Displacement vector A points due east and has a magnitude of 1.9 km. Displacement vector B points due north and has a magnitude of 2.08 km. Displacement vector C points due west and has a magnitude of 2.4 km. Displacement vector D points due south and has a magnitude of 2.8 km. Find the magnitude and direction (relative to due east) of the resultant vector A + B + C + D

Explanation:

We shall represent all the four displacement in vector form  in terms of unit vector i and j where i represents unit vector towards east , j represents unit vector towards north .

Displacement of A

D₁ = 1.9 i

Displacement of B

D₂ = 2.08 j

Displacement of C

D₃ = - 2.4 i

Displacement of D

D₄ = -2.8 j

Resultant displacement

= 1.9 i + 2.08 j - 2.4 i - 2.8 j

= - 0.5 i - 0.72 j

magnitude of resultant vector

= √ ( .5² + .72² )

=√ ( .25 + .5184 )

= √ .7684

= .876 km

Both i and j are negative of resultant displacement

hence its direction is towards south of west . Angle with west is Ф .

TanФ = .5184 / .25 = 2.0736

Ф = 64.25° .

From east direction is  = 180 + 64.25 = 244.25° .

car a and b are moving in opposite directions

plz mark.me the brainliest i rlly need this kind of yours in exchange

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures