In order to change the experiment vectors, A and B, as follows. Vector A has magnitude 0.855 N.(Ax = 0.255 N, with a negative y - component. Vector B is shown graphically below. Note that each small square represents 0.0250 N. (a) Calculate vertical component and direction of vector A. (b) How much mass, in grams, will student apply to the force table in order to produce vector A. (c) What are the values for the horizontal and vertical components of vector B? Calculate the magnitude of vector B. (d) Please calculate the direction of vector B, and amount of mass, in grams, student will apply to the force table in order to produce vector B.(e) What are the horizontal and vertical components of third vector that would produce equilibrium on the force table when vectors A and B are applied? (P) Please calculate the magnitude of the equilibrant vector. (g) At what angle third mass would be applied to the force table in order to produce equilibrium? (h) Please calculate the third mass, in grams, students will apply to the force table to produce equilibrium. Note: Angles reported in your answer should be measured counter-clockwise with respect to the positive x - axis, where the x-axis is zero degrees; i.e., all angles should be positive. Recall that g = 9.80 m/s. a ) (A x ) + ( Al ) * = 0.85 5 N (0.25 5 N) 2 + ( Ay) = 0.855 N (0. 255)# + Ay= = 0.731025 Ay= = 0. 731025- 10.25 Ay ? = 0.646 Ay = 0816N Of = tom Ay = - 0.816 N (down tan "' ( - 0.816 ) 17units 0.255 0- - 43. 6629 - Live 0. 255 -43.70 not 0.855 360 - 43.70 = 6) 0.855 N = ma d ) tan-1/0. 4500 316.330 0. 85 5 N = 141 /9.81) 0.4250 0. 0871 5 kg x 1060 = 8= 46. 640 02 87.155 - 87.291 0. 619 0 N = M ( 9. 81 ) 357 () BX = 17 ( 0 025 0 ) = 0.425ON 0.0631 kq * 1000 = By = 18 (0 . 0250 ) : 0.4500 N / -0.3 ( 1 x ) ? + By) : B B : 10. 42507+ (0,4500 ) 0. 61897 2 0.6190 N