Question

If a vector A has the following coponents, use trigonometry to find its magnitude and the and the counterclockwise angle it makes with the + x-axis.

Ax= 7.00lb, Ay= 5.00lb

Ax=7.00lb, Ay Please give your answer in the interval [0,360]?

Ax = -21.0m/s , Ay = -34.0m/s

Ax = -21.0m/s , Ay = -34.0m/s

Please give your answer in the interval [0,360]?

Ax = -1600km , Ay = 2500km

Ax = -1600km , Ay = 2500km

Please give your answer in the interval [0,360]?

Ax = 72.0N , Ay = -52.1N

Ax = 72.0N , Ay = -52.1N

Please give your answer in the interval [0,360]?.

Answer 1

TanA=5/7

A=35

TanB=34/21

B=58

TanC=2500/(-1600)

C=122.62

TanD=-52.1/72

D=144.11

Answer 2

Answer 3

The magnitude pf a vector is given by, The angle is given by, e- tan-1 Substitute given values in the equation A-A A 1지 1(71b)--(51b), 6.24 lb Therefore, the magnitude of vector A is 6.24 lb The angle is given by, θ = tan 15lb 7 lb 35.53 The counter clockwise angle is given by, θ= 35.53 +180 215.50 Therefore, the angle is 215.5° Substitute given values in the equation A-JA + A-21 m/s) +(-34 m/s) - 39.9 m/s

Therefore, the magnitude of vector A is 39.9 m/s The angle is given by, 0tan--34 m/s 21 m/s - 58.29 The counter clockwise angle is given by, a - 58.29+180 238.20 Therefore, the angle is 238.2 Substitute given values in the equation A-JA; +A 1600 km) +(2500 km) - 2968.16 km Therefore, the magnitude of vector A is 2968.16 km. The angle is given by, 6-tn-2500 km 1600 km -57.38 The counter clockwise angle is given by, e--57.38 +180 122.6 Therefore, the angle is 122.6

Substitute given values in the equation A-JA;+A; A72N) (-52.1N) -88.87 N Therefore, the magnitude of vector A is 88.87 N The angle is given by, 8- tan-52.1N 72 N -35.83 The counter clockwise angle is given by, 0 -35.830+180 144.16° Therefore, the angle is 144.16