Two airplanes leave an airport at the same time. the velocity of the first airplane is 740 m/h at a heading of 25.3◦. the velocity of the second is 570 m/h at a heading of 82◦.
how far apart are they after 1.5 h? answer in units of m.

They are 959.70 m apart after 1.5 h

Explanation:

Lets explain how to solve the problem

The given is:

→ The velocity of the 1st airplane is 740 m/h at a heading of 25.3°

→ The velocity of 2nd airplane is 570 m/h at a heading 82°

→ We need to find how far apart they are after 1.5 h

At first lets find the distance of each one after 1.5 h

→ d = v × t

→ = 740 × 1.5 = 1110 m

→ = 570 × 1.5 = 855 m

Assume that these two distance are two side of a triangle.

The angle between the two sides is the difference between their

heading.

The heading of the 1st airplane is 25.3° and the heading of the second

airplane is 82°

The angle between their distances = 82 - 25.3 = 56.7°

The angle between the two sides of the triangle is 56.7°

Lets use cosine rule to find the 3rd side of the triangle

→

→ = 1110 , = 855 , α = 56.7

Substitute these values in the rule

→ m

d represents the distance between the two airplanes after 1.5 h

They are 959.70 m apart after 1.5 h

kinetic

i'm so sorry if this is wrong i'm only in 6th grade xd

r=rho l/a where a is the area, l is the length and rho is resistivity so the greater the area the lesser the resistance thats why the earth pin is thicker bcos current follow the shortest path, so less resistive .