Answer to Question #243903 in Physics for manaa

Question #243903

A vector C has magnitude of 11 m/s and makes an angle 65° degree with the positive x-

axis, vector D has Cartesian co-ordinates Dx = 4m/s


2 and Dy = −3 m/s


2. Find the difference C⃗ -D�⃗ between


two vectors.


1
Expert's answer
2021-09-29T09:54:59-0400

Given:

Cx=11cos65=4.65m/sC_x=11*\cos65^{\circ}=4.65\:\rm m/s

Cy=11sin65=9.97m/sC_y=11*\sin65^{\circ}=9.97\:\rm m/s

Dx=4.00m/sD_x=4.00\:\rm m/s

Dy=3.00m/sD_y=-3.00\:\rm m/s



The difference between two vectors:


CD=(CxDx)i^+(CyDy)j^{\bf C}-{\bf D}=(C_x-D_x){\hat i}+(C_y-D_y){\hat j}=(4.654.00)i^+(9.97(3.00))j^=0.65i^+13.0j^=(4.65-4.00){\hat i}+(9.97-(-3.00)){\hat j}\\ =0.65{\hat i}+13.0{\hat j}


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