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The displacement of the train after 2.23 seconds is 25.4 m.
<h3>
Resultant velocity of the train</h3>
The resultant velocity of the train is calculated as follows;
R² = vi² + vf² - 2vivf cos(θ)
where;
- θ is the angle between the velocity = (90 - 51) + 37 = 76⁰
R² = 8.81² + 9.66² - 2(8.81 x 9.66) cos(76)
R² = 129.75
R = √129.75
R = 11.39 m/s
<h3>Displacement of the train</h3>
Δx = vt
Δx = 11.39 m/s x 2.23 s
Δx = 25.4 m
Thus, the displacement of the train after 2.23 seconds is 25.4 m.
Learn more about displacement here: brainly.com/question/2109763
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Answer:
The question is somewhat vague in that acceleration is not exactly defined:
Usually a = (v2 - v1) / t which would imply that
a = 32 / g = 32 / 9.8 = 3.27 the acceleration due to change in speed of the rocket
One can also say that the astronaut experiences an acceleration of 9.8 m/s^2 just by being motionless on the surface of the earth.
Then a = (32 - 9.8) / 9.8 = 2.27 due to the acceleration of the rocket
If we assume the first condition then
F = 65 kg * 3.27 * 9.8 m/s^2 = 2083 N
Answer:

Explanation:
Since the universal SI unit for velocity is meters/second, let's convert ft/s to m/s:

We can use the following kinematics equation to solve this question:

What we know:
- The initial velocity,
, is
- (physics concept) The final velocity must be equal in magnitude but opposite in direction to the initial velocity (
) - Acceleration,
, is acceleration due to gravity at about
Solving for
:

Answer:
605447.7066 kgm²/s
Explanation:
= Mass of sphere = 10000 kg
= Mass of rod = 10 kg
r = Radius of sphere = 2 m
l = Length of antenna = 3 m
Angular speed

Angular momentum is given by

Moment of inertia of the satellite is

Moment of antenna of the satellite is

The angular momentum of the system is

The angular momentum of the satellite is 605447.7066 kgm²/s