sc × t C D www.saplinglearning.com/ibiscms/mod/ibis/view.php?id=4373913 Sapling

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sc × t C D www.saplinglearning.com/ibiscms/mod/ibis/view.php?id=4373913 Sapling Learning macmillon laaning C) Jump to.. DeAngelo Vaughn-Allen Available From: Due Date: Points Possible: Grade Category: Description: Policies: 10/30/2017 08:00 AM 11/13/2017 10:00 PM 100 Graded Map Sapling Learning Two skydivers are holding on to eech other while falling straight down at a common terminal speed of 61.90 mls Suddenly, they push away from each other. Immediately after separation, the first skydiver who has a mass of 89.30 kg) has the following velocity components (with straight down' corresponding to the positive z-axis) Rubric 1 -4430 m s 4.750 m/s 1-61.90 m/s You can check your answers. You can view solutions after th due date. You can keep trying lo answer each questior until you get it right or give up. There is no penalty for incorrect answers. For multiple-choice questions, the penalty depends on the number of choices available What are thex- and y-components of the velocity of the second skydiver, whose mass is 57.70 kg immediately after separation? 10 Incorrect The initial momerntum in the x-direction is zero, so both skydivers cannot have a positive momentum 12 2 Im sin the x-direction after they separate 13 Numoer O eTextbook 7.351/ 14 2 15 D Help With This Topic What is the change in kinetic energy of the system? 16 2 Numocr O Web Help &Videos; 17 159030.64 Joules 18 O Technical Support and Bug Reports 19 3 Type here to search :25 PM 11/13/2017 : -

Explanation / Answer

Applying momentum conservation,

pi = pf

(89.30 + 57.70)(61.90)z^ = 89.30 (4.430 x^ + 4.750 y^ + 61.90 z^2) + (57.70)(v2x x^ + v2y y^2 + v2z z^)

along x^,

v2x = - 6.856 m/s

v2y = - 7.351 m/s

Ki = (89.30 + 57.70) (61.90^2)/2 = 281623.34 J

Kf = 89.30(4.430^2 + 4.750^2 + 61.90^2)/2 + 57.70(6.856^2 + 7.351^2 + 61.90^2)/2

Kf = 286422.06 J

delta(K) = Kf - Ki = 4800 J .......Ans

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