Answer:
2.85 rad/s
Explanation:
5 cm = 0.05 m
20 g = 0.02 kg
When dropping the 2nd object at a distance of 0.05 m from the center of mass, its corrected moments of inertia is:
So the total moment of inertia of the system of 2 objects after the drop is:
From here we can apply the law of angular momentum conservation to calculate the post angular speed
Answer:
Explanation:
When a standing wave is formed with six loops means the normal mode of the wave is n=6, the frequency of the normal mode is given by the expression:
Where is the length of the string and
the velocity of propagation. Use this expression to find the value of
.
The velocity of propagation is given by the expression:
Where is the desirable variable of the problem, the linear mass density, and
is the tension of the cord. The tension is equal to the weight of the mass hanging from the cord:
With the value of the tension and the velocity you can find the mass density:
This is a uniform rectilinear motion (MRU) exercise.
To start solving this exercise, we obtain the following data:
<h3><u>Data:</u></h3>To calculate distance, speed is multiplied by time.
We apply the following formula: d = v * t.
We substitute the data in the formula: the <u>speed is equal to 4.6 m/s,</u> the <u>time is equal to 10 s</u>, which is left as follows:
Therefore, the speed at 10 seconds is 46 meters.