1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
chubhunter [2.5K]
2 years ago
13

How do earths magnetic pole reversals provide evidence for plate tectonics

Physics
2 answers:
Alinara [238K]2 years ago
6 0
Rocks get older as you move away from the water surface
lianna [129]2 years ago
5 0

Answer:

Explanation:

As per the concept of plate tectonics, The continents are like rock plates moving on the magma underneath it and magnetism is a part of it.

As we know Earth spin on its axis, this interaction of Earth's spin and magnetic material inside the Earth creates a magnetic field from North to South pole.

When hot material inside the Earth cool, it solidify and the magnetic material form crystals which align themselves along the Earth's magnetic field.

Hence by this way Earth's magnetic pole reversal provide evidence for plate tectonics.

You might be interested in
Velocityis a vector quantity which has both magnitude and direction. Using complete sentences, describethe object's velocity. Co
mr_godi [17]

Answer:

Explanation:

A physical quantity which can be completely described by the magnitude and direction both are called vector quantities. For example, displacement, velocity, etc.

A physical quantity which can be completely explained by the magnitude only is called scalar quantity. For example, mass, time, etc.

8 0
3 years ago
A charge of -3.02 μC is fixed in place. From a horizontal distance of 0.0377 m, a particle of mass 9.43 x 10^-3 kg and charge -9
Andreyy89

Answer:

d = 0.0306 m

Explanation:

Here we know that for the given system of charge we have no loss of energy as there is no friction force on it

So we will have

U + K = constant

\frac{kq_1q_2}{r_1} + \frac{1}{2}mv_1^2 = \frac{kq_1q_2}{r_2} + \frac{1}{2}mv_2^2

now we know when particle will reach the closest distance then due to electrostatic repulsion the speed will become zero.

So we have

\frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{0.0377} + \frac{1}{2}(9.43 \times 10^{-3})(80.4)^2 = \frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{r} + 0

7.05 + 30.5 = \frac{0.266}{r}

r = 7.08 \times 10^{-3} m

so distance moved by the particle is given as

d = r_1 - r_2

d = 0.0377 - 0.00708

d = 0.0306 m

6 0
3 years ago
There is a seasaw that's holding two men. The seesaw has a length of 18m that can pivot from a point at its center. Man 1 has a
Alex

Answer:

Distance=  2.3864m

Explanation:

So that the balance is in equilibrium parallel to the floor, we must match the moment each man makes with respect to the pivot point.

In many cases the point of application of force does not coincide with the point of application in the body. In this case the force acts on the object and its structure at a certain distance, by means of an element that transfers that action of this force to the object.

This combination of force applied by the distance to the point of the structure where it is applied is called the moment of force F with respect to the point. The moment will attempt a rotation shift or rotation of the object. The distance from the force to the point of application is called the arm.

Mathematically it is calculated by expression:

M= F×d

The moment caused by the first man is:

M1= 75kg × (9.81m/s²) × 1.75m= 1287.5625 N×m

The moment caused by the second man must be equal to that caused by the first by which:

M2= 1287.5625 N×m= 55kg × (9.81m/s²) × distance ⇒

⇒distance= (1287.5625 N×m)/( (55kg × (9.81m/s²) )= 2.3864m

At this distance from the pivot point, the second should sit down so that the balance is balanced parallel to the ground.

3 0
2 years ago
if you have to apply 40 n of force on a crowbar to lift a 400 n rock what is the actual machanical andvatage of the crow bar
Eduardwww [97]
M.A. = 400/40

M.A. = 10

Hope this helps!
7 0
3 years ago
Derive the equation of motion of the block of mass m1 in terms of its displacement x. The friction between the block and the sur
Alenkasestr [34]

Answer:

the equivalent mass : m_e = m_1+m_2+\frac{I}{R^2}

the equation of the motion of the block of mass m_1 in terms of its displacement is = (m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)

Explanation:

Let use m₁ to represent the mass of the block and m₂ to represent the mass of the cylinder

The radius of the cylinder  be = R

The distance between the center of the pulley to center of the block to be = x

Also, the angles of inclinations of the cylinder and the block with respect to the ground to be \phi and \beta respectively.

The velocity of the block to be = v

The equivalent mass of the system = m_e

In the terms of the equivalent mass, the kinetic energy of the system can be written as:

K.E = \frac{1}{2} m_ev^2       --------------- equation (1)

The angular velocity of the cylinder = \omega  :  &

The inertia of the cylinder about its center to be = I

The angular velocity of the cylinder can be written as:

v = \omega R

\omega =\frac{v}{R}

The kinetic energy of the system in terms of individual mass can be written as:

K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I\omega^2

By replacing \omega with \frac{v}{R} ; we have:

K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I(\frac{v}{R})^2

K.E = \frac{1}{2}(m_1+ m_2+ \frac{I}{R} )v^2   ------------------ equation (2)

Equating both equation (1) and (2); we have:

m_e = m_1+m_2+\frac{I}{R^2}

Therefore, the equivalent mass : m_e = m_1+m_2+\frac{I}{R^2}    which is read as;

The equivalent mass is equal to the mass of the block plus the mass of the cylinder plus the inertia by  the square of the radius.

The expression for the force acting on equivalent mass due to the block is as follows:

f_{block }=m_1gsin \beta

Also; The expression for the force acting on equivalent mass due to the cylinder is as follows:

f_{cylinder} = m_2gsin \phi

Equating the above both equations; we have the equation of motion of the  equivalent system to be

m_e \bar x = f_{cylinder}-f_{block}

which can be written as follows from the previous derivations

(m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)

Finally; the equation of the motion of the block of mass m_1 in terms of its displacement is = (m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)

8 0
3 years ago
Other questions:
  • What would be most useful to help make a simple compass a nonmetal bar,a round metal can,a small iron nail,or a Quartz needle
    8·2 answers
  • 90 points whats the equation for how fast something will be traveling before it hits the ground. How to do the equation PLEASE E
    7·2 answers
  • g If the x-component of a force vector is 5.69 newtons and its y-component is 8.00 newtons, then what is its magnitude?
    8·2 answers
  • The rate at which correct flows is measured in _?
    7·1 answer
  • A 0.850 kg mass is placed on a
    14·1 answer
  • Can you find a vector quantity that has a magnitude of zero but components that are not zero? Explain. Can the magnitude of a ve
    7·1 answer
  • Which option is an element?
    12·2 answers
  • The bar graph shows energy data taken from a roller coaster at a theme park. analyze the data and assess its validity. 3-5 sente
    11·1 answer
  • How to change mass but keep the force the same?
    9·1 answer
  • The gravitation force between two masses (36n if the distance between masses is tripled, the force of gravity will be).
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!