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3 years ago

How is energy transfer related to the ozone layer and its temperature?

1 answer:

3 0

The ozone layer blocks around 97% of the ultraviolet radiation from the sun from being transferred to the surface, which affects how much energy is transferred to the earth. On one hand it’s part of he atmosphere that would by its presence increase the amount of energy the earth could hold in, but on he other hand it blocks radiation from reaching the earth, so I’m not sure how it would affect temperature.

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Spring #1 has a force constant of k, and spring #2 has a force constant of 2k. Both springs are attached to the ceiling. Identic

Gre4nikov [31]

Answer:

The ratio of the energy stored by spring #1 to that stored by spring #2 is 2:1

Explanation:

Let the weight that is hooked to two springs be w.

Spring#1:

Force constant= k

let x1 be the extension in spring#1

Therefore by balancing the forces, we get

Spring force= weight

⇒k·x1=w

⇒x1=w/k

Energy stored in a spring is given by $\frac{1}{2}kx^{2}$ where k is the force constant and x is the extension in spring.

Therefore Energy stored in spring#1 is, $\frac{1}{2}k(x1)^{2}$

⇒ $\frac{1}{2}k(\frac{w}{k})^{2}$

⇒ $\frac{w^{2}}{2k}$

Spring #2:

Force constant= 2k

let x2 be the extension in spring#2

Therefore by balancing the forces, we get

Spring force= weight

⇒2k·x2=w

⇒x2=w/2k

Therefore Energy stored in spring#2 is, $\frac{1}{2}2k(x2)^{2}$

⇒ $\frac{1}{2}2k(\frac{w}{2k})^{2}$

⇒ $\frac{w^{2}}{4k}$

∴The ratio of the energy stored by spring #1 to that stored by spring #2 is $\frac{\frac{w^{2}}{2k}}{\frac{w^{2}}{4k}}=$ 2:1

4 0

2 years ago

Hurryyyyyyyy

padilas [110]

The answer would be C. Gamma Rays and High Frequency EM waves travel at the speed of light and are transverse waves.

7 0

3 years ago

Read 2 more answers

A baseball thrown from the outfield to home plate does not have which of the following types of energy while it's in the air? A.

posledela

The ball does not have D. Radiant energy.

7 0

2 years ago

The basic building block of life is

snow_tiger [21]

<span>The basic building block of life is molecules, more specifically macromolecules. There are four different macromolecules which could all be described as the building blocks of life, namely carbohydrates, proteins, nucleic acids and proteins. </span>

8 0

3 years ago

A rocket is launched at an angle of 53.0° above the horizontal with an initial speed of 103 m/s. The rocket moves for 3.00 s alo

Serggg [28]

Before the engines fail $(0\le t\le3.00\,\rm s)$ , the rocket's horizontal and vertical position in the air are

$x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t^2$

$y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t^2$

and its velocity vector has components

$v_x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t$

$v_y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t$

After $t=3.00\,\rm s$ , its position is

$x=273\,\rm m$

$y=362\,\rm m$

and the rocket's velocity vector has horizontal and vertical components

$v_x=120\,\frac{\rm m}{\rm s}$

$v_y=159\,\frac{\rm m}{\rm s}$

After the engine failure $(t>3.00\,\rm s)$ , the rocket is in freefall and its position is given by

$x=273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)t$

$y=362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

and its velocity vector's components are

$v_x=120\,\frac{\rm m}{\rm s}$

$v_y=159\,\frac{\rm m}{\rm s}-gt$

where we take $g=9.80\,\frac{\rm m}{\mathrm s^2}$ .

a. The maximum altitude occurs at the point during which $v_y=0$ :

$159\,\frac{\rm m}{\rm s}-gt=0\implies t=16.2\,\rm s$

At this point, the rocket has an altitude of

$362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)(16.2\,\rm s)-\dfrac g2(16.2\,\rm s)^2=1650\,\rm m$

b. The rocket will eventually fall to the ground at some point after its engines fail. We solve $y=0$ for $t$ , then add 3 seconds to this time:

$362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=34.6\,\rm s$

So the rocket stays in the air for a total of $37.6\,\rm s$ .

c. After the engine failure, the rocket traveled for about 34.6 seconds, so we evalute $x$ for this time $t$ :

$273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)(34.6\,\rm s)=4410\,\rm m$

5 0

3 years ago