Various Math #22 - Q&A

1231231Question 1 Two 2.1cm diameter disks face each other, 2.9mm apart. They are charged to ±10nC. a) what is the electric field strength between two disks? b) a proton is shot from the negative disk toward the positive disk. What launch speed must the proton have to just barely reach the positive disk? Answer 1 Step 1 Given: Diameter of disk = 2.1cm Radius of disk = 1.05cm = 1.05 x 10⁻²m Distance between two disks, s = 2.9mm = 2.9 × 10⁻³m Charge on disks = ±10nC = ±10 × 10⁻⁹C Permitivity of free space, ξo = 8.854 × 10⁻¹² F/m Step 2 (a) Electric field strength can be calculated as E = σ / ξo Where, σ is the surface charge density σ = Q / A Where A is the area A = πr² Therefore, E = Q / (ξoπr²) = 10 × 10⁻⁹C / (8.854 × 10⁻¹² * (1.05 × 10⁻²)²) = 3.26 x 10⁶ N/C Step 3 Electric force = qE Also, Force = ma Since, both the force experienced by the same charge particle Therefore, qE = ma a = qE/m where m is the mass of the proton m = 1.67 × 10⁻²⁷kg and e is the charge of the proton a = (1.6 x 10⁻¹⁹ × 3.26 × 10⁶) / 1.673 × 10⁻²⁷ a = 3.12 x 10¹⁴ m/s² Step 4 Using the equation of motion s = ut + 0.5at² initial velocity of the particle, u = 0 s = 0.5at² t = √(2s/a) t = √(2 × 2.9 × 10⁻³ / 3.12 × 10¹⁴) = 4.3 x 10⁻⁹s Step 5 (b) Speed of the proton Again using the equation of motion v = u + at v = 0 + 3.12 × 10¹⁴ × 4.3 × 10⁻⁹ = 13.42 x 10⁵ m/s Step 6 Answer; (a) Electric field strength between two disks = 3.26 × 10⁶ N/C (b) launch speed of the proton to just barely reach the positive disk = 13.42 × 10⁵ m/s Answer 2 Explanation: The two disks form a parallel-plate capacitor, which will cause an electric field equal to: E = Q / (ε₀A) Where Q is the charge of the disks, A is the area of the disks and ε₀ is the vacuum permissivity equal to 8.8510⁻¹² C²/Nm²: E = (810⁻⁹C) / (π(0.02m)²) * 8.8510⁻¹² C²/Nm² E = 2.8810⁶ N/C Now, for the second part of the problem, we can use conservation of energy. The addition of potential and kinetic energy at launch point should be equal to the addition at the positive disk. Because the proton has positive charge, the potential energy of the proton will increase as its distance to the negative disk increases too. This is because the proton will be attracted towards the negative disk. The potential energy is given by: Ep = V * q Where V is the difference in potential (voltage) between the disks. In a parallel-plate capacitor: V = E * d, where d is the difference in position with the frame of reference. Our frame of reference will be the negative disk. q is the charge of the proton. The kinetic energy is given by: Ek = (1/2)mv² Then: Ep1 + Ek1 = Ep2 + Ek2 V₁ * q + (1/2)mv1² = V₂ * q + (1/2)mv2², Ep1 = 0 (1/2)mv1² = E * d * q v = √(2Edq/m) v = √((2 * 2.88 * 10⁶ * 0.001m * 1.6*10⁻¹⁹C) / 1.67 * 10⁻²⁷) v = 0.742 * 10⁶ m/s Answer 3 (A) electric field between to charged disk = E = Q/Aξo = charge on each disk = 9.0nC r = disk radius = 1.5e – 02m A = surface area of each disk = πr² = 0.0007069m² ξo = 8.854 × 10⁻¹²C²/N m² ⇒ E = Q/Aξo = 1.438e+06 N/C (B) kinetic energy required by the proton to move from negative disk to positive disk = K K = qΔV = qEd q = 1.602e – 19C E = 1.438e+06 d = gap between disk = 1.0mm ⇒ K = 2.304e – 016J if the launch speed = v K = √(2K/m); mass of proton = m = 1.673e - 027kg v = √(2K/m) = 5.249e + 05 m/s Question 2 A proton with an initial speed of 900,000 m/s is brought to rest by an electric field. What was the potential difference that stopped the proton? Express your answer to two significant figures and include the appropriate units. What was the initial kinetic energy of the proton, in electron volts? Express your answer using two significant figures. Answer 1 Step 1 Solution: a) (1/2)mpv² = qV ⇒ V = (mpv²)/(2q) V = (1.67 × 10⁻²⁷ × (9 × 10⁵)²) / (2 × 1.6 × 10⁻¹⁹) = 4227 * 10⁵ V b) Initial kinetic energy of proton = 4227 × 19eV Answer 2 Step 1 Initial speed of proton vi = 900,000 m/s final speedf problem vf = 0 Now, vi = 900,000 