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MAXImum [283]
3 years ago
6

What clues are useful in reconstructing pangaea

Physics
2 answers:
Gnoma [55]3 years ago
6 0
You can use map and notice one thinh. If you flipp over the edges of continents and put them together, you will get a big single continent that is called pangaea. Practically it's impossible but it could be imagined.
n200080 [17]3 years ago
6 0

Clues that are useful in reconstructing pangaea are the edges of the continents, glacial deposits, matching folded mountains, fossils found by scientist

<h3>Further explanation </h3>

Pangea is a supercontinent that existed during the late Paleozoic and early Mesozoic eras. Pangea assembled from earlier continental units approximately 335 million years ago, and it begin to break apart about 175 million years ago.

Clues that are useful in reconstructing pangaea are:

  1. Fossils found by scientist: Fossils of the same plant and animal species on some of the continents that the oceans separate.
  2. Matching folded mountains: If the continents were joined back, we can notice how the mountains match since they share, for example the same rocks sample
  3. Glacial Deposits: Some places have glacial deposits but the temperatures are not cold to have glaciers.
  4. Edge of Continents:   The edges of the continents are useful in reconstructing Pangaea. Some continents such as South America and Africa fit each other if joined together like. Aside from the fitting of edges of the continents, the evidence presence are found in the same continents made the reconstruction easier.
<h3>Learn more</h3>
  1. Learn more about reconstructing pangaea brainly.com/question/5959094
  2. Learn more about Fossils brainly.com/question/6454167
  3. Learn more about Glacial Deposits brainly.com/question/5632616

<h3>Answer details</h3>

Grade:  9

Subject:  physics

Chapter:  reconstructing pangaea

Keywords: Fossils, reconstructing pangaea, pangaea, clues, scientist

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Answer:

A) v3 = -[6.29 × 10^(6)]j^ - [7.06 × 10^(6)]i^

B) K_total = 373.08 × 10^(-15) J

Explanation:

We are given;

Mass of unstable atomic nucleus; M = 1.83 × 10^(-26) kg

Mass of first particle; m1 = 5.03 × 10^(-27) kg

Speed of first particle in y-direction; v1 = (6 × 10^(6) m/s) j^

Mass of second particle; m2 = 8.47 × 10^(-27) kg

Speed of second particle in x - direction; v2 = (4 × 10^(6) m/s) i^

Now, we don't have the mass of the third particle but since we are told the unstable atomic nucleus disintegrates into 3 particles, thus;

M = m1 + m2 + m3

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A) Applying law of conservation of momentum, we have;

MV = (m1 × v1) + (m2 × v2) + (m3 × v3)

Now, the unstable atomic nucleus was at rest before disintegration, thus V = 0 m/s.

Thus, we now have;

0 = (m1 × v1) + (m2 × v2) + (m3 × v3)

We want to find the velocity of the third particle v3. Let's make it the subject of the formula;

v3 = [(m1 × v1) + (m2 × v2)]/(-m3)

Plugging in the relevant values, we have;

v3 = [(5.03 × 10^(-27) × 6 × 10^(6))j^ + (8.47 × 10^(-27) × 4 × 10^(6))i^]/(-4.8 × 10^(-27))

v3 = [(30.18 × 10^(-21))j^ + (33.88 × 10^(-21))i^]/(-4.8 × 10^(-27))

v3 = -[6.29 × 10^(6)]j^ - [7.06 × 10^(6)]i^

B) Formula for kinetic energy is;

K = ½mv²

Now,total kinetic energy is;

K_total = K1 + K2 + K3

K1 = ½ × 5.03 × 10^(-27) × (6 × 10^(6))²

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K2 = ½ × 8.47 × 10^(-27) × (4 × 10^(6))²

K2 = 67.76 × 10^(-15)

To find K3, let's first find the magnitude of v3 because it's still in vector form.

Thus;

v3 = √[(-6.29 × 10^(6))² + (-7.06 × 10^(6))²]

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K3 = ½ × 4.8 × 10^(-27) × (9.46 × 10^(6))²

K3 = 214.78 × 10^(-15) J

K_total = (90.54 × 10^(-15)) + (67.76 × 10^(-15)) + (214.78 × 10^(-15))

K_total = 373.08 × 10^(-15) J

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