QUESTION 5 (OF) = (C) x 95 + 32 ( C) = ((OF)-32) x 59 Kelvin (K) = (C) + 273.15 (C) = (K) - 273.15 T = 34 OF T = ?K Enter your numeric result in the usual form with 3 significant digits. 12 points Save Answer QUESTION 6 (OF) = (C) x 95 + 32 ('C) = ((OF) - 32) x 5/9 Kelvin (K) = (C) + 273.15 (.C) = (K) - 273.15 T = 552 K T = ? 'C Enter your numeric result in the usual form with 3 significant digits. 13 points Save AnswerQUESTION 7 The bridge is exposed to temperatures ranging from 15C to 40C. The change in length between these temperatures is 0.429 m. What is the length of the main span of a Bridge in m?. Coefficient of linear expansion at: 12x10'6(1/"C) 12 points Save Answer QUESTION 8 The main span of a Bridge is 8855 m long at its coldest. The bridge is exposed to temperatures ranging from 15C to 40C. What is the change in length between these temperatures? Coefficient of linear expansion at: 12x10'6(1/"C) : 12 points Save Answer QUESTION 6' For the situation shown in Figure calculate FL , the magnitude of the force exerted by the left hand. The hands are 0.994 m apart, and the CG of the pole is 0.375 m from the left hand. The mass of the pole is 3.79 kg. Input your answer in N using 3 signicant gures. 9:980 m/s2 Center of gravity (CG) is the point where the total weight of the body is assumed to be concentrated. 10 points Save Answer QUESTION 1 (F) = (C) x 9/5 + 32 (\"Q = ((F) - 32) x 5/9 Kelvin (K) = (\"C) + 273.15 ('0) = (K) - 273.15 T=-83C T=?F Enter your numeric result in the usual form with 3 significant digits. 13 points Save Answer QUESTION 2 (F) = (\"C) x 9/5 + 32 (\"Q = ((F) 7 32) x 5'9 Kelvin (K) = (C) + 273.15 (C) = (K) 273.15 T=1C T=?K Enter your numeric result in the usual form with 3 significant digits. 13 points Save Answer QUESTION 3 (F) = (\"0) x 96 + 32 (\"C) = ((F) 32) x 5'9 Kelvin (K) = (C) + 273.15 (\"0) = (K) 273.15 T=583K T=?F Enter your numeric result in the usual form with 3 significant digits. 13 points Save Answer QUESTION 4 (\"H = (C) x 9'5 + 32 (\"C) = (W) 32) x 5/9 Kelvin (K) = (C) + 273.15 (\"0) = (K) 273.15 T=58F T=?C Enter your numeric result in the usual form with 3 significant digits. 1 2 points Save Answer QUESTION 3 1 An arrow is shot from a height h toward a cliff. It is shot with a velocity V at an angle of @ above the horizontal. It lands on the top edge of the cliff time t later. What is the height of the cliff? Oh+ tV sin+ 812/2, where g = 9.80 m/s2 Oh + t V cose - 81212, where g = 9.80 m/s2 Oh+ t V sino - 81212, where g = 9.80 m/s2 Oh + t V sino - 812/2, where g = -9.80 m/s2 10 points Save AnswerQUESTION 1' A 64.1-kg skier with an initial speed of 32.2 m/s coasts up a 4.30-m-high rise as shown in Figure. The angle 0 =35. Find her final speed at the top, given that the coefficient of friction between her skis and the snow is 0.0651. (Hint: Find the distance traveled up the incline assuming a straight-line path as shown in the figure.) Input your answer in m/s using 3 significant figures and exponential notation. VI = ? KE 10 points Save Answer QUESTION 2 A wave with the wave length ^ and the frequency f propagates with the speed v. A = 5.72 m, V = 46.9 m/s f = ? Hz, 10 points Save AnswerQUESTION 71 A mass that can oscillate without friction on a horizontal surface is attached to a horizontal spring and is pulled to the right and is released from rest at the time to=0. The period of the oscillation is T. The speed of the mass is V at the time t. What is the amplitude of the oscillation? T = 2.22s, V = 4.25m/s, t=2.72s. 10 points Save Answer QUESTION 81 The object moves in a straight line and makes two-step displacement: two consecutive displacements d, and d2. The object made the first displacement d, with the velocity V, and the second displacement d2 with the velocity V2. Find the average velocity for this total two-step displacement. O (d , + d ) 1 ( v , Id , + v 2 l d ) O ( v, / d, + v , / d ) 1 ( 1 / d , + 1 / d , ) O ( d, + d, ) 1 ( d , / v , + d , / v 2 ) O ( v , + v , ) 12 O ( d, v , + d, v2 ) I (d , + d, ) 10 points Save AnswerQUESTION 91 x =x, tv +-at2 *= 89.6 m v = 5.13 m/s 1= 2.34 s a = 1.41 m/s2 x = ? in m Input answer using 3 significant figures. 10 points Save Answer QUESTION 10 1/2mv, 2 + mgh, + w= 1/2mv,2+ mghz m = 2.65 kg V, = 5.20 m/s h = 2.33 m V, = 6.35 m/s h, = 8.41 m W = ? J Input your answer using 3 significant figures. 10 points Save AnswerQUESTION 4 ! 01 = 17, 02 = 16, T1 = 79.5 N, g=9.80 m/s2, m = ? kg Enter your answer using 3 significant figures. To T1 12 m W 10 points Save Answer QUESTION 5 Some point masses m1 and m2, located at a distance / =35 m from each other, are attracted to each other with a force of 5 N. What would be the distance between these masses to attract each other with a force of 86 N? Input your answer in N using scientific/exponential notation and 3 significant figures. 10 points Save