m/s vf = 0 Kinetic energy KEi = (1/2)mpvi² KEf = (1/2)mvf² = 0 Potential Energy PEi = 0 PEf =? Here, from Conservation of Energy. Kinetic Energy is Converted in to Potential Energy (1/2)mpvi² = PEf (1/2) × 1.67 × 10⁻²⁷ × (900000)² = PEf = 677 × 10⁻¹⁸ J Tunles [kg((m/s)²)] Now, Potential difference Δv = ΔPE/q = (PEf - PEi)/q Step 2 Potential difference to stop the proton given by Δv = PEf/q (.: PEi = 0) = 677 × 10⁻¹⁸ Ton < 1.6 x 10⁻¹⁹C Δv = 4228 V b) Now, initial kinetic energy KEi = (1/2)mpvi² KEi = 677.4 × 10⁻¹⁸T < v = 1.6 × 10⁻¹⁹T .. Kinetic energy is in ev is 4228 ev i.e. 4228 ev Question 3 A scalloped hammerhead shark swims at a steady speed of 1.0 m/sm/s with its 83 cm - cm-wide head perpendicular to the earth's 52µT magnetic field. What is the magnitude of the emf induced between the two sides of the shark's head? Express your answer using two significant figures. Answer 1 Step 1 Given: The velocity of hammerhead shark (v) = 1.0 m/s Length of hammerhead (L) = 83cm. the magnetic field of the earth (B) = 52µT the angle between the hammerhead and the earth's magnetic field = 90°. Step 2 the magnitude of electromotive force is given by, Femf = BL × v Femf = BLusinθ Femf = 52 × 10⁻⁶ x 83 × 10⁻² x1 Femf = 4.316 × 10⁻⁵N Step 3 Answer: electromotive force = 4.316 × 10⁻⁵N Answer 2 Given: The speed of scalloped hammerhead shark is: va = 1.0 m/s. The width of head is: wa = 83cm = 0.83m. The magnetic field is: Ba = 52µT = 53 × 10⁻⁶Τ. The expression for the change in area through which magnetic field moves is, ΔAa = Δlaw a Here Δla and wa are change in length and width respectively. The expression for the speed of scalloped hammerhead shark is, va = Δla/Δt Here Δt is time interval. The expression for the magnetic flux is, Δφa = BaΔA The expression for the induced emf by using the Faradays Answer 3 Answer: 7.4 × 10⁻⁵T. Explanation: Emf induced = BLv Where B is magnetic field, L is length of rod or any other object moving in magnetic field and v is velocity of moving rod. Here B = 56 × 10⁻⁶T L = 88 × 10⁻²m v = 1.5m/s emf induced = 56 × 10⁻⁶ x 88 × 10⁻² × 1.5 = 7.4 x 10⁻⁵т. Question 4 Consider the decomposition of a metal oxide to its elements, where M represents a generic metal. M2O3(s) ⇒ 2M(s) + (3/2)O2(g) info given for Gf (kJ/mol): M2O3-6.70 M(s) = 0 O2(g) = 0 what is the standard change in Gibbs energy for r x n as written in forward direction? (kJ/mol) What is the equilibrium constant (K) of this r × n, as written in forward direction at 298K? What is the equilibrium pressure of O2(g) over M(s) at 298K? (atm) Answer 1 Step 1 Given reaction: M2O3(s) ⇒ 2M(s) + (3/2)O2(g) Given: M2O3 = -6.70 kJ/mol M(s) = 0 O2(g) = 0 Calculation: ΔG°rxn = Σ ΔG°products - Σ ΔG°reactants = [(2×0)+(3/2×0)] - [1 × -6.70] ΔG°rxn = 6.70 kJ/mol Step 2 Given: T = 298K R = gas constant (8.314 J/mol K) ΔG° = 6.7 kJ/mol = 6.7kJ x 1000 J/kJ ΔG° = 6700 J/mol Calculation: ln K = -ΔG°/RT = -6700/(8.314 * 298K) ln K = -2.704 K = e^(-2.704) K = 0.0669 Step 3 M2O3(s) ⇒ 2M(s)+ (3/2)O2(g) K = p(O2(g))^3/2 p(O2(g)) = K^2/3 K = 0.0669 p(O2(g)) = (0.0669)^2/3 p(O2(g)) = 0.164atm Answer 2 Step 1 M2O3(s) ≤ 2M(s) + (3/2)O2(g) ΔG° = ΣηΔG°products - ΣηΔG°reactants ⇒ 2(0) + (3/2)(0) - (-6.7) ΔG° = 6.7 kJ/mol Step 2 ΔG° = -RTln(K) ln(K) = -ΔG°/RT ln(K) = -6.7 x 10³/ 8.314 * 298 = -2.704 K = 0.067 Step 3 ΔG° = -RTln(Kp) Kp = (Po₂)^3/2 ΔG° = -RTln[(Po₂)^3/2] ln[(Po₂)^3/2] = -ΔG°/RT ln[(Po₂)] = -2.704 (Po₂)^3/2 = e^-2.704 (Po₂) = 0.067 Po₂ = 0.017atm Question 5 Consider the following solvent pairs when mixed: Which are immiscible/miscible? If the solvents are immiscible, which solvent would be in the top layer? a. hexane and water b. methylene chloride and water c. ethanol and water d. ether and water Answer 1 a. hexane and water immiscible b. methylene chloride and water miscible c. ethanol and water miscible d. ether and water immiscible Answer 2 hexane and water immiscible methylene chloride and water immiscible c and d are miscible