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.
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Answer:
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Answer:
The correct answer is "6666.67 N".
Explanation:
The given values are:
Mass,
m = 0.100
Relative speed,
v = 4.00 x 10³
time,
t = 6.00 x 10⁻⁸
As we know,
⇒ 
On substituting the given values, we get
⇒ 
⇒ 
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