Well what’s the question?..
Taking the vertical component of the displacement
1.1 - 0.2 = 0.9 mile
The horizontal component of the displacement
-0.3 mile
The magnitude of the displacement is
√[ (0.9)² + (-0.3) ] = 0.95 mile
The direction is
θ = tan-1 (-0.3/0.9)
θ = 161.57 degrees.
Answer:
A. 30.38°
B 5.04N
Explanation:
Using
F= ILBsin theta
2 .55N= 8.4Ax 0.5mx 1.2T x sintheta
Theta = 30.38°
B. If theta is 90°
Then
F= 8.4Ax 0.5mx 1.2x sin 90°
F= 5.04N