Good evening! Kindly answer these three questions accurately:

(4%) Problem 23: The coordinate axes of reference frame S' are parallel to those of reference frame S, and, at time = 0.00 s, the origins coincide. The velocity of frame 5" with respect to frame S 15 constant and equal to rerg = 40.534 3345 rgrg=4053m Correct! gg = (116 m/s) i + (245 m/s) j + (5.48 m/s) k 5% Part (a) What is the distance, in meters, between the origins of frames and ' at = 6.63 s? 25% Part (b) Suppose that at time fy = 0.00 s. particle P is at the origin of both frames, and ', and its velocity in frame ' is constant and equal to (7.64 m/s) i . How far, in meters, 1s the particle from the origin of reference frame at time = 6.63 s7 rps =7.640 m 25% Part (c) What is the magnitude, in meters per second, of the velocitv of the particle in reference frame 57 wes = 2.000 le{i X Attempts Remain X Attempts Femain Ay, 25% Part (d) What 1s the acceleration, in meters per squared second, of the particle in reference frame S7 Degrees () Radians Submit Hint | I give up! Grade Summary Deductions Potential 18005 Late Work %0 100% Late Potential 100% Submissions Aftempts remaining: 000 (0% per attempt) detailed view (4%) Problem 9: A secret agent skis off a slope inclined at # = 32.6 degrees below horizontal at a speed of vy = 73.3 m/s. He must clear a gorge, and the slope on the other side of the gorge 1s h=11.7m below the edge of the upper slope. - l = 4 What 1s the maximum width, w, of the gorge (in meters) so that the agent clears 1t7 w= 19345 Degrees (' Radians ol i | e et ] Hints: | fora deduction. Hints remaining: 0 Feedback: | fora deduction -If upward is the positive y direction, then @, = -8.81 m - anda =0 Ussy=y,+ No meaningful feedback available for the corrent submission. Vit = (1 2]::[.!1 to calenlate the time it takes to descent vertically the given height. Then use x =x; = vy, to caleulate the horizontal distance covered in that time. Grade Summary Deductions Potential 10005 Late Work %0 100% Late Potential = 100% Submissions Atterapts remaining: 9073 (0% per attempt) detailed view 1 =l ba (4%) Problem 8: A car moves along a horizontal road with constant velocity - vy = Vpg i = (41.6 m/s) i until 1t encounters a smooth inclined hill. It climbs the hill with constant velocity - T = Uz ; + 'Ul,y.}' (28.2m/s) i + (3.23 m/s) j as indicated in the figure. The period of time during which the car changes its velocity is At=1.7T4s 30% Part (a) Enter an expression, in Cartesian umt-vector notation, for the average acceleration of the car during the given ime period using the symbols provided. Bavg ={ Viy - Vg ) UM+ (v, ) At & Correct! # 0% Part (b) What 1s the magmtude, in meters per squared second, of the car's average acceleration during the given time period. = Grade Summary Qavg = |1.955 m/s Deductions ) Potential 10044 Late Work %0 100% Late Potential ~ 100% Submissions Attempts remaining: 900 (0% per attempt) 5T detailed view 1 Degrees O Radians S s e | o