A 6.0-μF capacitor charged to 50 V and a 4.0-μF capacitor charged to 34 V are connected to each other, with the two positive pla
1 answer:
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Answer:
w = 0.943 rad / s
Explanation:
For this problem we can use the law of conservation of angular momentum
Starting point. With the mouse in the center
L₀ = I w₀
Where The moment of inertia (I) of a rod that rotates at one end is
I = 1/3 M L²
Final point. When the mouse is at the end of the rod
= I w + m L² w
As the system is formed by the rod and the mouse, the forces during the movement are internal, therefore the angular momentum is conserved
L₀ = L_{f}
I w₀ = (I + m L²) w
w = I / I + m L²) w₀
We substitute the moment of inertia
w = 1/3 M L² / (1/3 M + m) L² w₀
w = 1 / 3M / (M / 3 + m) w₀
We substitute the values
w = 1/3 / (1/3 + 0.02) w₀
w = 0.943 w₀
To finish the calculation the initial angular velocity value is needed, if we assume that this value is w₀ = 1 rad / s
w = 0.943 rad / s
Answer:
θ = 19.66°
Explanation:
To determine the angle that the rope makes with the vertical for the two people, you first take into account the potential energy of the first person before he swings on the rope:
h: distance to the ground
g: gravitational acceleration = 9.8m/s^2
m: mass of the first person = 55 kg
In the image attache below you can notice that the height h is:
Then, the potential energy is:
When the first person picks up the second person (when the rope is exactly vertical), all the potential energy becomes kinetic energy. Next, when both people reaches the maximum height h' the energy must be equal to the initial potential energy of the first person:
From the previous equation you can get h':
Finally, you obtain the angle between the rope at the height h,' and the vertical, by calculating the following:
hence, the angle between the rope and the vertical, when the two people are in the rope is 19.66°
