This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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Here are the solutions to your questions. 2. Temperature distribution along a metal bar For both cases, assume the heated end is at x=0 with temperature T_H, and the end dipped in melting ice is at x=L with temperature 0^ C. The temperature T is plotted on the y-axis and the distance x along the bar on the x-axis. a) Lagged and the other end dipped in melting ice: The graph will be a straight line connecting the point (0, T_H) to (L, 0^ C). This indicates a uniform temperature gradient along the bar because there is no heat loss to the surroundings, so the rate of heat flow is constant. b) Unlagged and the other end dipped in melting ice: The graph will be a curve connecting the point (0, T_H) to (L, 0^ C). Due to heat loss from the surface of the unlagged bar to the surroundings, the rate of heat flow decreases along the bar. This results in a temperature gradient that is steeper near the heated end and less steep near the cold end. The curve will be concave up (or convex down). 3. Photoelectric effect for tungsten Given: Wavelength of light = 248 nm = 248 × 10^-9 m Maximum kinetic energy KE_max = 8.6 × 10^-20 J Constants: Planck's constant h = 6.626 × 10^-34 J s Speed of light c = 3.00 × 10^8 m/s Charge of an electron e = 1.602 × 10^-19 C (1 eV = 1.602 × 10^-19 J) The photoelectric equation is KE_max = E_photon - , where E_photon = (hc)/() and is the work function. i) The work function of the tungsten in eV Step 1: Calculate the energy of the incident photon (E_photon). E_photon = (hc)/() E_photon = (6.626 × 10^-34 J s)(3.00 × 10^8 m/s)248 × 10^-9 m E_photon = 19.878 × 10^-26248 × 10^-9 J E_photon = 8.015 × 10^-19 J Step 2: Calculate the work function () in Joules. = E_photon - KE_max = (8.015 × 10^-19 J) - (8.6 × 10^-20 J) = (80.15 × 10^-20 J) - (8.6 × 10^-20 J) = 71.55 × 10^-20 J = 7.155 × 10^-19 J Step 3: Convert the work function from Joules to electron volts (eV). _eV = 7.155 × 10^-19 J1.602 × 10^-19 J/eV _eV = 4.466 eV The work function of tungsten is 4.47 eV. ii) The threshold frequency of the tungsten The threshold frequency (f_0) is related to the work function by = hf_0. Step 1: Calculate the threshold frequency using the work function in Joules. f_0 = ()/(h) f_0 = 7.155 × 10^-19 J6.626 × 10^-34 J s f_0 = 1.080 × 10^15 Hz The threshold frequency of tungsten is 1.08 × 10^15 Hz. That's 2 down. 3 left today — send the next one.