Question Solved1 Answer Problem 2. (20 pts) A closed surface with dimensions a = b =0.300 m and e=0.500 m is located as in the figure below. The left edge of the closed surface is located at position x=a. The electric field throughout the region is non-uniform and given by E-(2.0+ 3.0x){N/C, where x is in meters. Calculate the net electric flux leaving the closed surface. What net charge is enclosed by the surface?

Given the wavefunctions \[ \psi_{1}=5 \sin 2 \pi(0.4 x+2 t) \] and \[ \psi_{2}=2 \sin (5 x-1.5 t) \] determine in each case the values of (a) frequency, (b) wavelength, (c) period, (d) amplitude, (e) phase velocity, and (f) direction of motion. Time is in seconds and \( x \) is in meters.

Problem 4.11 Part A Use the node-voltage method to find v1 in the circuit(Figure 1) if v 130 V, i 2.2 A Express your answer with the appropriate units. 四? Value Units Submit Request Answer Part B Use the node-voltage method to find 2 in the circuit. Express your answer with the appropriate units. Figure 1 of 1 oz= Value Units Submit Request Answer 4Ω 80Ω Problem 4.11 Part A Use the node-voltage method to find v1 in the circuit(Figure 1) if v 130 V, i 2.2 A Express your answer with the appropriate units. 四? Value Units Submit Request Answer Part B Use the node-voltage method to find 2 in the circuit. Express your answer with the appropriate units. Figure 1 of 1 oz= Value Units Submit Request Answer 4Ω 80Ω Provide Feedback t) 102 5Ω

5. Evaluate the following integrals: a. Sg(+ b)8(t – t)dt -00 b. $(2t? + 8)8(2 – t)dt c. S. cos (2x – 1)]8(3x – 2)dx - Hint: 8(x) is located at x = 0. For example, 8(1-t) is located at 1 – t=0, that is, at t = 1, and so on.

How much heat is released when 36,0 g of water at 20.0°C is cooled to form ice at 20.0°C? Specific heat capacity of H2O(s) - 2.09 J/g.°C Specific heat capacity of H2O(l) - 4.18 J/g.°C Specific heat capacity of H2O(g) - 1.84 J/g.°C Molar mass of H2O - 18.0 g/mol AHtus -6.02kJ/mol AHvap 40.7kJ/mol . . 14.3 kJ 16.6 kJ 18.8 kJ 21.1 kJ 23.3 kJ

Please assist me with the bottom part ONLY! 5Br (aq) + BrO3(aq) + 6H+(aq) + 3Br2(1) + 3H20(1) The above reaction is expected to obey the mechanism: BrO3(aq) + H+(aq) HBrO3(aq) Fast equilibrium HBrO3(aq) + H+ (aq) H2BrO3+ (aq) Fast equilibrium Slow H2BrO3+ (aq) + Br (aq) (Br-BrO2)(aq) + H20 (1) (Br-Bro2)(aq) + 4H+(aq) + 4Br"(aq) kes , products Fast Choose, from the list below, correct expressions for the overall rate law which are completely consistent with the above mechanism. Correct: - d[Br"]/dt = k[Br"][H+]2[BrO3-] Correct: -d[BrO3-]/dt = k[BrO3"][Br"][+]2 Incorrect: d[Br-Bro2]/dt = k[Br"][BrO3"][H+]2 Incorrect: k[BrO 3 ) = -d[Br"]/dt Correct: -d[H+]/dt = k[H+]2[BrO 3 ][Br"] Incorrect: -d[Br"]/dt = k[Br"][BrO 3 ][H+] You are correct. Previous Tries Your receipt no. is 151-3441 For each of the given rate expressions choose the correct expression for the rate constant (k) from the list below. A. K = 5x[(K1+ka+k3)/(k-1+k-2)] B. K = 5x(k1k2k3-k-1k-2) C. K = 6xk3 D. k = [(k1k2k3)/(k-1k-2)] E. K = 6x([(k1k2k3)/(k-1k-2)]) F. none of the above -d[BrO3']/dt -d[Br"]/dt -d[H+]/dt

Point charge q1 = -5.00 nC is on the x-axis at x = -0.400 m. Point P is on the x-axis at x=0.200m. Point charge q2 is at the origin. What are the sign and magnitude of q2 if the resultant electric field at point P is zero? Express your answer in nanocoulombs.