How to deal with physics in an easy way with these short summaries created througout the semester? PDF

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The summary notes to help you with physics. Written during the full semester. All about physics and how to deal with it in an easy way with these short summaries. Easy explanations using dot points and easy english....

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Components Of Motion Horizontal Component - Range V x = Ux /\x = Uxt V x2 = U x2 Vertical Component - Height Vy = Uy+at /\y = Uyt+1/2at2 Vy2 = Uy2+2a/\y

1) A tennis ball is struck at 5.0m/s, 55o above the horizontal. What is the velocity of the tennis ball 1.2 seconds later. U=5m/s Theta=55o Ux=5cos55= 2.9m/s Ux=Vx=2.87m/s Uy=5sin55= 4.10m/s Vy=Uy+at=4.10+(-9.8x1.2)=-7.66m/s =7.66m/s upwards V= sqrt[2.872+7.662] V= 8.2m/s 2) A rifle is fired horizontally at a target 81m away and the bullet takes 0.35 seconds to strike it. What was the velocity of the bullet? T=0.35 Uy=0 Ux=? /\x=81m /\x=UT Ux=/\x/t =81/0.35 Ux=231m/s 3) A rocket is fired with an initial velocity of 50m/s at an angle of 30o to the horizontal. Calculate: a) Its time of flight U=50m/s Theta=30o Uy=Usin[theta] = 50sin30 = 25m/s Vy=Uy+at = 0=25+(-9.8t) = 9.8t=25 = 25/9.8 = 2.6s = 5.1s T=5.1s b) Maximum height Vy2=Uy2+2a/\y (Vy2-Uy2)/2a=/\y 0-252/2*9.8 c) Range 4) An athlete in a long jump competition leaps with a velocity of 7.50ms at an angle of 30o to the horizontal. a) What is his velocity at the highest point? b) What is the max height gained in the jump? c) What is the total flight time? a) 6.5m/s b) 0.72m

c) 0.77s 5) Calculate the height, time & velocity of the ball thrown vertically upward with a velocity of 45km/h. Uy=45km/h = 12.5m/s Vy=0 a) Peak Height: /\y=(Vy2-Uy2)/2a = -12.52/(2x-9.8) = 7.97m b) Flight Time: T=2(Uy/a) = 2(12.5/9.8) = 2.55s c) Velocity after 0.5s: Vy=Uy+at = 12.5+(-9.8x0.5) = 7.6m/s upwards d) Velocity after 1.5s: Vy=Uy+at = 12.5+(-9.5x1.5) = 2.2ms downwards 6) A rock is thrown horizontally out to sea from the top off a cliff with an initial velocity of 20m/s and it reaches water after 3.0s. Find: /\y=? /\x=? Ux=20m/s=Vx Uy=0 U=? T=3.0 a) The height of the cliff: /\y = at2/2 = (9.8x32)/2 = 44.1m b) The horizontal distance from the cliff: /\x=Uxt=20x3=60m c) The velocity just before it hits the water: Vy=Uy+at = 9.8x3 = 29.4m/s V=sqrt[Vx2+V2] = 35.56m/s Theta= tan-1(20/29.4) = 34o Velocity= 35.6m/s at an angle of 34o 7) An object is dropped from a height of 90m. Find: a) Impact Velocity: V=sqrt[Uy2+2a/\y] = sqrt[2x9.8x90] = 42m/s b) Time to hit the ground: T=(Vy-Uy)/a = 42/9.8 = 4.3s

Circular Motion Newtons 3 laws of motion Linear speed = distance/time (v=s/t) Centripetal force: Fc=mv2/r F=ma Mv2/r = f Ac=v2/r Uniform circular motion: motion in a circle of constant radius & constant speed. Instantaneous velocity is always tangent to the circle. For an object to be in uniform circular motion, there must be a net force acting on it. Fc=mac=m(v2/r) Circular Motion Satellite in orbit around Earth Car moving around a flat curve Car moving around a banked exit Toy plane tied to a rope and moving in a circle Astronaut in a rotating space station

Centripetal Force Gravitational force of Earth Static frictional force Static frictional force & normal force The rope’s tension Normal force by the surface/floor

Rider on a roller coaster

Weight and/or normal force

We define this inward acceleration as the centripetal acceleration. Centripetal means ‘centre seeking’. To find the magnitudes of each we have: Vc=2pir/t & ac=v2/r

