5. a)The angular velocity is 500x2x3.143=524 rads-1.
b) Angular velocity of the earth is 2x3.142/24x3600= 7.27 x 10-5 rads-1.
centripetal acc = w2R = 0.0337 ms-2.
c) The acceleration due to gravity is diff as there is centripetal acc due to the rotation due to Earth at the equator but not at the poles.
ii) The difference is small because the centripetal acc is small compare to acceleration due to gravity.
d) force per unit mass is g. Hence g = GM/R2
e) w2R3 = GM ..(1)but M is not known . But 9.81 = GM/Re2, subs GM = 9.81x (6.38 x 107)2 into the equation (1). Hence R3 = GM/w2. R = 4.23 x 107m
ii) The satellite must be above the equator and the rotate from West to East.
f) The polar orbit - obtain data from the whole Earth surface and stronger/ no delay in data transmission.
The geostat orbit - no tracking required and greater area coverage for each satellite.
6. Must use 273.15K and leave answer to two dec places i.e 195.85 C
The absolute scale is a theoretical scale that does not depend on any thermometric property.
The internal energy of a gas is the sum of ke and pe of all the molecules.
The ideal gas obeys the ideal gas equation at all pressure , volume and temperature.
Using pV=nRT , n = 0.3 mole.
ii) The portable supply can deliver until the pressure is 3.23 x 105 Pa which is the pressure of the tyre when filled. Hence the portable supply has n = 2.6 mol of gas available to fill the tyres. But 4 tyres only need 1.2 moles. Hence the portable supply has more than enough air and the pressure will still be above 3.23 x 105 Pa.
Alternatively find the pressure after n=1.2 mol is used. The pressure of supply is still above 3.23 x 105Pa.
e) Internal energy is 3/2 kT.
ii) The Internal energy of one mole = Na x Internal energy of one molecule.
Hence U = 3720 J
iii) increase in internal energy =3/2 x (increase in n) R T = 1100 J
f) Increase in internal energy is the sum of heat supplied and work done on the system.
ii) Hence increase in internal energy is equal to work done on the system as no heat is supplied.
W = 1100 J.
7a) Photoelectric effect refers to the emission of electrons from a metal surface when irradiated with light of sufficiently high frequeny.
Evidence : 1. No emission is possible no matter how high the intensity when f is below threshold freq. This can only be attributed to the fact that light is a photon (particle) . The photon energy is too small due to low freq hence no emission . The intensity increase the energy but the photon energy did not increase. Increasing the intensity only increases the rate of photons incident on surface. When freq is above the threshold , the KE of the electron is independant of the intensity but dependant on the freq, ie. energy of the photon. Increasing the intensity increases the rate of photoelectrons emitted but the KE of the photoelectron depends only of freq.
b) Equation: E = Work function + KE max
E is the photon energy
Work function is the energy required to release the electron from the metal surface.
and KEmax is the maximum kinetic energy of the photoelectron.
c) work function = hf - KE = 3.06 x 10-19 J.
d) length of pulse = vt = 3000 m
ii) uncertainty in position= 3000 m
iii) uncertainty in momentum = 2 x 10-38 kgms-1
e) Draw the wavefunction as it passes through 3 regions I, II and III
where E is the energy of the particle and U is the potential barrier. The amplitude of the wavefunction reduces in region II and III as it passes the potential barrier . This means the probability of finding the particle in region II and III is non -zero but decreased ( probability = square of wavefunction)
