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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.
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Paleomagnetism, magnetic field placement in the layers of the rock are present.</span>

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.

You might be interested in

Which energy transformation occurs in the core of a nuclear reactor?

romanna [79]

The correct answer among the choices given is option B. The energy transformation that occurs in the core of a nuclear reactor is from nuclear energy to thermal energy. In a power plant nuclear fission which involves nuclear energy to heat up water around it. This part is the core of the process.

5 0

3 years ago

In comparison with other ocean basins, major sedimentary features such as continental rises and abyssal plains are relatively ra

Maksim231197 [3]

Answer:

Why are continental rises and abyssal plains relatively rare in the Pacific? This is because the extensive system of trenches along the active margins of the Pacific, trap much of the sediments flowing off the continents, preventing them from building the broad, flat abyssal plains typical of the Atlantic ocean basins.

3 0

2 years ago

In mammals, the weight of the heart is approximately 0.5% of the total body weight. Write a linear model that gives the heart we

hammer [34]

Answer:

1201 lbs

Explanation:

Given that in mammals, the weight of the heart is approximately 0.5% of the total body weight.

Let the weight of the heart of a mammal be H

And the weight of the total body be B

The linear model that can gives the heart weight in terms of the total body weight will be:

H = 0.005B

B.) To find the weight of the heart of a whale whose weight is 2.402 × 105 lbs, substitute the whole weight in the formula.

H = 0.005 × 2.402 × 10^5

H = 1201 lbs

Therefore, the weight of the heart of the whale is 1201 lbs

8 0

3 years ago

Which of the following statements about X-rays and radio waves is not true? Which of the following statements about X-rays and r

Lapatulllka [165]

Answer:

X-rays travel through space faster than radio waves.

Explanation:

Electromagnetic waves consist of oscillations of the electric and the magnetic field in a plane perpendicular to the direction of motion the wave.

All electromagnetic waves travel in a vacuum always at the same speed, the speed of light, whose value is:

$c=3.0\cdot 10^8 m/s$

Electromagnetic waves are classified into 7 different types, according to their wavelength/frequency. From shortest to longest wavelength (and so, from highest to lowest frequency), we have:

Gamma rays

X rays

Ultraviolet

Visible light

Infrared radiation

Microwaves

Radio waves

Now we can analyze the 4 statements:

X-rays and radio waves are both forms of light, or electromagnetic radiation --> TRUE. They are both types of electromagnetic waves.

X-rays have higher frequency than radio waves. --> TRUE, as we can see from the table above.

X-rays have shorter wavelengths than radio waves. --> TRUE, as we can see from the table above.

X-rays travel through space faster than radio waves. --> FALSE: all electromagnetic waves travel in space at the same speed, the speed of light.

8 0

2 years ago

Consider a cyclotron in which a beam of particles of positive charge q and mass m is moving along a circular path restricted by

Ulleksa [173]

A) $v=\sqrt{\frac{2qV}{m}}$

B) $r=\frac{mv}{qB}$

C) $T=\frac{2\pi m}{qB}$

D) $\omega=\frac{qB}{m}$

E) $r=\frac{\sqrt{2mK}}{qB}$

Explanation:

When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:

$\Delta U = qV$

where

q is the charge of the particle (positive)

On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):

$\Delta K=\frac{1}{2}mv^2$

where

m is the mass of the particle

v is its final speed

According to the law of conservation of energy, the change (decrease) in electric potential energy is equal to the increase in kinetic energy, so:

$qV=\frac{1}{2}mv^2$

And solving for v, we find the speed v at which the particle enters the cyclotron:

$v=\sqrt{\frac{2qV}{m}}$

When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

$F=qvB$

where B is the strength of the magnetic field.

This force acts as centripetal force, so we can write:

$F=m\frac{v^2}{r}$

where r is the radius of the orbit.

Since the two forces are equal, we can equate them:

$qvB=m\frac{v^2}{r}$

And solving for r, we find the radius of the orbit:

$r=\frac{mv}{qB}$ (1)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

It can be calculated as the ratio between the length of the circumference ( $2\pi r$ ) and the velocity of the particle (v):

$T=\frac{2\pi r}{v}$ (2)

From eq.(1), we can rewrite the velocity of the particle as

$v=\frac{qBr}{m}$

Substituting into(2), we can rewrite the period of revolution of the particle as:

$T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}$

And we see that this period is indepedent on the velocity.

The angular frequency of a particle in circular motion is related to the period by the formula

$\omega=\frac{2\pi}{T}$ (3)

where T is the period.

The period has been found in part C:

$T=\frac{2\pi m}{qB}$

Therefore, substituting into (3), we find an expression for the angular frequency of motion:

$\omega=\frac{2\pi}{(\frac{2\pi m}{qB})}=\frac{qB}{m}$

And we see that also the angular frequency does not depend on the velocity.

For this part, we use again the relationship found in part B:

$v=\frac{qBr}{m}$

which can be rewritten as

$r=\frac{mv}{qB}$ (4)

The kinetic energy of the particle is written as

$K=\frac{1}{2}mv^2$

So, from this we can find another expression for the velocity:

$v=\sqrt{\frac{2K}{m}}$

And substitutin into (4), we find:

$r=\frac{\sqrt{2mK}}{qB}$

So, this is the radius of the cyclotron that we must have in order to accelerate the particles at a kinetic energy of K.

Note that for a cyclotron, the acceleration of the particles is achevied in the gap between the dees, where an electric field is applied (in fact, the magnetic field does zero work on the particle, so it does not provide acceleration).

6 0

3 years ago