First we need to find the acceleration of the skier on the rough patch of snow.
We are only concerned with the horizontal direction, since the skier is moving in this direction, so we can neglect forces that do not act in this direction. So we have only one horizontal force acting on the skier: the frictional force,
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. For Newton's second law, the resultant of the forces acting on the skier must be equal to ma (mass per acceleration), so we can write:
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Where the negative sign is due to the fact the friction is directed against the motion of the skier.
Simplifying and solving, we find the value of the acceleration:
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Now we can use the following relationship to find the distance covered by the skier before stopping, S:
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where
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is the final speed of the skier and
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is the initial speed. Substituting numbers, we find:
1 cubic cm is the same as 1 mL, so the answer would be C.
Answer:
We kindly invite you to read carefully the explanation and check the image attached below.
Explanation:
According to this problem, the rocket is accelerated uniformly due to thrust during 30 seconds and after that is decelerated due to gravity. The velocity as function of initial velocity, acceleration and time is:
(1)
Where:
- Initial velocity, measured in meters per second.
- Final velocity, measured in meters per second.
- Acceleration, measured in meters per square second.
- Initial time, measured in seconds.
- Final time, measured in seconds.
Now we obtain the kinematic equations for thrust and free fall stages:
Thrust (
,
,
,
)
(2)
Free fall (
,
,
,
)
(3)
Now we created the graph speed-time, which can be seen below.
Explanation:
A) Use Hooke's law to find the spring constant.
F = kx
40 N = k (0.4 m)
k = 100 N/m
B) Period of a spring-mass system is:
T = 2π √(m / k)
T = 2π √(2.6 kg / 100 N/m)
T = 1 s
Frequency is the inverse of period.
f = 1 / T
f = 1 Hz