Question

4 of 10 Item 4 Submit Request Answer Part E Find the magnitude of the vector difference A-B. VAED ΑΣΦ ? Request Answer Submit Part F Find the direction of the vector difference A-B. Express your answer as counterclockwise angle from x-axis to the vector. Πνα ΑΣφ Request Answer Submit Part G Figure 1 of 1 Find the magnitude of the vector difference B-A. nν ΑΣφ ? B(15.0 m) 30.0 D (10.0 m) 53.0 Request Answer Submit 25.00 Part H Complete previous part(s) C120 m) A (8.00 m) Next> Provide Feedback

Answer 1

From the figure, sin 0= X=B sin e X= 15 msin 30º=7.5m this is x component of vector B, similarly, for y component of vector B, cos e= Y=Bcos e y=15 cos 30º = 12.99 m now, its vector form,

B=7.5mî+12.99mì A-BB-8.0mj-7.5mi-12.99mj À-B=-7.5m î–20.99 m Î now, lets apply the phythagoras theorem for its magnitude, JÄ-B}= (–7.5)? +(–20.99)'m |A-BI=22.29m Ans.

From the figure, a=tan" 20:59 a=tan-'(2.798) A-B a = 70.34 degree with negative x axis, now, angle with positive x axis, 180°+70.34°=250.34° Ans. part G, magnitude of B-À would be same as the magnitude of À-B

Thus, magnitude of B-A=20.29 m Ans.