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# Physics 1315 Exam 1 Study Guide PHYS 1315

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This 11 page Study Guide was uploaded by Bridget Davis on Thursday September 29, 2016. The Study Guide belongs to PHYS 1315 at University of North Texas taught by Tilo Reinart in Fall 2016. Since its upload, it has received 62 views.

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Date Created: 09/29/16

Physics 1315 Exam 1 Study Guide Chapter 2: Describing Motion Know the following equations: v=v 0at ∆ v=v fv i d s= t v a= t d=vt 1 2 d= 2t d=v t+ at 2 0 2 F=ma Know the following concepts: 1) Speed vs Velocity a. Velocity is magnitude and direction b. Speed is just magnitude c. Average velocity is equal to half the final velocity if starting from rest 2) Vectors vs Scalars a. Vectors are units involving magnitude and direction i. Velocity, acceleration, force, displacement b. Scalars are units just involving magnitude i. Speed, mass, volume, distance 3) Acceleration a. If vectors change, there is acceleration b. If velocity is constant, there is no acceleration 4) Graphing motion a. Distance vs time graph i. Distance is vertical component and time is horizontal ii. The slope of a distance vs time graph is the velocity iii. The steeper the slope, the larger the speed b. Velocity vs time graph i. Velocity is vertical component and time is horizontal ii. The slope of a velocity vs time graph is the acceleration iii. If there is no acceleration the slope will be a horizontal line iv. “change of the change of the distance” 1. accelerationvelocitydistance 5) Uniform acceleration a. Constant force b. If uniform acceleration is positive, velocity is increasing i. Slope is upward c. If uniform acceleration is negative, velocity is decreasing i. Slope is downward Practice: 1) A traveler covers a distance of 800 miles in a time of 10 hours. What is the travelers average speed? (s=?) 2) A hiker walks with an average speed of 3 m/s. What distance in kilometers does the hiker travel after 1.5 hours? (Don’t forget to convert from meters to kilometers. 1km=1000m or 1m=.001km) (d=?) 3) A car traveling with an initi2l velocity of 18 m/s accelerates at a constant rate of 4.5 m/s for 5 seconds. a. What is its velocity at the end of this time? (final velocity) (v =?) f b. What distance does the car travel during this time? (d=?) 4) If a sprinter ran a distance of 150 meters at a constant speed of 10 m/s, how long will it take him to travel the distance? (t=?) 2 5) Starting at rest, a car accelerates at a rate of 5 m/s for a time of 2.5 seconds. What is its velocity at the end of this time? (v =?) (If an object is starting at rest v=i) Chapter 3: Falling Objects and Projectile Motion Know the following equations: 2 g=9.8m/s 1 2 d= a2 d=v t+ 1 at2 0 2 v=v 0at v=at 1 2 d=d 0v 0+ at 2 Know the following concepts: 1) Acceleration due2to gravity a. 9.8 m/s b. also called the gravitational acceleration c. if velocity changes, there is acceleration d. Galileo was the first to accurately measure it e. Does not depend on weight i. Heavy objects do not fall faster than lighter objects 2) Tracking a falling object a. If you need to find distance and the starting velocity is 0 use the equation d= at 2 2 b. The effects of air resistance increase as velocity increases c. v=at 3) Throwing a ball downward a. Starting velocity is NOT zero i. v=v +at 0 1. Use (+) for positive acceleration and (–) for negative acceleration b. The distance traveled by a uniformly accelerating object can be calculated using the equation d=v 0+ 1 at2 2 c. If mass decreases, acceleration increases 4) Throwing a ball upward a. Acceleration is opposite of the original upward velocity b. Velocity will be 0 at the highest point i. The highest point is