Determine the vector sum of the displacements Δd1 = 2.4 m [32° S of W]; Δd2 = 1.6 m [S]; and Δd3 = 4.9 m [27° S of E].

3.44 metres

Explanation:

To determine the vector sum of the displacements Δd1 = 2.4 m [32° S of W]; Δd2 = 1.6 m [S]; and Δd3 = 4.9 m [27° S of E], resolve the given parameters into x - component and y - component.

Resolving into x - component

- 2.4cos32 + 4.9cos27 = 2.3306

Resolving into y - component

- 2.4sin32 - 4.9sin27 - 1.6 = - 2.553

The vector sum of the displacement will be

Sqrt( 2.3^2 + 2.6^2) =

Sqrt ( 11.81)

3.44 m

Therefore, the vector sum of the displacements is 3.44 metres

it is called kinetic energy

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.