Answer:
c = 204 x 5 = 1020 m/s so it travels 1020 meters in 1 second.
Explanation:
Answer:
Acceleration = 1.428m/s2
Tension = 102.85N
Explanation:
The detailed solution is attached
Answer:
Here are 5:
Distance from source to receiver
Wind speed and direction
Wind gradients
Temperature gradients
Atmospheric attenuation
and there are many more...
Hope that was helpful.Thank you!!!
The final velocity of the train at the end of the given distance is 7.81 m/s.
The given parameters;
- initial velocity of the train, u = 6.4 m/s
- acceleration of the train, a = 0.1 m/s²
- distance traveled, s = 100 m
The final velocity of the train at the end of the given distance is calculated using the following kinematic equation;
v² = u² + 2as
v² = (6.4)² + (2 x 0.1 x 100)
v² = 60.96
v = √60.96
v = 7.81 m/s
Thus, the final velocity of the train at the end of the given distance is 7.81 m/s.
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