Question

Two airplanes leave an airport at the same

Answer 1

All you have to do is add up the numerators (numbers on top) and then simplify your ... They have no common factors, so let's just multiply to create two new equivalent fractions. ... Add whole numbers: 2 + 5 = 7

Answer 2

For the given problems, both the planes are 1486 m apart after 2.4 h.

Explanation:

Let us consider two airplanes as A1 and A2. We know that the

           $\text { Distance covered in a time }=\text { velocity of airplanes } \times \text { given time }$

The distance covered by airplane A1 is

     $\text { Distance covered by } A 1=750 \times 2.4=1800 \mathrm{m}$

Similarly, the distance covered by airplane A2 is

     $\text { Distance covered by } A 2=620 \times 2.4=1488 \mathrm{m}$

As the airplanes take off at an angle from the base, so they will be exhibiting X and Y components of displacements.

X components of both the airplanes

         = $(1800 \times \sin 51.3)-(1488 \times \sin 163)=1404.77-435.04=969.7$

Y components of both the airplanes

        = $=(1800 \times \cos 51.3)-(1488 \times \cos 163)=1125.43+0.956=1126$

The net displacement after 2.4 hours will be equal to the distance at which both planes are airplanes to each other at 2.4 hours.

                             $\text { Distance apart }=\sqrt{X^{2}+Y^{2}}$

Therefore, Net displacement is

$=\sqrt{(969.7)^{2}+(1126)^{2}}=\sqrt{940318.09+1267876}=\sqrt{2208194.09}=1485.99 \mathrm{m}$

So, both the planes are 1486 m apart from each other after 2.4 hours.