Atrain is moving at a constant speed on a surface inclined upward at 15.0° with the horizontal and travels 300 meters in 5 seconds. calculate the horizontal velocity of the train at the end of 3 seconds

It would be helpful to draw the system to fully understand the situation. We would see that a right triangle can be made. We can use trigonometric functions so as to relate terms.

velocity, v = 300/5 = 60 m/s
horizontal component of velocity = vcos(theta) = 60cos15 = 57.96 m/s

It will have the same horizontal velocity at the end of 3 seconds which is 57.96 m/s since it is moving at a constant speed. Hope this answers the question. Have a nice day.

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.