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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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marishachu [46]

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Hari didn't plan to go abroad

Explanation:

Abroad planned to go to hari.

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What is the total energy equation?
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6 0
4 years ago
If the distance of a galaxy is 2,000 Mpc, how many years back into the past are we looking when we observe this galaxy
ruslelena [56]

The age of the galaxy when we look back is 13.97 billion years.

The given parameters:

  • <em>distance of the galaxy, x = 2,000 Mpc</em>

According Hubble's law the age of the universe is calculated as follows;

v = H₀x

where;

H₀ = 70 km/s/Mpc

T = \frac{x}{V} \\\\T = \frac{x}{xH_0} \\\\T = \frac{1}{H_0} \\\\T = \frac{1}{70 \ km/s/Mpc} \\\\T = \frac{1 \ sec}{70 \times 3.24 \times 10^{-20} } \\\\T = 4.41 \times 10^{17} \ sec\\\\T = \frac{4.41 \times 10^{17} \ sec\  \times \ years}{3600 \ s \ \times\  24\ h\  \times \ 365.25 \ days} \\\\T = 1.397  \times 10^{10} \ years\\\\T = 13.97 \ billion \ years

Thus, the age of the galaxy when we look back is 13.97 billion years.

Learn more about Hubble's law here: brainly.com/question/19819028

8 0
3 years ago
We know that the magnitudes of the negative charge on the electron and the positive charge on the proton are equal. Suppose, how
TEA [102]

Answer:3,45 x 10^9 N

Explanation: We have considered the total charge for each coin , this is the total atoms x 29 electrons for cooper and multiplier by electron charge, the total charge for each coin is 0,464 C

Finally we use the Coulomb law,

F=k Q/ (r)^2

6 0
3 years ago
A test charge of 13 mC is at a point P where an external electric field is directed to the right and has a magnitude of 4 3 106
LenKa [72]

Answer:

The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C, but the direction is still to the right.

Explanation:

From coulomb's law, F = Eq

Thus,

F = E₁q₁

F = E₂q₂

Then

E₂q₂ = E₁q₁

E_2 = \frac{E_1q_1}{q_2}

where;

E₂ is the external electric field due to second test charge = ?

E₁ is the external electric field due to first test charge = 4 x 10⁶ N/C

q₁ is the first test charge = 13 mC

q₂ is the second test charge = 23 mC

Substitute in these values in the equation above and calculate E₂.

E_2 = \frac{4*10^6*13}{23} = 2.26 *10^6 \ N/C

The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C when 13 mC test charge is replaced with another test charge of 23 mC.

However, the direction of the external field is still to the right.

8 0
3 years ago
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