A truck weighing 3.0 × 104 newtons was driven up a
hill that is 1.6 × 103
meters long to a level area that
is 8.0 × 102
meters above the starting point. If the
trip took 480 seconds, what was the minimum power
required?

A truck weighing 3.0 × 104

newtons was driven up a

hill that is 1.6 × 103

meters long to a level area that

is 8.0 × 102

meters above the starting point. If the

trip took 480 seconds, what was the minimum power

required?

Explanation:

the answer is 1292.8

brainliest pl

5.0x10^4

Explanation:

Im gunna geuss that you didn't add the exponents so...

P=power

F=force

d=displacment

P=Fd

equation:

3.0x10^4(1.6x10^3 - 8.0x10^2)/480

=30000(800)/480

=50000 or 5.0x10^4

the components of a vector are the parts of the vectors that are to each other.

b perpendicular