<p>Formula Sheet for Stage 6 Physics</p><p>Preliminary Course H.S.C. Course - Core H.S.C. Course - Options v = fl Z = ru m1m2 Ep = - G ж 1 ц r з I ө ч 2 2 Z - Z и d ш Ir [ 2 1]  r = 2 F = mg Io [Z2 + Z1] v1 sini = 2 2 v2 sinr vx = ux d r 2 2 M = m - 5log( )  F vy = uy + 2ayDy 10 E = q Dx = uxt I A = 100(M B - MA ) / 5 1 2 V Dy= u t+ a t IB R = y 2 y I 2 3 3 4p r r GM m1 + m 2 = Energy = VIt 2 = 2 GT T 4p 1 ж 1 1 ц = R 2 - 2 P=VI 2 з ч v l n f ni L = L 1- и ш r v 0 c2 h v = D l = av m mv t m = 0  v 2 1- Vout  V c 2 a av  V t t in t  0   v v2 V0  v  u 1  2 a  c A 0  av V  V t  + -    F = BIℓ  F  ma Constants F I I Speed of light c 3 X 108 ms-1  k 1 2 1 2 l d in vacuum E  mv k 2  -2   Fd Acceleration g 9.81 ms   due to gravity p  mv    nBIAcos -11 2 -2 Impulse  F t Universal Gravitation G 6.67 X 10 Nm kg</p><p> v p np  9 -1 -1 m1m2 v n Electric constant k 9 X 10 NC m F  G s s d2 -7 2 -2 F  qvB sin  Magnetic constant k 2 X 10 Ns C r3 GM  -34 T 2 4 2 V Planck’s Constant h 6.626 X 10 Js E  d Ryberg’s Constant R 1.10 X 107 m-1</p><p>E  hf -27 mass of proton mp 1.67 X 10 kg</p><p> c  f -27 mass of neutron mn 1.68 X 10 kg</p><p>-31 mass of electron me 9.11 X 10 kg</p><p> charge on an electron e 1.602 X 10-19 C How to Use the Formulas for Stage 6 Physics</p><p>Preliminary Course Formula Name Comments Typical Problem Typical Answer v  f Wave v= velocity (m/s) Calculate the wavelength v 3 v  f     0.5m Equation f = frequency (hz) of a water wave travelling f 6 8.2.1  = wavelength (m) at 3 m/s whose frequency is 6 Hz.  1  Intensity Law I = intensity (no units) A globe is viewed from 2   I    1  1 1  2 8.2.3 d = distance (m) metres at a certain I  d  3d,  thus it appears 9 times less d  2  2 9 brightness. How bright  d  d does it appear 6 metres bright. away?</p><p>Snell’s Law v1 = speed in first medium A light ray travelling at 3 1 sin i sin i v sin i v 8 v1 2  (1 n2 ) 8.2.4 v = speed in second medium X 10 m/s in air enters a  sin r  sin r 2 sin r v v2 sin i = angle in first medium pool of water at 45 to the v2 1 8 ° sin r = angle in second medium 2.4 10 sin 45 ° normal. If it slows down   38  Angles are always to 2.4 X 108 m/s, what is 3108 measured to the normal its angle in the water?     F Electric Field E = Electric field (N/C, V/m) What is the force exerted F 6 3 E  -6 E  F  qE  2 10 X 5 10  0.01N q 8.3.2 F = Force (N) on a 2 X 10 C charge q q = charge (C) moving in an electric field of size 5 X 103 V/m? V Ohm’s Law R = resistance () What is the current V V 18 R  R  I    3A I 8.3.2 V = voltage (V) through a 6 resistor I R 6 I = current (A) when a voltage of 18 V is applied across it? Energy = VIt Electrical Energy (J) How much energy is Energy = VIt = 240 X 2 X 3 X 60 (sec) Energy V = voltage (V) delivered to an electric = 86400 = 86.4 kJ 8.3.4 I = current (A) kettle if a current of 2A is T = time (s) used for 3 minutes plugged into the mains (240V)? P=VI Electrical P = power (W) What current is required P 2400 P VI I   10A Power V = voltage (V) by a 2400W electric heater V 240 8.3.4 I = current (A) plugged into the mains (240V)? Formula Name Comments Typical Problem Typical Answer</p><p> r Average vav = average velocity (m/s) How much distance is r vav   v   r  v  t 16  4  60  384m t Velocity r = distance covered (m) covered by a car travelling av t av 8.4.1 t = time (s) for 4 mins at an average speed of 16 m/s?     