Question Solved1 Answer \( * 12-64 \). The boat is originally traveling at a speed of \( 8 \mathrm{~m} / \mathrm{s} \) when it is subjected to the acceleration shown in the graph. Determine the boat's maximum speed and the time \( t \) when it stops.

23Urgent. Please help answer fast. Please do not round up the decimal, and also, please write three or four decimals, including zero. Thank you Two identical cylindrical vessels with their bases at the same level each contain a liquid of density 1.00 g/cm². The area of each base is 4.76 cm?, but in one vessel the liquid height is 0.822 m and in the other it is 1.69 m. Find the work done by the gravitational force in equalizing the levels when the two vessels are connected. Number i Units

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.