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3 years ago

Can someone Plzzz help meee.

Physics

1 answer:

mario62 [17]3 years ago

6 0

solid...............

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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

$\alpha =$ Angular acceleration

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 $2.45*10^{-3}N\cdot m$

3 0

3 years ago

Which statement below concerning the photoelectric effect is true? View Available Hint(s) Which statement below concerning the p

kakasveta [241]

There is a threshold frequency for each metal, and only light of a frequency higher than this threshold causes electrons to be emitted from the metal surface.

8 0

3 years ago

Two speakers, one directly behind the other, are each generating a 240-Hz sound wave. What is the smallest separation distance b

AveGali [126]

Answer:

The smallest separation distance between the speakers is 0.71 m.

Explanation:

Given that,

Two speakers, one directly behind the other, are each generating a 240-Hz sound wave, f = 240 Hz

Let the speed of sound is 343 m/s in air. The speed of sound is given by the formula as :

$v=f\lambda\\\\\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{343\ m/s}{240\ Hz}\\\\\lambda=1.42\ m$

To produce destructive interference at a listener standing in front of them,

$d=\dfrac{\lambda}{2}\\\\d=\dfrac{1.42}{2}\\\\d=0.71\ m$

So, the smallest separation distance between the speakers is 0.71 m. Hence, this is the required solution.

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3 years ago

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Answer:C

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0.546 $\hat k$