1) A rock of mass 2.5kg is being swung on a 1.2m string so that it is in circular motion with an orbital velocity of 15m/s. Calculate the centripetal force applied by the string. M=2.5kg R=1.2m V=15m/s F=mv2/r = (2.5x152)/1.2 = 468.75N 2) Calculate the centripetal force acting on a space shuttle travelling at 27,700km/h as it orbits Earth at an altitude of 350km asl. M=82 tonne = 82000kg RE=6378km Ro=6728km = 6728000m V=27,700km/h = 7694.4m/s Fc=mv2/r = (8200x7694.42)/6728000 = 7.2x104

Angular Velocity Angular Velocity, w, of an object travelling through an angle of O in a period of time, t, is: W=/\O/t In circular motion, angular displacement /\O for an object travelling through an arc length ‘s’ of radius ‘r’ is given by: /\O=s/r Magnitude: ac = v2/r = w2r

1) Convert to radians: a) 30o = 0.52 b) 90o = 1.57 c) 45o = 0.79 2) Convert the following to degrees: a) pi/5 = 36o b) 3pi/4 = 135o

Apple faced potato brained carrots are evil 1) A conical pendulum is made of a 10g mass suspended from a string with a length of 20cm. The mass makes a horizontal circle of radius 5cm as it moves. Determine: - Angle between the string and vertical: theta=sin-1(0.05/0.2) = 14.47o - String Tension: T=w/costheta = 0.098/cos14.47 = 0.10N - Centripetal force: Fnet=sqrt[0.12-0.0982] = 0.02N - Avg. speed: V=sqrt[rFnet/m] = sqrt[0.05x0.02/0.01] = sqrt[0.1] = 0.32m/s 2) A 150g mass attached to a string of length 3.0m is rotating in a horizontal circle with a period of 0.8 5s a) Calculate the centripetal acceleration b) Calculate string tension M=150g=0.15kg T=0.85s L=3m Ac=? Ft=? a) b)

3) Example in textbook. a) Radius = lstringsintheta = 2sin50 = 1.53m b) c) Wtantheta = 1.96tan50 = 2.34N to the left d) Ft=sqrt[Fnet2+Fw2] = sqrt[1.532+1.962] = 3.05N

1) A car has mass of 1500kg It is travelling horizontally at 80km/h around a bend that is banked at 10o to the horizontal. M=1500 V=80km/h=22.2m/s Theta=10o Fnet=? R=? a) Fnet=tanthetafg = tan10(9.8x1500) = 2592.01N towards the centre of the circle b) r=v2/tanthetag = 22.22/tan10(9.8) = 285.2m 2) A curve in a road has a radius of 4.5m and a banking angle of 12o. At what speed will no friction be required between the car’s tyres and the road surface for a car on this curve? R=45m Theta=12o V=? V=sqrt[rgtantheta] = sqrt[(45x9.8)tan12o] = sqrt[93.7] = 9.7m/s

Exercise 3.2: 1) a) R=lcostheta = 2.4cos60 = 1.2m b) forces: gravity, tension, centripetal c) inwards, towards the centre of her swinging d) Fnet = mgtantheta = (30x9.8)tan60 = 509.2N

Work done in uniform circular motion Horizontal Circle Object rotates at a constant height, therefore potential energy stays the same. Vertical Circle & Cone Pendulum Object changes position 9height0 above earth’s surface, therefore potential energy changes – hence work is done. Upon completing each cycle, objects return to an initial position and initial potential energy – hence total work done per cycle is 0. Work done on an object is equal to the energy transferred to the object. So for an object undergoing uniform circular motion in a horizontal plane the total energy is constant so no work is done. W=force.displacement 9towards the force0 In circular motion the force towards the centre of the circle and the displacement is at right angles to the force. So w=fs = f.0 = 0

Torque & Circular Motion Torque (t) is the turning effect of the force. It is a vector quantity measured in Nm T=force*displacement T=Fr Where f=force in newtons, r=force arm in meters, t=torque in Nm. T=rfsintheta

