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2 years ago

Drag the tiles to the correct boxes to complete the pairs.

Physics

1 answer:

klio [65]2 years ago

7 0

Considering the given evidences for Alfred Wegener’s theory of continental drift;

Scientists find fossils of species in the Arctic that typically lived in warmer areas is a form of climatic evidence
Scientists spot fossils on different continents that look similar and prove to be the same age is a form of biological evidence
Scientists observe that some mountain ranges on different continents tend to look alike is a form of geological evidence

<h3>What is Alfred Wegener’s theory of continental drift?</h3>

Alfred Wegener’s theory of continental drift is the theory that the whole continents were once one supercontinent which as a result of geological activities broke up into the different continents which then drifted apart.

Considering the given evidences for Alfred Wegener’s theory of continental drift;

Scientists find fossils of species in the Arctic that typically lived in warmer areas: It is a form of climatic evidence
Scientists spot fossils on different continents that look similar and prove to be the same age: It is a form of biological evidence
Scientists observe that some mountain ranges on different continents tend to look alike: It is a form of geological evidence

In conclusion, Alfred Wegener’s theory of continental drift describes how the continents of today were formed.

Learn more about theory of continental drift at: brainly.com/question/7350119

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Note that the complete question is given below:

Drag the tiles to the correct boxes to complete the pairs. The examples provide evidence for Alfred Wegener’s theory of continental drift. Determine whether each example is a form of biological, geological, or climatic evidence.

1) biological

2)geological

3)climatic

*Scientists find fossils of species in the Arctic that typically lived in warmer areas.

*Scientists spot fossils on different continents that look similar and prove to be the same age.

*Scientists observe that some mountain ranges on different continents tend to look alike.

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andreev551 [17]

Answer:

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Explanation:

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3 years ago

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Explanation:

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3 years ago

g initial angular velocity of 39.1 rad/s. It starts to slow down uniformly and comes to rest, making 76.8 revolutions during the

MrRa [10]

Answer:

Approximately $-1.58\; \rm rad \cdot s^{-2}$ .

Explanation:

This question suggests that the rotation of this object slows down "uniformly". Therefore, the angular acceleration of this object should be constant and smaller than zero.

This question does not provide any information about the time required for the rotation of this object to come to a stop. In linear motions with a constant acceleration, there's an SUVAT equation that does not involve time:

$v^2 - u^2 = 2\, a\, x$ ,

where

$v$ is the final velocity of the moving object,
$u$ is the initial velocity of the moving object,
$a$ is the (linear) acceleration of the moving object, and
$x$ is the (linear) displacement of the object while its velocity changed from $u$ to $v$ .

The angular analogue of that equation will be:

$(\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta$ , where

$\omega(\text{final})$ and $\omega(\text{initial})$ are the initial and final angular velocity of the rotating object,
$\alpha$ is the angular acceleration of the moving object, and
$\theta$ is the angular displacement of the object while its angular velocity changed from $\omega(\text{initial})$ to $\omega(\text{final})$ .

For this object:

$\omega(\text{final}) = 0\; \rm rad\cdot s^{-1}$ , whereas
$\omega(\text{initial}) = 39.1\; \rm rad\cdot s^{-1}$ .

The question is asking for an angular acceleration with the unit $\rm rad \cdot s^{-1}$ . However, the angular displacement from the question is described with the number of revolutions. Convert that to radians:

$\begin{aligned}\theta &= 76.8\; \rm \text{revolution} \\ &= 76.8\;\text{revolution} \times 2\pi\; \rm rad \cdot \text{revolution}^{-1} \\ &= 153.6\pi\; \rm rad\end{aligned}$ .

Rearrange the equation $(\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta$ and solve for $\alpha$ :

$\begin{aligned}\alpha &= \frac{(\omega(\text{final}))^2 - (\omega(\text{initial}))^2}{2\, \theta} \\ &= \frac{-\left(39.1\; \rm rad \cdot s^{-1}\right)^2}{2\times 153.6\pi\; \rm rad} \approx -1.58\; \rm rad \cdot s^{-1}\end{aligned}$ .