Formula Sheet for Stage 6 Physics

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How to Use the Formulas for Stage 6 PhysicsHSC Course (Core)Formula Name Comments Typical Problem Typical AnswerE p= - G m 1m 2 Gravitationalm m2r Potential= − GrEnergy9.2.1rF = mgrv x 2 = u x2v 2 y = u 2 y + 2ay∆y∆x = u x t∆y = u y t + 1 2 a yt 2r 3T 2= GM4π 2L v= L 01− v 2t v=t 01 − v2c 2c 2GravitationalForce9.2.1Newtons’ Lawsof Motion9.2.2Kepler’s thirdlaw9.2.2RelativisticLengthContraction9.2.4Relativistic timedilation9.2.4E p = Potential energyG = universal gravitation constant(6.67 X 10 -11 Nm 2 kg -2 )m 1 = mass of body 1 (kg)m 2 = mass of body 2 (kg)r = separation between the twobodies from infinity to r (m)F = force (N)M = mass (kg)g = gravitational constant at thesurface of the Earth 9.81 ms -2u x = initial speed in x direction (m/s)v x = final speed in x direction (m/s)u y = initial speed in y direction (m/s)v y = final speed in y direction (m/s)a = constant acceleration (ms -2 )δx, δy = change in displacement (m)t = time (s)r = radius of motion (m)T = period of motion (s)G = universal gravitation constant(6.67 X 10 -11 Nm 2 kg -2 )M = mass of system (kg)L v = apparent length (m)L o = “rest” length (m)v = relative velocity (m/s)c = speed of light (3 X 10 8 m/s)t v = apparent time (s)t o = “rest” time (s)v = relative velocity (m/s)c = speed of light (3 X 10 8 m/s)What is thegravitational potentialenergy between two100 kg massesthrough a distance of2000 m?What is the weight ofa 100 kg person?What is the maximumheight of a projectilelaunched at 45° to thehorizontal at 50 m/s?What is the period ofa 10 12 kg comet thatorbits at 5 X 10 8 m?What is the apparentlength of a spaceshipof rest length 150mtravelling at 0.9c?How much slowerdoes an astronauttravelling at 0.9cappear to an observer“at rest”−1116.67X10X100X100−12E p== 3.3X10J2000rF = mgr = 100 X 9.81 = 98. 1NLet up be positive.At max. height, v y =0, thus2 2 2 o0 = u + 2ay∆y= u sin 45 − 2X9. 81y∆yv2u sin 45∴ y =2X9.81rTg50=X sin19.6452 o 2 2 oGM∴T=24π24πrGM4π=6.67X10X 5X10332= =2−112L v(0.9c)= L01−= 150 1−2cc2t2v=tv1−c1(0.9c)1−2c0v= == =22210.1965.4m8X1012= 17203s2.29 times slower
Formula Name Comments Typical Problem Typical AnswerF = BIlMagnetic forceon a currentcarryingwire oflength l in amagnetic field9.3.1F = force (N)B= Magnetic Field (T)I = current (A)l = length (m)Calculate the force on2m of wire carrying acurrent of 4A in amagnetic field of0.1T.F = BIl= 0.1X 4X2 = 0. 8NFl= k I 1I 2dτ = rF dForce per unitlength9.3.1torque9.3.1τ = nBIAcosθ torque on a coilimmersed in amagnetic field9.3.1vpvs= n pnsTransformerequation9.3.4F = force (N)l = length (m) per unitI 1 , I 2 = two currentsparallel=repulsive,antiparallel=attractived = separation of the two currents(m)k=magnetic constant (2 X 10 -7 NC -1 m -1 )τ =torque (Nm)F =force (N)d=distance (m)τ =torque (Nm)n =number of turns of coilB=magnetic field (T)I = current (A)A = area of coil immersed inmagnetic field (m 2 )cos Θ=angle between the coil andthe magnetic fieldV p = primary voltage (V)V s = secondary voltage (V)N p = number of turns in the primarycoilN s = number of turns in thesecondary coilWhat is the force perunit length on twowires, both carrying10A, separated by adistance of 3m?What is the torque ona nut when a 0.6 mspanner has a force of80 N applied on it?What is the torque ona 0.20 m 2 coil of 200turns immersed in amagnetic field of 0.2T carrying a currentof 3 A?A transformer isrequired to step downmains voltage (240V)to 12 V. If theprimary coil has 960turns, how many turnsare required in thesecondary coil?FlI I= kd= 2X1010X10X = 6.67X1031 2−7−6τ = F rd = 80 X 0.6 = 48Nmτ = nBIAcosθ= 200 X 0.2X3X0.20 = 24Nmvvpsn=nps∴nssnpv=vps960X12= = 48 turns.240N / m
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Formula Sheet for Stage 6 Physics

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