Aspacecraft is launched from the surface of the earth with a velocity of 0.600c at an angle of 50.0 above the horizontal, positive x-axis. another spacecraft is moving past with a velocity of 0.700c in the negative x direction. determine the magnitude and direction of the velocity of the first spacecraft

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.

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