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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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What is the charge on 1.0 kg of protons? (e = 1.60 × 10-19 c, mproton = 1.67 × 10-27 kg)?
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First, we need to find the number of protons, which is the total mass divided by the mass of one proton:
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Then, the total charge is the number of protons times the charge of a  single proton:
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Find the velociity of a car which travels 35 m to the right over a period of 40 seconds
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Explanation:

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8 0
3 years ago
A proton is projected toward a fixed nucleus of charge Ze with velocity vo. Initially the two particles are very far apart. When
11111nata11111 [884]

Answer:

The value is R_f =  \frac{4}{5}  R

Explanation:

From the question we are told that

   The  initial velocity of the  proton is v_o

    At a distance R from the nucleus the velocity is  v_1 =  \frac{1}{2}  v_o

    The  velocity considered is  v_2 =  \frac{1}{4}  v_o

Generally considering from initial position to a position of  distance R  from the nucleus

 Generally from the law of energy conservation we have that  

       \Delta  K  =  \Delta P

Here \Delta K is the change in kinetic energy from initial position to a  position of  distance R  from the nucleus , this is mathematically represented as

      \Delta K  =  K__{R}} -  K_i

=>    \Delta K  =  \frac{1}{2}  *  m  *  v_1^2  -  \frac{1}{2}  *  m  *  v_o^2

=>    \Delta K  =  \frac{1}{2}  *  m  * (\frac{1}{2} * v_o )^2  -  \frac{1}{2}  *  m  *  v_o^2

=>    \Delta K  =  \frac{1}{2}  *  m  * \frac{1}{4} * v_o ^2  -  \frac{1}{2}  *  m  *  v_o^2

And  \Delta  P is the change in electric potential energy  from initial position to a  position of  distance R  from the nucleus , this is mathematically represented as

          \Delta P =  P_f - P_i

Here  P_i is zero because the electric potential energy at the initial stage is  zero  so

             \Delta P =  k  *  \frac{q_1 * q_2 }{R}  - 0

So

           \frac{1}{2}  *  m  * \frac{1}{4} * v_o ^2  -  \frac{1}{2}  *  m  *  v_o^2 =   k  *  \frac{q_1 * q_2 }{R}  - 0

=>        \frac{1}{2}  *  m  *v_0^2 [ \frac{1}{4} -1 ]  =   k  *  \frac{q_1 * q_2 }{R}

=>        - \frac{3}{8}  *  m  *v_0^2  =   k  *  \frac{q_1 * q_2 }{R} ---(1 )

Generally considering from initial position to a position of  distance R_f  from the nucleus

Here R_f represented the distance of the proton from the nucleus where the velocity is  \frac{1}{4} v_o

     Generally from the law of energy conservation we have that  

       \Delta  K_f  =  \Delta P_f

Here \Delta K is the change in kinetic energy from initial position to a  position of  distance R  from the nucleus  , this is mathematically represented as

      \Delta K_f   =  K_f -  K_i

=>    \Delta K_f  =  \frac{1}{2}  *  m  *  v_2^2  -  \frac{1}{2}  *  m  *  v_o^2

=>    \Delta K_f  =  \frac{1}{2}  *  m  * (\frac{1}{4} * v_o )^2  -  \frac{1}{2}  *  m  *  v_o^2

=>    \Delta K_f  =  \frac{1}{2}  *  m  * \frac{1}{16} * v_o ^2  -  \frac{1}{2}  *  m  *  v_o^2

And  \Delta  P is the change in electric potential energy  from initial position to a  position of  distance R_f  from the nucleus , this is mathematically represented as

          \Delta P_f  =  P_f - P_i

Here  P_i is zero because the electric potential energy at the initial stage is  zero  so

             \Delta P_f  =  k  *  \frac{q_1 * q_2 }{R_f }  - 0      

So

          \frac{1}{2}  *  m  * \frac{1}{8} * v_o ^2  -  \frac{1}{2}  *  m  *  v_o^2 =   k  *  \frac{q_1 * q_2 }{R_f }

=>        \frac{1}{2}  *  m  *v_o^2 [-\frac{15}{16} ]  =   k  *  \frac{q_1 * q_2 }{R_f }

=>        - \frac{15}{32}  *  m  *v_o^2 =   k  *  \frac{q_1 * q_2 }{R_f } ---(2)

Divide equation 2  by equation 1

              \frac{- \frac{15}{32}  *  m  *v_o^2 }{- \frac{3}{8}  *  m  *v_0^2  } }   =  \frac{k  *  \frac{q_1 * q_2 }{R_f } }{k  *  \frac{q_1 * q_2 }{R } }}

=>           -\frac{15}{32 } *  -\frac{8}{3}   =  \frac{R}{R_f}

=>           \frac{5}{4}  =  \frac{R}{R_f}

=>             R_f =  \frac{4}{5}  R

   

7 0
3 years ago
A ray of light incident in air strikes a rectangular glass block of refractive index 1.50, at an angle of incidence of 45°. Calc
balandron [24]

Answer:

Approximately 28^{\circ}.

Explanation:

The refractive index of the air n_{\text{air}} is approximately 1.00.

Let n_\text{glass} denote the refractive index of the glass block, and let \theta _{\text{glass}} denote the angle of refraction in the glass. Let \theta_\text{air} denote the angle at which the light enters the glass block from the air.

By Snell's Law:

n_{\text{glass}} \, \sin(\theta_{\text{glass}}) = n_{\text{air}} \, \sin(\theta_{\text{air}}).

Rearrange the Snell's Law equation to obtain:

\begin{aligned} \sin(\theta_{\text{glass}}) &= \frac{n_{\text{air}} \, \sin(\theta_{\text{air}})}{n_{\text{glass}}} \\ &= \frac{(1.00)\, (\sin(45^{\circ}))}{1.50} \\ &\approx 0.471\end{aligned}.

Hence:

\begin{aligned} \theta_{\text{glass}} &= \arcsin (0.471) \approx 28^{\circ}\end{aligned}.

In other words, the angle of refraction in the glass would be approximately 28^{\circ}.

7 0
2 years ago
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