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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Answer: In the 5th dimension, they who claim to know, say that there is only one time, including the past and the future.
Answer:
it's depent on height and gravity
Answer
given,
mass of people = 65 kg
number of people =
diameter of the merry-go-round = 4.2 m
radius = 2.1 m
angular velocity = 0.8 rad/s
moment of inertia = 1760 kg m²
Using the law of conservation of angular momentum, we have
L₁= L₂
I₁ω₁ =I₂ω₂
I₁ω₁= ( I₁+ 4 m r²)ω²
( 1760 x 0.8) = ( 1760 + 4 x 65 x 2.12)ω²
1408 = 2311.2 ω²
ω² = 0.609
ω = 0.781 rad/s
b) If the people jump of the merry-go-round radially, they exert no torque.
hence, it will not change the angular momentum of the merry-go-round. It will continue to move with the same ω .
