Memorandum for Physics 2019 SOLUTIONS:

kQq From Coulomb’s Law: F = the magnitude of the original force 2 d k×12×6 72k F = = and after contact the magnitude of the new force 0 2 2 d d k× 3×3 9k F = = So the new force is 1/8th of the original. [...] So in effect the voltmeter is measuring the PD across the terminals of the cell, ie 4V (maybe 3.99999999V) 38 With S closed, the current will flow through the two resistors and the voltmeter will measure the PD across the 3Ω resistor. [...] G R 100 So the resistance of the shunt will such that the ratio of the resistors in parallel will be such that the shunt S will carry the bulk of the current: 5A G -4 -4 S 5×10 5×10 ×100 = so S = = 0.01Ω 100 100 5 5 S 40 The RMS current = to the equivalent DC current. [...] h 49 de Broglie wavelength = λ = As both the alpha particle and the proton have been P accelerated through the same PD, the alpha particle picks up twice the energy, (q = 2) So Eα = 2EP and mα = 4mP 2 So for the alpha particle P = √and for the proton we have P = 2m E P P 2 P Now E = so PP = √(2mE) and Pα = 2 √(2 x 2mE) = 2√2PP 2m 1 λα = λP 2 2 50 Frequency of blue light is ~ 7.5 x 1014 Hz Frequenc. [...] The bond length of the C – I bond is greater than that of the C- Cℓ bond which also causes the latter to be stronger, but the primary factor is the difference in electronegativity allowing the bond between C and Cℓ to be shorter and stronger.