The route followed by a hiker consists of three displacement vectors a with arrow, b with arrow, and c with arrow. vector a with arrow is along a measured trail and is 1230 m in a direction 36.0° north of east. vector b with arrow is not along a measured trail, but the hiker uses a compass and knows that the direction is 41.0° east of south. similarly, the direction of vector vector c is 37.0° north of west. the hiker ends up back where she started, so the resultant displacement is zero, or a with arrow + b with arrow + c with arrow = 0. find the magnitudes of vector b with arrow and vector c with arrow

B= 5,655 m  C= 5,890 m.

Explanation:

We are told that the hiker ended up back where she started, so the total displacement was 0.

As displacement is a vector, if the magnitude is 0, this means that their components, along any axis must be 0 too.

If we choose the W-E direction as our x axis (with origin in the start and end point of the travel), being the E direction the positive, and we do the same with the N-S direction (we make it our y axis, with N as positive), we can write the following:

Δx = Δy=0

Next, we sum all the components of A, B and C along x axis, based on the question premises, as follows:

Δx = 1,230 m* cos 36º + B*sin 41º - C*cos 37º =0 (1)

Δy = 1,230 m*sin 36º -B*cos 41º + C* sin 37º = 0

We have a system of 2 linear equations with 2 unknowns (B and C) which can be solved by any suitable method: substitution, sum and substraction, determinants, etc).

Using determinants, we find first the determinant of the unknowns, as follows:

Δ = sin 41º*sin 37º - (-(cos 37º)*(-cos 41º) = -0.208

Now we find the numerator in order to get the value of B, replacing the B column by the independent terms'column:

ΔB = -1230*cos 36º*sin 37º - (-1230*sin 36º*(-cos 37º) = -1,177

⇒ B = ΔB / Δ = -1,177/-0.208 = 5,655 m

Repeating the same process, we get the value for ΔC, as follows:

ΔC = -1230*sin 36º*sin 41º - (-1230*cos 36º*(-cos 41º)) = -1,225

⇒ C = ΔC / Δ = -1,225/-0.208 = 5,890 m

1 a

2 b

3 b

4 d

5 d

6 d

explanation:

the electric field is a “force field” around a charged object that illustrates the direction the electric force would push an imaginary