 x 10 -5 s Distance = speed x time 1600m/s x 1.5x 10 -5 s =.024 m = 24 mm 1.Slide 13: Joe Cool, a famous track star, ran 600 meters at 10m/sec. How long did it take Joe to run the race? Distance= 600m Speed = 10 m/s Time = distance/speed 600m / 10 m/s = 60s 2.Slide 14: Q= 14Km/12 min=1.17km/min R= 12km/8min= 1.50 km/min S = 15km/9 min =1.67 km/min- This is the greatest speed T= 11km/15min < 1.00 km/min 3.Slide 15: Distance moved = 3 cm Time = 2 seconds Speed = 3cm/2seconds = 1.5 cm/s (Be careful to measure front-bumper-to-front-bumper distance.) 4. What is the speed of the toy car shown above if it moves through the distance shown in two seconds? 1.5 cm/s 2 cm/s 2 cm/s 2 3 cmAcceleration: Acceleration Acceleration is change in an object’s velocity (speed or direction ) over time. - Acceleration can be positive or negative . (Negative acceleration is sometimes called deceleration .) The unit of acceleration is meters/second 2 Equation : Acceleration = Final Velocity – Initial Velocity Time a = v f -v i tSlide 17: Plotted on a distance vs. time graph, acceleration is a curve ( a parabola ). The slope of a velocity/time graph represents the ACCELERATION . Velocity-versus-Time graphs are used to show the acceleration of an object . - Time is ALWAYS the x axis. - Velocity is ALWAYS the y axis. Velocity/Time GraphsSlide 18: What is the acceleration of the car during the following time intervals? (m/s 2 ) From 0s – 2s From 2s – 5s From 5s – 8s (3-0)/2 = 1.5 m/s 2 (3-3)/3 = 0 m/s 2 (0-3)/3 = -1 m/s 2Slide 19: A fighter jet landing on aircraft carrier’s flight deck must reduce its speed from 153 m/s to exactly 0 m/s in 2 s. What is the jet’s acceleration? A = V final – V initial time 0 m/s – 153 m/s = -76.5 m/s 2 2 s 5.Slide 20: 6. a = v final – v initial Between 6s and 10s, the time velocity did not change! 6.5m/s-6.5m/s = 0 m/s 2 4 sSlide 21: A Force is a push or a pull . The unit for force is Newtons (N). The net force is the result of multiple forces acting on an object. There can be many forces, but the object will act as if there is only one force: the net force. - To find the net force, add together all forces acting on the object. ForceBalanced Forces: Balanced Forces Forces are balanced if they are equal in magnitude (amount) and opposite in direction. When forces are balanced , the net force is zero.Slide 23: Forces are unbalanced if one of them is greater . - Can be the same or opposite directions. - Only unbalanced forces change an object’s velocity . + = Unbalanced Forces + =Slide 24: Unbalanced , Accelerates Balanced Unbalanced , accelerates Unbalanced , accelerates Unbalanced , accelerates Unbalanced , accelerates Balanced/Unbalanced Examples Is it balanced? Or is the object accelerating? … If the forces are balanced, it will not accelerate.Slide 25: Answer: D. An opposing force acted on the puck. (Kinetic friction prevented the puck from sliding forever.) 7.Slide 26: An object in motion stays in motion and an object at rest stays at rest unless acted on by an outside force. Newton’s 1 st Law (Law of Inertia )Inertia: Inertia Inertia is the resistance of any physical object to a change in its state of motion or rest. It is proportional to an object’s mass.Inertia Examples : Inertia Examples What happens to you or objects in your car when you slam on the brakes? Will it take longer for a train or a car to stop? You and the objects in your car will continue to move in the same direction (forward). Train, more mass = greater inertiaNewton’s 2nd Law (F = m x a) : Newton’s 2 nd Law (F = m x a) The force required to change an object’s motion is dependent on the mass and acceleration of that object. The unit of force is Newtons (N) = kg m/s 2 Force = Mass * Acceleration F = m * aSlide 30: Weight (pull of gravity) is a commonly measured force (measured in N), calculated by F=m*g - g is the acceleration due to gravity, 9.8 m/s 2 - Mass stays the same, weight changes depending on gravity. Newton’s 2 nd Law (continued)Acceleration is produced when an unbalanced force acts on a mass. The greater the mass (of the object being accelerated) the greater the amount of force needed (to accelerate the object).: Acceleration is produced when an unbalanced force acts on a mass. The greater the mass (of the object being accelerated) the greater the amount of force needed (to accelerate the object). Newton’s 2 nd LawSlide 32: If the man pushed the car with a force of 100N,what would the new acceleration be? F = ma So, if the mass (m) is 1000 kg and the required acceleration (a) is .05 m/s 2 then: F = 1000 kg x .05 m/s 2 = 50 N What force is needed to push the car? a=F/m 100N/1000kg= 0.1m/s 2Newton’s 3rd Law: Newton’s 3rd Law For every action there is an equal and opposite reaction. When the rocket pushes hot gases downward, hot gases push the rocket upward. Which is the action – ignition or take off? Which is the reaction – ignition or take off? Action! Reaction!Slide 34: Momentum is used to measure transfer of movement from one object to another The unit for momentum is kg * m/s Equation: Momentum = mass * velocity p = m * v MomentumSlide 35: The total momentum of two objects before a collision is equal to total momentum of two objects after a collision. The system neither gains nor loses total momentum , it is simply transferred between the two objects. Law of Conservation of MomentumSlide 36: Momentum = mass * velocity Mass = momentum / velocity 15kg.m/s / 30m/s = 0.5 kg 8.Slide 37: 9.Slide 38: 10.Slide 39: Force = m x a = 1300 kg X 1.5m/s 2 = 1950 N 11.Slide 40: P = m*v P = 100 kg * 6.3 m/s P = 630 kg*m/s 12. (Make sure to practice filling in this answer correctly!)Slide 41: D. Standing motionless, the total momentum is zero. If the ball goes forward, the student must roll backward to conserve momentum. 