Exercise 3.4 1) a) It is possible to move a heavy door from the handle rather than the middle of the door as the torque is equal to the force multiplied by the force arm length. By pushing in the middle of the door, you decrease the amount of torque, making it harder to move. b) It is possible to move a heavy rock with a long crowbar rather than without as the torque is equal to the force multiplied by the force arm length. By pushing with a crowbar, you increase the amount of torque, making it easier to move. 2) a) t=fr = 100x2 = 200Nm counter clockwise around the pivotal point b) 0Nm 3) t=fr, r=t/f = 15/30 = 0.5m 4) t=fr, f=t/r = 9/0.5 = 18N 5) t=fr = 225x0.4 = 90Nm 6) a) t=fr = (1x9.8)x0.5 = 4.9Nm Downwards b) t=fr = 9.8x1 = 9.8Nm Downwards c) t=frsintheta = 9.8sin30 = 4.9Nm Downwards 7) t=frsintheta = (300x0.3)sin30 = 45Nm 8) a) 3.4x104N b) the weight stays the same when the crane moves the mass closer to the top, so the torque remains constant. c) t=frsintheta = [(3.4x104)x25]sin37 = 5.1x105Nm towards the horizontal

Motion In a Gravitational Field Newton’s law of universal gravitation Ever mass attracts another mass F (directly proportional) m1m2 F (inversely proportional) r2 F=G(m1m2)/r2 G=6.67x10-11Nm2kg-2 F=GMm/r2 Where F=gravitational Force (N) M=mass1 (kg) m=mass2 (kg) r=distance between the centre of M&m G=gravitational constant, 6.67x10-11Nm2kg-2 G=GM/r2

1) M1=90kg M2=75kg R=0.8m F=Gm1m2/r2 = [(6.67x10-11)90x75]/0.82 = 7.03x10-7Nm2kg-2 2) M=5.97x1024kg R=6.38x106 a) 9.8N/kg b) 9.1N/kg c) 7.3N/kg

Exercise 3.5

1) The proportionalities of Newtons law of universal gravitation is that the masses is directly proportional and is inverse to the squared distance between the two masses. 2) Fg=m1m2/r2 = (6.67x10-11)(2x1030)(6.4x1023)/(2.2x1011)2 = 1.8x1021N 3) F=ma, a=F/m = (1.8x1021)/6.4x1023) = 2.81x10-3m/s 4) a) Fem = GMeMm/rem 2 = (6.67x10-11)(6x1024)(6.4x1023)/(9.3x1010)2 = 3.0x1016N b) Fes = GMeMs/res2 = (6.67x10-11)(6x1024)(2x1030)/(1.53x1011)2 = 3.4x1022N c) The force of Earth on Mars is 0.00009% of that of the Sun on Earth. 5) If the distance from the centre of a celestial body is increased from 400km to 1200km, then the strength of the field will decrease by a factor of 32 or increase b 1/9 6) 0.0002N

Orbital Velocity & Period Of Orbit. Fg=GmM/r2 = Fc =mv2/r Fg=Fc GM/r2=mv2/r v2=GM/r v=sqrt(GM/r)

Altitude (km) Orbital Period Uses

Other Characteristics

Field Of View

Leo 100-1000 (sub inner Van Allen) 60-90minutes @250km alt. - Military - Earth Observation - Weather Monitoring - Shuttle Missions - HUBBLE Space Telescope - Frequent Coverage Of Specific/Varied Locations - Orbits sub 400km are difficult due to satellite drag Smallest

Determine the orbital velocity required by a satellite at altitudes: a) 250km: b) 400km: c) 40000km: V=sqrt[GM/R] = sqrt[(6.67x10-11)(5.97x1024)/6.63x106] = 7750m/s (a) = sqrt[(6.67x10-11)(5.97x1024)/6.78x106] = 7664m/s (b) = sqrt[(6.67x10-11)(5.97x1024)/4.64x107] = 2929m/s (c)

Kepler’s Law of Periods Kepler’s first law: Orbits Satellites move in elliptical orbits with the central body at one focus of the eclipse. Kepler’s second law: Areas When a planet moves in orbit, it’s radius vector sweeps out an equal area in an equal interval of time. Kepler’s third law: Periods

Geo >35800 (6.6xearth radius) (above outer Van Allen) ~24hours - Communications - Mass-Media - Weather Monitoring

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Largest

Allows tracking of a stationary point on Earth AUSSAT & OPTUS

The square of a planet’s orbital period, T, is proportional to the cube of the average distance, r, of the planet from a common central mass. r3/T2 = GM/4pi2 T2 = kr3

1) Calculate the orbital period of the moon. It’s average distance from the Earth is 406676km and has a diameter of 3467km T2=r34pi2/GM T2= (406676x103)4pi2/G(5.97x1024) T2= 6.67x1012 T=2582275

Ex 4.2 1) C 2) B 3) 506m/s2...