The final velocity of the train after 8.3 s on the incline will be 12.022 m/s.
Answer:
Explanation:
So in this problem, the initial speed of the train is at 25.8 m/s before it comes to incline with constant slope. So the acceleration or the rate of change in velocity while moving on the incline is given as 1.66 m/s². So the final velocity need to be found after a time period of 8.3 s. According to the first equation of motion, v = u +at.
So we know the values for parameters u,a and t. Since, the train slows down on the slope, so the acceleration value will have negative sign with the magnitude of acceleration. Then
v = 25.8 + (-1.66×8.3)
v =12.022 m/s.
So the final velocity of the train after 8.3 s on the incline will be 12.022 m/s.
Answer:
vp = 0.94 m/s
Explanation
Formula
Vp = position/ time
position: Initial position - Final position
Position = 25 m - (-7 m) = 25 m + 7 m = 32 m
Then
Vp = 32 m / 34 seconds
Vp = 0.94 m/s
Boron Group
elements have three valence electrons and are fairly reactive. All of them are solids at room temperature. Boron is a very hard, black metalloid with a high melting point.
The answer is D, it falls between the infrared and ultraviolet.
Answer:
Explanation:
Remark
The only thing that might trip you up is what to do with the angle. The vertical component of the 15 degrees does no work against anything. So the 15 degrees limits the horizontal force.
The formula is
Work = F * d * cos(15)
The givens are
F = 2000 N
d = 30 m
Cos(15) = 0.9659
Solution
Work = 2000 * 30 * cos(15)
Work = 57,955
Rounded to two places would be 5.8 * 10^4
C