13.Slide 42: Force = mass x acceleration …By Newton’s Third Law, the two pushes are equal and opposite – we can compute either one. Mass of smaller skater = 40 kg Acceleration on smaller skater = 3.0 m/s – 0 m/s = 2.5 m/s 2 1.2 s Push on smaller skater = 40 kg x 2.5 m/s 2 = 100N 14.Slide 43: B. For every action there is an equal and opposite reaction. The shrinking rubber pushes the air out, and the escaping air pushes the balloon the other direction. 15.Slide 44: D. The scallop will move in the opposite direction. 16.Slide 45: The 500 g cart is moving in a straight line at a constant speed of 2 m/s. Which of the following must the 250 g toy car have in order to maintain the same momentum as the cart? A An acceleration of 5 m/s2 for 2 seconds B A potential energy of 20 J C A constant velocity of 4 m/s D An applied force of 5 N for 5 seconds 17.Slide 46: A 0.50 kg ball with a speed of 4.0 m/s strikes a stationary 1.0 kg target. If momentum is conserved, what is the total momentum of the ball and target after the collision? A 0.0 kgm/s B 0.5 kgm/s C 1.0 kgm/s D 2.0 kgm/s 18.Slide 47: A woman lifts a 57-newton weight a distance of 40 centimeters each time she does a particular exercise. It takes her 0.60 second to lift the weight. How much power does she supply for lifting the weight one time? A 24 W B 34 W C 38 W D 95 W SKIP 19.Slide 48: The frog leaps from its resting position at the lake’s bank onto a lily pad. If the frog has a mass of 0.5 kg and the acceleration of the leap is 3 m/s 2 , what is the force the frog exerts on the lake’s bank when leaping? A 0.2 N B 0.8 N C 1.5 N D 6.0 N 20.Slide 49: How many Newtons of force does a 50.0 kg deer exert on the ground because of gravity? 21. Acceleration of gravity = 9.8 m/s 2 F = m * a F = 50.0 x 9.8 = 490 NEnergy: Energy Energy is the ability to do work . - Two Types of Energy: Kinetic Energy (KE) & Potential Energy (PE) - The unit for energy is Joule (J).Slide 51: Kinetic energy is also known as the energy of motion . The amount of KE an object has depends on its mass (in kg) and its velocity in (m/s) Equation: KE = 1/2 * mass * (velocity) 2 KE = ½ m * v 2 Kinetic EnergyPotential Energy: Potential Energy Potential Energy is also known as stored energy, or energy of position . The amount of GPE an object has depends on its mass (in kg), height (in meters) and the acceleration due to gravity (a constant 9.8 ). (GPE = gravitational potential energy) Equation: GPE = mass * gravity * height PE = m * 9.8 * hSlide 53: Circle the one that has more kinetic energy A 25 kg mass or a 30 kg mass going 5 m/s. Two 10 kg masses, one going 75 m/s, one going 45 m/s. A car at rest or a car rolling down a hill. A heavy bike or a light bike (both at 8m/s).Slide 54: K.E. = ½ mv 2 50 J= ½ (1 kg) v 2 100= V 2 V= 10 m/s 22.Slide 55: When an object is at rest at some height, all of the energy is potential . Once that object moves , then some of the energy is kinetic . - Where the PE (or GPE) is greatest , the KE is smallest . - Where the KE is greatest , the PE/GPE is smallest . This is because total energy is always conserved.Slide 56: Which two letters represent points of greater potential energy ? Which points have greater kinetic energy ? Where is potential energy the highest? Where is kinetic energy the highest? W & Y X & Z W XSlide 57: GPE =m x g x h 2.0kg x 9.8m/s 2 x 3.0 m =58.8 J 2.0kg x 9.8 m/s 2 x 1 m =19.6 J 59.8-19.6= 39.2 J 23.Slide 58: An inventor claims to have created an internal combustion engine that converts 100 kJ of chemical energy from diesel fuel to 140 kJ of mechanical energy. This claim violates the law of conservation of A. momentum B. inertia C. energy D. mass 24.Slide 59: Which of the following is an example of solar energy being converted into chemical energy? A. Plants producing sugar during the day B. Water evaporating and condensing in the water cycle C. The sun unevenly heating Earth’s surface D. Lava erupting from volcanoes for many days 25.Slide 60: As a rocket rises, its kinetic energy changes. At the time the rocket reaches its highest point, most of the kinetic energy of the rocket has been – A. Permanently destroyed B. Transformed into potential energy C. Converted to friction D. Stored in bonds between its atoms 26.Slide 61: G.P.E = m x g x h = 95kg x 9.8m/s 2 x 100m = 93,100 J 27.Slide 62: A motor produces less mechanical energy than the energy it uses because the motor A. Gains some energy through motion B. Stores some energy as electrons C. Converts some energy into heat and sound D. Uses some energy to increase in mass 28.Activity -Transformation of Energy: Activity -Transformation of Energy In this activity, you will identify examples of 8 energy transformations. Fill in the blank column of the pink table mat with the yellow energy transformation cards. You do not have the permission to view this presentation. In order to view it, please contact the author of the presentation.