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by: Breanab

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# Physics 151 week three notes PHYC 151 001

Marketplace > University of New Mexico > Physics 2 > PHYC 151 001 > Physics 151 week three notes
Breanab
UNM
GPA 4.1

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These notes cover everything on the exam
COURSE
General Physics
PROF.
Dr. Dave Cardimona
TYPE
Class Notes
PAGES
3
WORDS
CONCEPTS
Physics
KARMA
25 ?

## Popular in Physics 2

This 3 page Class Notes was uploaded by Breanab on Monday February 8, 2016. The Class Notes belongs to PHYC 151 001 at University of New Mexico taught by Dr. Dave Cardimona in Winter 2016. Since its upload, it has received 9 views. For similar materials see General Physics in Physics 2 at University of New Mexico.

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Date Created: 02/08/16
Chapter 3: Vectors and Motion in Two Dimensions 3.1 Using Vectors recall discussion in Section 1.5 3.2 Using Vectors on Motion Diagrams For example: circular motion 3.3 Coordinate Systems and Vector Components Vector Components Looking for resultant vectA+ B = R Given A and B vectors. Then findR x A +xB xand Ry= A y B y “SOH – CAH – TOA” Find Ax, Ay, x ,y. Finally finR and θ using the above. R 3.4 Motion on a Ramp X 3.5 Relative MotionX 3.6 Motion in Two Dimensions: Projectile Motion A projectile is an object that moves in two dimensions under the influence of gravity and nothing else. Projectile motion is made up of two independent motions: uniform motion at constant velocity in the horizontal direction and free-fall motion in the vertical direction. 0 height The horizontal and vertical components of θ projectile motion are independent of each other. 0 BUT … They are connected by the time. range A projectile follows a parabolic trajectory because it "falls" a distance ½ g t below a straight-line trajectory. horizontal vertical x = v t y = v t − 12g t2 0x 0y a = 0 a = −g x y v x v 0x v = v − g t y 0y (time, t, connects the two directions) where v = v cosθ and v =v sinθ 0x 0 0 0y 0 0 € Two €pecial formulas: height and range At x = R , v x v 0xand vy= −v 0y Aty = h ,2v y2 (top of mot2on) Then: x = v0x ⇒ t = R v 0x Then: v yv 0y =2a y ⇒ −v 0y =2 ( )h and: vy= v 0ygt ⇒ −v 0y= v0y g R(v 0x) € € 2 2 2 2 v0y v0sin θ 0 € 2v0x €y v0sin2θ 0 OR: h = = € OR: R = = € 2g 2g g g € € For horizontally-projected objects (wherv = v and v = 0 ), take ‘down’ to be positive and use: 2 0x 0 0y € x = v0x y = 2g t € € € y If one ball is dropped and one € € is projected horizontally from the same height, which hits the ground first? x 3.7 Projectile Motion: Solving Problems EXAMPLE 3.11: Dock jumping If a dog runs at 8.5 m/s straight off the end of a dock that is 0.61 m above the water, how far will the dog go before splashing into the water? EXAMPLE 3.12: Checking the feasibility of a Hollywood stunt In the movieSpeed:A50 ft (15 m) section of a freeway overpass is missing.Abus must jump over the gap. The road leading up to the break has an angle of about 5°.Aview of the speedometer just before the jump shows that the bus is traveling at 67 mph (30 m/s). The movie bus makes the jump and survives. Is this realistic, or movie fiction? Range of a projectile: Trajectories of a projectile for different launch angles 3.8 Motion in Two Dimensions: Circular Motion Uniform Circular Motion: Constant speed, but constantly changing direction results in a non-zero acceleration – but NOT in the direction of motion. It is ‘center-seeking’ or ‘centripetal’. ! ! Δv = v −v! Δv −v 1 2 1 2 v a = a = v 2 c r r € € € “Period” =T = time for one revolution € “Frequency = f = number of revolutions per second =1/T Speed in uniform circular motion =v= distance/time =2 π r / T EXAMPLE 3.15: Acceleration in the turn Female world-class speed skaters can cover a 500 m track in 45 s. The most challenging aspect of the race are the turns. What is the magnitude of the centripetal acceleration experienced by the skaters if the tight turns have a radius of 11 m?

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