Two airplanes leave an airport at the same time. the velocity of the first plane is 670 m/h at a heading of 48°. the velocity of the second is 560 m/h at a heading of 132°. how far apart are they after 2.2h?

   First plane is 1819.58 miles is far from second plane after 2. hours.

Explanation:

     Let east represents positive x- axis and north represent positive y - axis. Horizontal component is i and vertical component is j.

 Velocity of the first plane is 670 m/h at a heading of 48° = (670 cos 48 i + 670 sin 48 j)      

                                = (448.32 i + 497.91 j) m/h

 Velocity of the second is 560 m/h at a heading of 132° = ( 560 cos 132 i + 560 sin 132 j)

                                 = (-374.71 i + 416.16 j) m/h

We know the equation displacement = Velocity * Time taken

Displacement of first plane after 2.2 hr = (448.32 i + 497.91 j)*2.2

                               = (986.304 i + 1095.402 j) miles.

 Displacement of second plane after 2.2 hr = (-374.71 i + 416.16 j)*2.2

                               = (-824.362 i + 915.552 j) miles.

 Displacement between two planes = (986.304 i + 1095.402 j) - (-824.362 i + 915.552 j)

                                            = (1810.666 i + 179.868 j) mi

 Distance between two planes after 2.2 hours =  

  So first plane is 1819.58 miles is far from second plane after 2. hours.

the body's situation won't change the forces are equal opposit and on the same body

connect the batteries in series and hope