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2 years ago

How does the temperature affect the phase of water

Physics

1 answer:

Sever21 [200]2 years ago

8 0

Answer:

Temperature affects phase change by slowing down the movement in between the atoms, thus causing a change in kinetic energy, which in turn causes the atoms to undergo forms of combining or a type of disepersion.

Explanation:

Kinetic energy while being the reason phase changes are constant, Kinetic Energy can be caused by other means. Pressure and temperature can affect many other states kinetic energy, which in turn can affect each state of matter. Making a group of atoms or compounds compacts will force the atoms to move closer together thus with a lower net kinetic energy energy. Reducing temperature also works along the same lines. Colder temperatures can slow down atomic movements which in turn will naturally make each atom move close to each other.

With all of the information provided, it is only feasible that pressure and temperature are directly corresponding with the matter and atomic phase change

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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

$\tau = I\alpha$

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