also called the turnaround c. Acceleration is unrelated to the size of the velocity d. v=v 0at e. the larger the original velocity, the greater the time to reach the highest point f. the time it takes the object to travel back to the starting point once it has reached the highest point is the same amount of time it took the object to reach the highest point 5) Projectile motion a. Two things occur: i. Object accelerates downward with the influence of gravity ii. The object is moving sideways with constant horizontal velocity b. Trajectory i. The path that a moving object follows ii. Combining the two motions from 5a 1. Horizontal and vertical motion c. Object will accelerate downward with constant gravitational acceleration i. 9.8 m/s 2 d. The vertical motion is independent of the horizontal motion e. Horizontal distance is determined by time and initial velocity 1 2 i. d=d +0 t0 at2 1at 2 ii. For the horizontal distance d =00and 2 =0 f. Acceleration of the horizontal motion is 0 i. Object moves in constant horizontal velocity g. Vertical distance is determined by the equation i. d=d +v t+ at1 2 0 0 2 ii. d =height, v t=0, and a=9.8m/s (downward) 0 0 h. Total velocity is a vector sum of the horizontal and vertical components i. Horizontal velocity is constant with no acceleration j. Vertical velocity is gets larger while acceleration is constant 6) Angle of maximum distance a. The angle used to achieve the max distance is 45° i. The horizontal component is equal to the vertical component ii. Object stays in air longer and travels further Practice: 1) A ball is thrown upward with an initial velocity of 16 m/s. Using 2 g=10m/s , what are the magnitude and direction of the ball’s velocity? a. 1 second after (t=1) b. 2 seconds after (t=2) 2) A ball is thrown downward with an initial velocity of 10 m/s. Using g=10m/s , what is the velocity of the ball after 1 second? (t=1) 3) A ball rolls off a table that is 5 meters tall. The ball’s horizontal velocity as it leaves the table is 7 m/s. a. How much time does it t2ke for the ball to hit the ground? (assuming g=10m/s ) (t=?) b. How far from the base of the table does the ball hit the floor? (horizontal distance) 4) A steel ball is dropped with an initial velocity of zero. Using g=10m/s 2 a. What is the velocity of the ball 0.8 seconds after the release? (v=?) b. What is its velocity 1.5 seconds after the release? (v=?) 5) A bullet is fired horizontally with an initial velocity of 850 m/s at a target located 200 m away. a. How much time is required for the bullet to reach the target? (t=?) b. Using g=10m/s , how far does the bullet fall in this time? Chapter 4: Newton’s Laws Know the following equations: Fneta F =F +F net 1 2 Fnet a= m W=mg F2=−F 1 2 1newton=1kg∙m/s Know the following concepts: 1) Newton’s first Law a. An object remains at rest, or in uniform motion in a straight line, unless it is compelled to change by an externally imposed force b. Tells what happens in the absence of force c. Velocity will not change without a force acting on an object 2) Newton’s second law a. The acceleration of an object is directly proportional to the magnitude of the imposed force and inversely proportional to the mass of the object b. Acceleration is directly related to force, not velocity c. Total force determines acceleration 3) Inertia a. Tendency of matter to remain at rest b. Tells how much resistance an object has to change in motion c. An object with a greater mass needs a bigger force to move 4) Forces a. Units of force are Newtons b. Vector quantity c. Force due to gravity, force due to friction d. If net force=0 the object moves with constant velocity e. Force causes change in velocity f. If the forces are the same, acceleration is zero g. If net force is 0 it is still possible for the object to