A train 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.

57,9 m/s

Explanation:

A train is moving at a constant speed on a surface inclined upward at 15.0° with the horizontal and travels 300 meters in 5 seconds. The horizontal velocity of the train at the end of 3 seconds is 100m/s.

The first formula we will use is:

v = d/t

300m ÷ 5s  = 60m/s

The last formula we will use is:

vh = v·cosx

60m/s·cos15.0°

60m/s · 0,965  = 57,9 m/s

Therefore, the answer is 57,9 m/s.

Hope this helped!

57,9 m/s

velocity formula

v = d/t

= 300m/5s

= 60m/s

horizontal velocity formula

vh = v·cosx

= 60m/s×cos15°

= 60m/s×0,965

= 57,9 m/s

the answer would be 32*f

the chart show the masses and velocities of two colliding objects that stick together after a collision.

according to the law of conservation of momentum, the momentum of the object after the collision is  

answer

1,500

kg • m/s.