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3 years ago

What statement is true about earth tectonic plates

2 answers:

3 0

Continent jig-saw shapes when puzzled and combined together, formed one big continent - Pangea, and was separated by drifts.

Fossil comparisons of different species were discovered into two different, separated continents in which when you combine them, they were one in the past.

Seismic, volcanic, and geothermal activity are found along imagined plate boundaries.

Plates were actually rubbing against each other as evidence is seen on the formed mountain ranges.

Paleomagnetism, magnetic field placement in the layers of the rock are present.

sveta [45]3 years ago

3 0

Answer:

Fossil comparisons of different species were discovered into two different, separated continents in which when you combine them, they were one in the past.

Seismic, volcanic, and geothermal activity are found along imagined plate boundaries.

Plates were actually rubbing against each other as evidence is seen on the formed mountain ranges.

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The Pangaea theory supports the theory of plate tectonics because _____.

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It is based on the idea that all the present continents were on supercontinent.

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3 years ago

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What is the relationship between matter and energy as it changes states of matter (phase changes?)

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These energy exchanges are not changes in kinetic energy. They are changes in bonding energy between the molecules. If heat is coming into a substance during a phase change, then this energy is used to break the bonds between the molecules of the substance. The example we will use here is ice melting into water.

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3 years ago

If a circle was flattened by pushing down on it, it would most likely form into the shape of ----------- which has ----------- f

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The question is looking for "ellipse" and "two" to fill in the blanks.

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3 years ago

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Two 1.0 cm * 2.0 cm rectangular electrodes are 1.0 mm apart. What charge must be placed on each electrode to create a uniform el

kvv77 [185]

Answer:

The number of electrons that must be moved from one electrode to the other to accomplish this is 1.4 X 10⁹ electrons.

Explanation:

Step 1: calculate the charge on each electrode

Given;

Electric field strength = 2.0 X 10⁶ N/C

The distance between the electrode = 1mm = 1 X 10⁻³ m

Electric field strength (E) = Force (F)/Charge (q)

$E =\frac{Kq}{r^2}$

where;

E is the electric field strength = 2.0 X 10⁶ N/C

K is coulomb's constant = 8.99 X 10⁹ Nm²/C²

r is the distance between the electrodes = 1 X 10⁻³ m

q is the charge in each electrode = ?

$q = \frac{Er^2}{K} = \frac{(2X10^6)(1X10^{-3})^2}{8.99 X10 ^9}$ = 0.2225 X 10⁻⁹ C

The charge on each electrode is 0.2225 X 10⁻⁹ C

Step 2: calculate the number of electrons to be moved from one electrode to the other.

1 electron contains 1.602 X 10⁻¹⁹ C

So, 0.2225 X 10⁻⁹ C will contain how many electrons ?

= (0.2225 X 10⁻⁹)/(1.602 X 10⁻¹⁹)

= 1.4 X 10⁹ electrons

Therefore, the number of electrons that must be moved from one electrode to the other to accomplish this is 1.4 X 10⁹ electrons.

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3 years ago

A particle (mass = 2.0 mg, charge = −6.0 μC) moves in the positive direction along the x axis with a velocity of 3.0 km/s. It en

Virty [35]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The acceleration would be $a = 0.003* 6 =0.018\ m/s^2$

Explanation:

The objective of this solution is to obtain the acceleration of the particle

Now looking at Newton law which is mathematically represented as

$F = ma$

Where F is the force experience by a particle

m is the mass of the particle

a is the acceleration of the particle

And also this force is equivalent to magnetic force in a magnetic field which is mathematically represented as

$F = qvB$

Where q is the charge of the particle

v is the velocity of the charge

B is the magnetic field the charge is under it influence

Now equating this two formulas

$ma = qvB$

Making a the subject we have

$a = \frac{qvB}{m}$

In the question the direction of the is in the positive x-axis which is $i$ hence the direction would be in the $i$ direction

So substituting $(2.0i +3.0j+4.0k)mT = (2.0i +3.0j+4.0k)*10^{-3}T$ for B

$a = \frac{q}{m} * v (2.0i +3.0j +4.0k)*10^{-3}$

Substituting $3.0 Km /s = 3.0*10^{3}\ m/s$ for v and $-6.0 \muC = -6.0*10^{-6} C$ for q

$a = \frac{-6.0*10^{-6}}{2.0*10^{-3}} * 3.0*10^{3} *(2i+3j+4k) *10^{-3}$

$a = 0.003 * 3i(2i+3j+4k)$

$a = 0.003 *((3*2)i \ \cdot i \ +(3*3) i \ \cdot \ j \ + (3*4)i \ \cdot \ k)$

According to vector multiplication

$i \cdot i = j \cdot j = k\cdot k = 1\\\\and \ i\cdot j = i\cdot k = 0$

$a = 0.003* 6 =0.018\ m/s^2$

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3 years ago