be moving h. Reaction force is an oppositely directed force in response to the initial force in an interaction 5) Mass and weight a. Mass is the property of matter that determines how much an object resists a change in motion i. The greater the mass the greater the inertia b. The force required is proportional to mass c. Weight is the gravitational force acting on an object i. Measured in Newtons ii. W=mg iii. In absence of gravity, there is no weight d. Gravitational acceleration is independent of mass i. a= mg =g m 6) Newton’s third law a. If object A exerts a force on object B, B exerts a force on object A that is equal in magnitude but opposite in direction to the force exerted on B. i. Action/reaction principle ii. F 2−F 1 b. Force from mass is inertia c. Two forces always act on different objects d. Normal force is perpendicular force a surface exerts on an object (N) e. F2= f −ma 7) Applications of Newton’s Laws a. Air resistance (R) b. R increases with velocity c. Terminal velocity is velocity at its max d. W=force of gravity due to interaction with earth e. N=upward force exerted f. P=outside force g. f=frictional force h. P must be larger than f in order for there to be acceleration i. P and f determine horizontal acceleration j. When a ball is thrown 3 forces are involved i. Initial push, downward pull of gravity, air-resistance k. No force is needed to keep a ball moving once it has been thrown 8) Motion of connected objects a. If the masses of the two objects and the magnitude of the forces are known, acceleration can be find with Newton’s second law i. F=ma Practice: 1) A single force of 20 N acts upon a 10 kg block. What is the magnitude of the acceleration of the block? (a=?) 2) A 5 kg block being pulled across a table by a horizontal force of 60 N also experiences a frictional force of 5 N. What is the acceleration of the block? (a=?) 3) Two forces, one of 30 N and the other of 10 N, act in opposite directions on a box. What is the mass of the box if its acceleration is 4 m/s ? (m=?) 4) A 1 kg book rests on a table. A downward force of 10 N is exerted on the top of the book by a hand pushing down on it. a. What is the magnitude of the gravitational force? (W=?) b. What is the magnitude of the upward (normal) force exerted by the table on the book? (N=?) 5) Amanda has a weight of 120 lbs. a. What is her weight in Newtons? (1lb=4.45N) b. What is her mass in kg? Practice solutions: Chapter 2: 1) S=? S=d/t800m/10s= 80m/s 2) d=? Convert 3 m/s to km/h km/h=m/s(3.6) 3m/s(3.6)= 10.8km/h S=d/t(t)s=d d=(10.8km/h)(1.5h) d=16.2 km 3) a) v=v 0at (18m/s) + (4.5m/s )(5s) 18m/s + 22.5m/s vf=40.5m/s b) d=v t+1/2at 2 0 2 =(18m/s)(5s) + ½(4.5m/s )(5s) =90m + 11.25m d=101.25m 4) s=d/t (t)(s)=d(t) st/s=d/s t=d/s =150m÷10m/s t=15s 5) v=v 0at =0 + (5m/s )(2.5s) v=12.5m/s Chapter 3: 1) a) t=1 v=? v=v 0 at 2 =16m/s +(-10m/s )(1s) =16m/s – 10m/s v=6m/s b) t=2 v=? v=v + at 0 2 =16m/s +(-10m/s )(2s) =16m/s – 20m/s v=-4m/s 2) v=v 0 at 10m/s + 10m/s (1s) v=20m/s 3) a) t=? 2 d=1/2at t =d/(1/2)a =5m/(1/2)(10m/s ) 2 2 t =1s √t =√1s t=1s b) dhorizontal d=d +v t+ at 2 0 0 2 =0 + v 0 + 0 d=v 0 =7m/s(1s) d=7m 4) a) v=? if t=0.8s v=at 2 =(10m/s )(0.8s) v=8m/s b) v=? if t=1.5s v=at =(10m/s )(1.5s) v=15m/s 5) a) t=? d=v 0 d/v0=t/v0 t=d/v0 2 =200m/(850m/s ) t=.235s 2 b) dverticalf g=10m/s dverticalat 2 =1/2(10m/s )(.235 ) dvertical68m=27.68cm Chapter 4: 1) F=ma a=F/m =20N/10kg a=2m/s 2 2) Fnet +1 2 =60N + (-5N) =55N F=ma F/m=ma/m a=F/m =55N/5kg a=11m/s 2 3) Fnet +1 2 =30N + (-10N) =20N F=ma F/a=ma/a m=F/a =20N/(4m/s )2 m=5kg 4) a) W=? W=mg 2 =1kg(9.8m/s ) W=9.8N b) N=? N=P + W =10N + 9.8N N=19.8N 5) a) Weight in Newtons 1lb = 4.45N 120lb(4.45N) W=534N b) Mass in kg 1lb = 0.45359237kg 120lb(0.45359237kg) m=54.4kg

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