Various Math #59 - Q&A

Question 1 The unstable nucleus uranium-236 can be regarded as auniformly charged sphere of charge Q = +92e and radius R = 7.4 × 10⁻¹⁵ m. In nuclear fission, this can divide into twosmaller nuclei, each of 1/2 the charge and 1/2 the voume of theoriginal uranium-236 nucleus. This is one of the reactionsthat occurred n the nuclear weapon that exploded over Hiroshima, Japan in August 1945. A. Find the radii of the two "daughter" nuclei of charge+46e. B. In a simple model for the fission process, immediatelyafter the uranium-236 nucleus has undergone fission the "daughter"nuclei are at rest and just touching. Calculate the kineticenergy that each of the "daughter" nuclei will have when they arevery far apart. C. In this model the sum of the kinetic energies of the two daughter" nuclei is the energy released by the fission of oneuranium-236 nucleus. Calculate the energy released by thefission of 10.0 kg of uranium-236. The atomic mass ofuranium-236 is 236 u, where 1 u = 1 atomic mass unit = 1.66 × 10⁻²⁷ kg. Express your answer both in joules and in kilotonsof TNT (1 kiloton of TNT releases 4.18 × 10¹² J when itexplodes). Answer a.) The two daughter nuclei have half the volume of the originaluranium nucleus, so their radii are smaller by a factor of ∛{2} r = 7.4 × 10⁻¹⁵/∛2 = 5.9 x 10⁻¹⁵ m b) U = k(46e)² / 2r = (k(46)²(1.6 × 10⁻¹⁹)² / 2r ) = 1.17 × 10⁻¹⁴ / 2r = 4.14 × 10⁻¹¹ J Each daughter has half of the potential energy turn into itskinetic energy when far from each other, so K = U/2 = 4.14 × 10⁻¹¹/2 = 2.07 × 10⁻¹¹ J c) If we have 10.0 kg of uranium, then the number of nucleiis n = 10 / (236u * 1.66 × 10⁻²⁷) = 2.55 x 10²⁵ nuclei And each releases energy: E = nU = (2.55 × 10²⁵) * (4.14 × 10⁻¹¹) = 1.06 × 10¹⁵ J = 253 kilotons of TNT d.) We could call an atomic bomb an "electric" bomb sincethe electric potential energy provides the kinetic energy of the particles Question 2 The x and y components of a vector r are rx = 14 m and ry = -95 m respectively. Find the direction and the magnitude of the vector r. Answer Magnitude of r = √(rx² + ry²) = √(14² + (−95)²) = 96.02m Direction Angle = tan⁻¹(ry/rx) = tan⁻¹(-95/14) = -81.61 Question 3 The normal boiling point of benzene is 80 degrees celsius. What is the vapor pressure of benzene at 80 degress celsius? Answer according to raoults law the vapour pressure of benzene Po = P * x Po = vapoure pressure P1 is the vapour pressure x = mole fraction of solvent =1 Po = P1 = 760 [mmHg] Question 4 A person stands in a stationary canoe and throws a 5.22 kgstone with a velocity of 8.20 m/s at an angle of 34.0 degrees abovethe horizontal. The person and the canoe have a combined massof 105 kg. Ignoring air resistance and effects of the water,find the horizontal recoil velocity (magnitude and direction) of the canoe. Answer Conservation of linear momentum: (m₁ + m₂)vᵢ = m₁v₁f + m₂v₂f 0 = (5.22)(8.2)cos 34 + 105v₂f Solve for v₂f. The answer will be negative, meaning the velocity of the canoe will be in the direction oppositethat of the stone. Question 5 Calculate the following limits using the factorization formula xⁿ - aⁿ = (x – a)(xⁿ⁻¹ + axⁿ⁻² + a²xⁿ⁻³ +…+ aⁿ⁻²x + aⁿ⁻¹), where n is a positive integer a is a real number. limx→2 (x⁵ - 32) / (x - 2) Answer Step 1 To find: limx→2 (x⁵ - 32) / (x - 2) Using the formula: xⁿ - aⁿ = (x – a)(xⁿ⁻¹ + axⁿ⁻² + a²xⁿ⁻³ +…+ aⁿ⁻²x + aⁿ⁻¹) Step 2 limx→2 (x⁵ - 32) / (x - 2) x⁵ - 32 can be expanded as follows: x⁵ - 32 = x⁵ - 2⁵ ⇒ x⁵ - 32 = (x - 2)(x⁴+2x³ + 2²x² + 2³x + 2⁴) ⇒ x⁵ - 32 = (x - 2)(x⁴+2x³ + 4x² + 8x + 16) Hence we get: ⇒ limx→2 (x⁵ - 32) / (x - 2) = limx→2 (x - 2)(x⁴+2x³ + 4x² + 8x + 16) / (x - 2) Step 3 Cancelling the like terms, we get: ⇒ limx→2 (x⁵ - 32) / (x - 2) = limx→2(x⁴+2x³ + 4x² + 8x + 16) Plugging the limit, we get: ⇒ limx→2 (x⁵ - 32) / (x - 2) = 2⁴ + 2(2³) + 4(2²) + 8(2) + 16 ⇒ limx→2 (x⁵ - 32) / (x - 2) = 16 + 16 + 16 + 16 + 16 ⇒ limx→2 (x⁵ - 32) / (x - 2) = 80