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AVprozaik [17]
3 years ago
13

Electrical outlets in a house maintain a steady voltage, even when the amount of resistance on them changes. How is this possibl

e?
Physics
1 answer:
makvit [3.9K]3 years ago
8 0
The situation presented above is possible because the outlets could be operating in a parallel circuit. <span>Electrical outlets in a house maintain a steady voltage, even when the amount of resistance on them changes because it operates with a parallel circuit wherein voltage is constant even if resistance changes.</span>
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330 Hz

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When the play button is pressed, a CD accelerates uniformly from rest to 450 rev/min in 3.0 revolutions. If the CD has a radius
Marina CMI [18]

To solve this problem it is necessary to apply the kinematic equations of angular motion.

Torque from the rotational movement is defined as

\tau = I\alpha

where

I = Moment of inertia \rightarrow \frac{1}{2}mr^2 For a disk

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The angular acceleration at the same time can be defined as function of angular velocity and angular displacement (Without considering time) through the expression:

2 \alpha \theta = \omega_f^2-\omega_i^2

Where

\omega_{f,i} = Final and Initial Angular velocity

\alpha = Angular acceleration

\theta = Angular displacement

Our values are given as

\omega_i = 0 rad/s

\omega_f = 450rev/min (\frac{1min}{60s})(\frac{2\pi rad}{1rev})

\omega_f = 47.12rad/s

\theta = 3 rev (\frac{2\pi rad}{1rev}) \rightarrow 6\pi rad

r = 7cm = 7*10^{-2}m

m = 17g = 17*10^{-3}kg

Using the expression of angular acceleration we can find the to then find the torque, that is,

2\alpha\theta=\omega_f^2-\omega_i^2

\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}

\alpha = \frac{47.12^2-0^2}{2*6\pi}

\alpha = 58.89rad/s^2

With the expression of the acceleration found it is now necessary to replace it on the torque equation and the respective moment of inertia for the disk, so

\tau = I\alpha

\tau = (\frac{1}{2}mr^2)\alpha

\tau = (\frac{1}{2}(17*10^{-3})(7*10^{-2})^2)(58.89)

\tau = 0.00245N\cdot m \approx 2.45*10^{-3}N\cdot m

Therefore the torque exerted on it is 2.45*10^{-3}N\cdot m

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3 years ago
A comparison of a machine’s work output with its work input is called __??__.
Aloiza [94]
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3 years ago
A 0.660 kg ball is dropped from rest from the top of a building and falls for 5.65 seconds. How tall was the building ?
kirill115 [55]

Answer:

The height of the building is approximately 156.58 m

Explanation:

The mass of the ball dropped from rest from the building top = 0.660 kg

The time in which the ball falls, t = 5.65 seconds

The height, h, of the building is given from the following equation of motion;

h = u·t + ¹/₂·g·t²

Where;

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Plugging in the values, we have;

h = 0 × 5.65 + ¹/₂ × 9.81 × 5.65² ≈ 156.58 m

The height of the building, h ≈ 156.58 m.

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