V Average a = average acceleration (ms-2) A train accelerates from  V 18 12 a  av a    0.5ms 2 av t Acceleration v = change in velocity (m/s) 12 m/s to 18 m/s in 12 s. av t 12 8.4.2 t = change in time (s) What is the value of this acceleration?     F  ma Newton’s  F = sum of all forces (N) Second Law m = mass (kg) 8.4.2  a = acceleration (ms-2) 1 2 Kinetic Ek = kinetic energy (J) A ball possesses 10 J of 1 2 2Ek 2 10 Ek  mv E  mv m    5kg 2 Energy m= mass (kg) energy. What is its mass if k 2 v 2 22 8.4.3 v = speed (m/s) it is moving at 2 m/s?     p  mv Momentum P = momentum (Ns, kgm/s) What is the momentum of p  mv 1400  6  8400kgms 1 8.4.4 m = mass (kg) 1400 kg car moving at 6  v = velocity (m/s) m/s? Impulse Impulse = change in momentum A ball of mass 0.5 kg Impulse  F t Impulse  p f  p i  m(v f  v i )  0.5(2  (2)) 8.4.4 (Ns, kgm/s) travelling at 3 m/s hits  Impulse 2 wall and bounces back at Impulse  Ft  F    20N the same speed. If it is in t 0.1 contact with the wall for 0.1 s, what is the force exerted by the wall? 11 24 m1m2 Universal F = force (N) What is the gravitational m1m2 6.6710 X 2X 6X10 F  G 2 F  G   19.6N Gravitation G = universal gravitation force between the Earth 2 6 2 d d (5.810 m) 8.5.4 constant (6.67 X 10-11 Nm2 kg-2) (m=6X1024 kg) and 2 kg m1 = mass of body 1 (kg) ball given that the radius 6 m2 = mass of body 2 (kg) of the Earth is 5.8X10 m d = separation between the two bodies (m) 3 r GM 3 2 3 2 8  Kepler’s Third r = radius of motion (m) What is the period of a r GM 4 r 4 X 5X10 T 2 4 2 Law T = period of motion (s) 1012 kg comet that orbits at  T    17203s 2 2 GM 11 12 8.5.4 G = universal gravitation 5 X 108 m? T 4 6.67X10 X10 constant (6.67 X 10-11 Nm2 kg-2) M = mass of system (kg) How to Use the Formulas for Stage 6 Physics</p><p>HSC Course (Core) Formula Name Comments Typical Problem Typical Answer 11 m1 m2 Gravitational Ep = Potential energy What is the m1 m2 6.67X10 X100X100 12 Ep = - G E   G   3.3X10 J r Potential Energy G = universal gravitation constant gravitational potential p r 2000 9.2.1 (6.67 X 10-11 Nm2 kg-2) energy between two m1 = mass of body 1 (kg) 100 kg masses m2 = mass of body 2 (kg) through a distance of r = separation between the two 2000 m? bodies from infinity to r (m)     F  mg Gravitational F = force (N) What is the weight of F  mg  100X 9.81  98.1N Force M = mass (kg) a 100 kg person? 9.2.1 g = gravitational constant at the surface of the Earth 9.81 ms-2 2 2 vx  ux Newtons’ Laws ux = initial speed in x direction (m/s) What is the maximum Let up be positive. 2 2 of Motion vx = final speed in x direction (m/s) height of a projectile At max. height, vy=0, thus vy  uy  2ayy 2 2 2 ° 9.2.2 uy = initial speed in y direction (m/s) launched at 45 to the 0  u  2a yy  u sin 45  2X 9.81yy x u t v g   x vy = final speed in y direction (m/s) horizontal at 50 m/s? 2 2 ° 2 2 ° -2 u sin 45 50 X sin 45 1 a = constant acceleration (ms )  y    y  u t  a t2 2X 9.81 19.6 y 2 y x, y = change in displacement (m) t = time (s) 3 r GM 3 2 3 2 8  Kepler’s third r = radius of motion (m) What is the period of r GM 4 r 4 X 5X10 T 2 4 