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2 years ago

12

The ozone layer is an important component of Earth’s atmosphere because it helps to block UV radiation from reaching Earth. The

ozone layer absorbs most of the harmful UV radiation from the Sun and stores it as heat within the stratosphere. If the ozone layer were completely removed, what effect would that change have on temperature?
The temperature would remain unchanged.
The temperature of Earth would increase.
The temperature of the upper atmospheric layers would decrease.
The temperature of Earth would decrease.
The temperature of the stratosphere would decrease.
THERE ARE MULTIPLE ANSWERS

2 answers:

kolbaska11 [484]2 years ago

3 0

The temperature on the Earth would increase.

The temperature of the stratosphere would decrease.

emmainna [20.7K]2 years ago

3 0

Answer:

The temperature on the Earth would increase.

The temperature of the stratosphere would decrease.

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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

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

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2 years ago

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ankoles [38]

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2 years ago

A disk of radius 25 cm spinning at a rate of 30 rpm slows to a stop over 3 seconds. What is the angular acceleration? B. How man

Ne4ueva [31]

Answer:

A. α = - 1.047 rad/s²

B. θ = 14.1 rad

C. θ = 2.24 rev

Explanation:

A.

We can use the first equation of motion to find the acceleration:

$\omega_f = \omega_i + \alpha t$

where,

ωf = final angular speed = 0 rad/s

ωi = initial angular speed = (30 rpm)(2π rad/1 rev)(1 min/60 s) = 3.14 rad/s

t = time = 3 s

α = angular acceleration = ?

Therefore,

$0\ rad/s = 3.14\ rad/s + \alpha(3\ s)$

<u>α = - 1.047 rad/s²</u>

B.

We can use the second equation of motion to find the angular distance:

$\theta = \omega_it +\frac{1}{2}\alpha t^2\\\theta = (3.14\ rad/s)(3\ s) + \frac{1}{2}(1.04\ rad/s^2)(3)^2$

<u>θ = 14.1 rad</u>

C.

θ = (14.1 rad)(1 rev/2π rad)

<u>θ = 2.24 rev</u>

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2 years ago