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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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280 N

Explanation:

Applying Newton's third second law of motion,

F = m(v-u)/t................... Equation 1

Where F = Magnitude of the average force on the ball during contact, v = final velocity of the ball, u = initial velocity of the ball, t = time of contact of the ball and the wall.

Note: Let the direction of the initial velocity of the ball be positive

Given: m = 4 kg, u = 3.0 m/s, v = -4.0 m/s (bounce off), t = 0.1 s

Substitute into equation 1

F = 4(-4-3)/0.1

F = 4(-7)/0.1

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A 0.10 g honeybee acquires a charge of +23 pC while flying.
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Answer:

a) \frac{F}{w} =2.347\times 10^{-6}\ N

b) E=4.2609\times 10^7\ N.C^{-1} parallel to the earth surface.

  • In this case according to the Fleming's left hand rule the direction of movement of bee must be in a direction parallel to the earth surface and perpendicular to the electric field at the same time.

Explanation:

Given:

mass of the bee, m=10^{-4}\ kg

charge acquired by the bee, q_2=23\times 10^{-12}\ C

a.

Electrical field near the earth surface, E=100\ N.C^{-1}

Now the electric force on the bee:

we know:

F=\frac{1}{4\pi.\epsilon_0} \times \frac{q_1.q_2}{r^2}

F=E.q_2

F=100\times 23\times 10^{-12}

F=23\times 10^{-10}\ N

The weight of the bee:

w=m.g

w=10^{-4}\times 9.8

w=9.8\times10^{-4}\ N

Therefore the ratio :

\frac{F}{w} =\frac{23\times 10^{-10}}{9.8\times10^{-4}}

\frac{F}{w} =2.347\times 10^{-6}\ N

b.

The condition for the bee to hang is its weight must get balanced by the electric force acing equally in the opposite direction.

So,

F=9.8\times10^{-4}\ N

E.q_2=9.8\times10^{-4}\ N

E\times 23\times 10^{-12}=9.8\times10^{-4}\ N

E=4.2609\times 10^7\ N.C^{-1} parallel to the earth surface.

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Answer:

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