Answer:
Տᗩᗰᗴ ᵐᵃᵗᵉ Տᗩᗰᗴ
If a stone with an original velocity of 0 is falling from a ledge and takes 8 seconds to hit the ground, what is the final veloc
2 answers:
You might be interested in
Answer:
<em>The output power is greater in the interval from 1.0 s to 2.5 s</em>
Explanation:
<u>Physical Power </u>
It measures the amount of work W an object does in certain time t. The formula needed to compute power is
Work can be computed in several ways since we are given the motion conditions, we'll use this formula, for F= applied force, x=distance parallel to F
The second Newton's law gives us the net force as
being m the mass of the object and a the acceleration it has for a given period of time. In our problem, we have two different behaviors for each interval and we must calculate this force since the acceleration is changing. Let's calculate the acceleration in the first interval. We can use the formula for the final speed vf knowing the initial speed vo (which is 0 because the sprinter starts from rest), the acceleration a, and the time t:
Solving for a
The distance traveled in the interval is given by
Since vo=0
The force is given by
We don't know the value of m, so the force is
Computing the work done by the sprinter
The power is finally computed
During the second interval, from t=1 sec to 1.5 sec, the speed changes from 5.4 m/s to 9.8 m/s. This allows us to compute the second acceleration
The distance is
The net force is
The work done by the sprinter is now computed as
At last, the output power is
By comparing both results, and being m the same for both parts, we conclude the output power is greater in the interval from 1.0 s to 2.5 s
The period of a simple pendulum is given by
where
L is the pendulum length
g is the acceleration of gravity
If we move the same pendulum from Earth to the Moon, its length L remains the same, while the acceleration of gravity g changes. So we can write the period of the pendulum on Earth as:
where
is the acceleration of gravity on Earth, while the period of the pendulum on the Moon is
where
is the acceleration of gravity on the Moon.
If we do the ratio of the two periods, we get
but the gravity acceleration on the Moon is 1/6 of the gravity acceleration on Earth, so we can write
and we can rewrite the previous ratio as
so the period of the pendulum on the Moon is
where
L is the pendulum length
g is the acceleration of gravity
If we move the same pendulum from Earth to the Moon, its length L remains the same, while the acceleration of gravity g changes. So we can write the period of the pendulum on Earth as:
where
where
If we do the ratio of the two periods, we get
but the gravity acceleration on the Moon is 1/6 of the gravity acceleration on Earth, so we can write
so the period of the pendulum on the Moon is