During takeoff, an airplane climbs with a speed of 130 m/s at an angle of 47 degrees above the horizontal. the speed and direction of the airplane constitute a vector quantity known as the velocity. the sun is shining directly overhead. how fast is the shadow of the plane moving along the ground? (that is, what is the magnitude of the horizontal component of the plane's velocity? )

130cos(47deg) = 88.6598 m/s

Answerthe energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.explanationwe define electric potential as the ratio electric potential energy to the charge,    then the following gives us the potential difference,      the formula  may be on your formula sheet.the change in electric potential energy would be the energy that would needed to move it from point a to point b. solving for that, we get    using q = 8.00  × 10⁻¹⁹ c and  δv = 2.00 v, we get    the energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.