2 law T = period of motion (s) a 1012 kg comet that  T    17203s 2 2 GM 11 12 9.2.2 G = universal gravitation constant orbits at 5 X 108 m? T 4 6.67X10 X10 (6.67 X 10-11 Nm2 kg-2) M = mass of system (kg)</p><p>2 v Relativistic Lv = apparent length (m) What is the apparent 2 2 L  L 1 v (0.9c) v 0 2 L  L 1  150 1  65.4m c Length Lo = “rest” length (m) length of a spaceship v 0 2 2 Contraction v = relative velocity (m/s) of rest length 150m c c 9.2.4 c = speed of light (3 X 108 m/s) travelling at 0.9c?</p><p> t0 Relativistic time tv = apparent time (s) How much slower t t  0 1 1 v v2 dilation t = “rest” time (s) does an astronaut t v     2.29 1  o v 2 (0.9c) 2 0.19 times slower 2 9.2.4 v = relative velocity (m/s) travelling at 0.9c 1 1 c 2 2 c = speed of light (3 X 108 m/s) appear to an observer c c “at rest” Formula Name Comments Typical Problem Typical Answer F = BIℓ Magnetic force F = force (N) Calculate the force on F  BIℓ  0.1X 4X 2  0.8N on a current- B= Magnetic Field (T) 2m of wire carrying a carrying wire of I = current (A) current of 4A in a length l in a l = length (m) magnetic field of magnetic field 0.1T. 9.3.1 F I1I2 Force per unit F = force (N) What is the force per F I I 10X10  k  k 1 2  2X10 7 X  6.67X10 6 N / m l d length l = length (m) per unit unit length on two l d 3 9.3.1 I1, I2 = two currents wires, both carrying parallel=repulsive, 10A, separated by a antiparallel=attractive distance of 3m? d = separation of the two currents (m) k=magnetic constant (2 X 10-7 NC- 1m-1)     Fd torque  =torque (Nm) What is the torque on   Fd  80X 0.6  48Nm 9.3.1 F =force (N) a nut when a 0.6 m d=distance (m) spanner has a force of 80 N applied on it?   nBIAcos torque on a coil  =torque (Nm) What is the torque on   nBIAcos  200X 0.2X 3X 0.20  24Nm immersed in a n =number of turns of coil a 0.20 m2 coil of 200 magnetic field B=magnetic field (T) turns immersed in a 9.3.1 I = current (A) magnetic field of 0.2 A = area of coil immersed in T carrying a current magnetic field (m2) of 3 A? cos =angle between the coil and the magnetic field v n p p Transformer Vp = primary voltage (V) A transformer is v p n p n p vs 960X12   n    48 v n equation Vs = secondary voltage (V) required to step down ss turns. s s v s n s v p 240 9.3.4 Np = number of turns in the primary mains voltage (240V) coil to 12 V. If the Ns = number of turns in the primary coil has 960 secondary coil turns, how many turns are required in the secondary coil? Formula Name Comments Typical Problem Typical Answer F  qvB sin  Magnetic force F = force (N) What is the force on F  qvB sin  1.6X10 19 X105 X 3X sin 30°  2.4X10 14 N on a charge in a q = charge (C) an electron travelling magnetic field v = velocity (m/s) at 105 m/s in a 9.4.1 B = magnetic field (T) magnetic field of 3 T Sin  = angle between the velocity at an angle of 30 to and the magnetic field the field? V Electric Field E = Electric Field (N/C, V/m) What is the electric V 240 E  E    1.33X105 V / m d 9.4.1 V = Voltage (V) field between the d 18X10 3 d = distance (m) prongs of a mains outlet (240V) if its separation is 18 mm? E  hf Energy of a E = Energy (J) What is the energy of E  hf  6.67X10 34 X 5.1X1015  3.2X10 18 J Photon h = Planck’s constant 6.626 X 10-34 a