0.00325 x 10^-7 their separation 1.00 s after the iaterval of 1.20 s C) 18.8 m a
ID: 1777521 • Letter: 0
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0.00325 x 10^-7 their separation 1.00 s after the iaterval of 1.20 s C) 18.8 m apart Whsat in D) 19.8 m 7. Two obiects are droppod from a bridge, an E) 23.7 m A) 4.90 m B) 9.80 m particO is given 8. Refer to Fig. 2. The magnitades of the forces as shown in the figure arei Fr 80.0N, F 50.0 N, and F 60.0 N. The resultant force acting on A) 18.0 N at an angle 60.0 with respect to vx-axis B) 20.7 N at an angle 47.1° with respect to +x-axis. C) 13.7Nat an angle 47.1 with respect to x-axis D) 25.5 N at an angle 30.0 with respect to +x-axis E) 20.0 N at an angle 90.0° with respect to "x-axis nspire cx Fig. 2 9. Vector A 2001+200j and vector B-3.001 +4.00j. What is vector C-3.00 A-1.00B ? A)-3.00i -18.00j B)9.00 10.0j C) 3.00i+ 2.00 j )-3.00i -2.00 j 3.00 18.0j A child throws a ball with an initial speed of 8.00 m's at an angle of 50.0° above the horizontal. The ball nd? es her hand 1.00 m above the ground. How far from where the child is standing does the ball hit the t1g 22 m B)S.14m C)7.18 m D)7.46 m E)1.58 m wimmer heading directly across a river 200 m wide reaches the opposite bank in 6 min 40 s. She is ownstream 480 m. How fast can she swim in still water? n/s B) 0.80 m/s C) 1.2 m/s D) 1.4 m/s E) 1.8 m/s A.Explanation / Answer
7. distance travelled by the first object in (1.20 s + 1 s = 2.2 s) is s1 = ut + 0.5at2
= 0.5 a t2 since initial velocity u = 0m/s
= 0.5 x 9.8 x 2.2 x 2.2
= 23.716 m
distance travelled by the second object in 1s is
s2 = 0.5 a t2
= 0.5 x 9.8 x 1 x 1
= 4.9 m
separation = s1 - s2 = 23.716 - 4.9 = 18.8m
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