3 questions:

Question 13 The magnitude of the impulse experienced by the airbag landing system during its first collision with the surface of the Red Planet was Not yet answered Select one: Marked out of O a. 1.4 x 10* N.5 1.0 O b. 8.8 x 103 N.S Flag question O c. 5.5 x 105 N.S O d. 3.4 x 103 N.S Question 14 The collision of the airbag landing system with the surface of the Red Planet is classified as an _i_collision because _ii_ conserved. Not yet answered Select one: Marked out of O a. i = elastic, ii = momentum is conserved 1.0 O b. i = elastic, ii = kinetic energy is P Flag question O c. i = inelastic, ii = momentum is not conserved . i = inelastic, ii = kinetic energy is not conservedQuestion 15 The following diagram illustrates a collision of Ball A and Ball B. Not yet answered Marked out of Before Collision 5.0 After Collision Flag question MA= 5.00 kg MA= 5.00 kg V'A = 1.7 m/s VA = 3.0 m/s A Z A WE S MB= 6.00 kg B VB = 2.5 m/s V'B B Determine the final velocity of Ball B, sketch a vector addition diagram consistent with the vector analysis method you choose, and state all necessary physics principles and formulas. B E U S X2 x2 I O C C > Do not type in this area but rather upload a picture of your work Do not type in this area but rather upload a picture of your work Do not type in this area but rather upload a picture of your work Do not type in this area but rather upload a picture of your work Maximum file size: 2GB, maximum number of files: 1EQUATIONS Kinematics Waves Atomic Physics Ad hi -di he Wave= W = hfo E =hf = At d = val - bar T = 2 m = T do dave = Av T = 2AND Ek de stop N = NO(4) + d = vi+ car ve = v.2+ 2ad n12 sine T = Quantum Mechanics and Nuclear Physics sine, AE = Amcz E = pc T2 n2 VI v = fx = V2 = M P = Dynamics A = h (1 - cose) mic(I a= - Inet F = Gmm2 f = (viv|Js 1 = d sine Trigonometry and Geometry 8 = Gm opposite Line 1 = xd sine = Ay nl hypotenuse M= - AX F, = -kx Fg g = m adjacent COSO = y = mx + b Electricity and Magnetism hypotenuse Momentum and Energy Fel = kq192 opposite Area 12 AV = AP tand = _ adjacent p = m Ek = =mv2 Rectangle = /w FAI = MAV E = 14 1 = 4 c' = a2+ 62 Ep = mgh Triangle = zab W = Fl la| cose Ep = -Kx2 Fe b Circle = 1/2 E = q IFm| = 1 B C sin A sin B sin C W = AE P = . W IFml = qVB c' = a2 + b2 - 2ab cos C Circumference E = AV Circle = 2ar