photon of yellow 9.4.2 Js light (f=5.1X1015 f = frequency (Hz) Hz) ? c  f Wave Equation c = speed of light 3 X 108 m/s What is the frequency v 3X108 c  f  f    5.1X1015 Hz 9.4.2 f = frequency (Hz) of yellow light given  590X10 9  = wavelength (m) that its wavelength is 590 nm? HSC Course (Options) Formula Name Comments Typical Problem Typical Answer Z =  Acoustic Z = Acoustic impedance (kgm2s-1 What is the acoustic Z =   0.003X 650  1.95kgm 2 s 1 Impedance =Rayls) impedance of vaseline given 9.6.1  = acoustic density (kgm) that its acoustic density is  = speed of sound in medium 0.003 kgm and the speed of (m/s) sound in vaseline is 650 m/s? 2 2 2 I Z  Z  Reflection Ii=initial intensity What is the reflection intensity I Z  Z  (1.95  0.5) r 2 1 ( R) r 2 1 0.35  2  Intensity Io= output intensity at the interface of air (acoustic  2  2  I o Z 2  Z1  2 -1 I o Z 2  Z1  (1.95  0.5) 9.6.1 Z2 = acoustic impedance impedance = 0.5 kgm s ) and (medium 1) vaseline (Z=1.95 kgm2s-1)? Z1 = acoustic impedance (medium 2) mM d Astronomical M = absolute magnitude How far away is a star that d 1 M  m  5log( ) Distance m = relative magnitude appears m=4.5 on Earth while M  m  5 log( )d  10 5.02 10 10 9.7.4 d = distance in parsecs its absolute magnitude is –3.4? 4.5(3.4) 1 5  5 100  5.02 d  10 5.02  143par sec s</p><p>IA Ratio of intensity IA,B = intensity of objects A and B How much dimmer is Sirius A IA  100(MB  MA) /5  100(MB  MA) /5 IB 9.7.4 MA,B = absolute magnitude of A (M=-4.3) compared to Echelon IB & B (M=-2.1)? 5  5 100  5.02 4 2r 3 Kepler’s Third r = radius of motion (m) Sirius A and B orbit each other T = 61 X 365.25 X 24 X 60 X 60 m  m  1 2 GT Law T = period of motion (s) every 61 years. What is the = 1925013600 s 9.7.5 G = universal gravitation constant radius of this orbit if Sirius A 2 3 4 r GT(m1  m2 -11 2 -2 27 m m r 3 (6.67 X 10 Nm kg ) has a mass of 10 kg and 1  2    2 29 GT 4 M1+M2 = total mass of system Sirius B is 10 kg? (kg) 6.67X10 11 X1925013600(10 27 10 29 ) r  3 4 2 r = 6.9 X 109 m 1  1 1  Ryberg’s  = wavelength (m) What wavelength of light is 1  1 1  1 1  R     7 2 2 equation ni,f = quantum states (shells) produced in hydrogen by a  R 2  2  1.1X10 X ( 2  2 )  nf ni  7    4 1 9.8.1 R = Ryberg’s constant 1.10 X 10 transition between the first and  n f ni  m-1 fourth orbitals (shells)? 1    9.7X10 8 m 1 1 1.1X10 7 (  ) 4 2 12 Formula Name Comments Typical Problem Typical Answer h 34 Wavelength of a  = wavelength (m) What is the wavelength h 6.626X10 37   -     1.2X10 m mv particle h = Planck’s constant 6.626 X 10 associated with an elephant mv 1400X 4 9.8.2 34 Js (m=1400 kg) moving at 4 m/s? m = mass of particle (kg) v = speed of particle (m/s)</p><p>V Amplifier Gain Vout = output voltage (V) A BC547 transistor has a gain V out h  o V  V h  120X 60X10 3  0.72V V 9.9.6 Vin = input voltage (V) of 120. What is the output FE V o i FE in voltage if the input voltage is i 60 mV? V 0 Open Gain Loop Ao = Amplifier gain A 741 op-amp has an output V0 12 A 0  A    10.9 V  V  9.9.6 Vo = output voltage (V) voltage of 12 V when its V+ is 0 + - V+  V-  (2.3 1.2) V+ = positive input voltage (V) 2.3 V and its V- is 1.2 V. V- = negative input voltage (V) What is its gain in this configuration?</p>