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2 years ago

Why is it sometimes difficult for scientists to identify a a fold or fault?

1 answer:

4 0

Folds and faults are difficult to identify because they occur in the interior of rocks and also due to the dense nature of the materials.

<h3>What are faults and folds?</h3>

Faults are lines of weakness are present in materials dues to uneven positioning of the particles of the material.

Folds occurs when infolds occur in materials.

Faults and folds usually occur in rocks.

Folds and faults are difficult to identify because they occur internally and also due to the dense nature of the materials.

Learn more about faults and folds at: brainly.com/question/14240712

You might be interested in

30) A force produces power P by doing work W in a time T. What power will be produced by a force that does six times as much wor

schepotkina [342]

Answer:

A) 12P

Explanation:

The power produced by a force is given by the equation

$P=\frac{W}{T}$

where

W is the work done by the force

T is the time in which the work is done

At the beginning in this problem, we have:

W = work done by the force

T = time taken

So the power produced is

$P=\frac{W}{T}$

Later, the force does six times more work, so the work done now is

$W'=6W$

And this work is done in half the time, so the new time is

$T'=\frac{T}{2}$

Substituting into the equation of the power, we find the new power produced:

$P'=\frac{W'}{T'}=\frac{6W}{T/2}=12\frac{W}{T}=12P$

So, 12 times more power.

4 0

3 years ago

A wall, acted upon by a force of 20 N, does not move. The work done on the wall in this process is

prisoha [69]

Explanation:

A wall that is acted upon by a force of 20N does not move suggests that no work has been done on the wall.

Work done is the force applied to move in the direction of the applied force.

Work done = force x distance.

Since the distance is 0, the work done on the body too is zero.

Work is only done when forces moves a body through a given distance.

Learn more:

Work done brainly.com/question/9100769

#learnwithBrainly

5 0

3 years ago

A magnet in the form of a cylindrical rod has a length of 7.30 cm and a diameter of 1.5 cm. It has a uniform magnetization of 5.

yulyashka [42]

Answer:

Magnetic dipole moment is 0.0683 J/T.

Explanation:

It is given that,

Length of the rod, l = 7.3 cm = 0.073 m

Diameter of the cylinder, d = 1.5 cm = 0.015 m

Magnetization, $M=5.3\times 10^3\ A/m$

The dipole moment per unit volume is called the magnetization of a magnet. Mathematically, it is given by :

$M=\dfrac{\mu}{V}$

$\mu=M\times \pi r^2\times l$

Where

r is the radius of rod, r = 0.0075 m

$\mu=5.3\times 10^3\ A/m\times \pi (0.0075)^2\times 0.073\ m$

$\mu=0.0683\ J/T$

So, its magnetic dipole moment is 0.0683 J/T. Hence, this is the required solution.

6 0

3 years ago

Read 2 more answers

Save

Gelneren [198K]

Answer:

response

Explanation:

Acceleration is your changing Velocity. An object that is ACCELERATING is experiencing a change in velocity. usually positive. if an object such as a car reduces velocity, it is called deceleration

3 0

3 years ago

A disk of radius a has a total charge Q uniformly distributed over its surface. The disk has negligible thickness and lies in th

sleet_krkn [62]

The electric potential V(z) on the z-axis is : V = $(\frac{Q}{a^2} ) [ (a^2 + z^2)^{\frac{1}{2} } -z$

The magnitude of the electric field on the z axis is : E = kб 2 $\pi$ ( 1 - [z / √(z² + a² ) ] )

<u>Given data :</u>

V(z) =2kQ / a²(v(a² + z²) ) -z

<h3>Determine the electric potential V(z) on the z axis and magnitude of the electric field</h3>

Considering a disk with radius R

Charge = dq

Also the distance from the edge to the point on the z-axis = √ [R² + z²].

The surface charge density of the disk ( б ) = dq / dA

Small element charge dq = б( 2πR ) dr

dV $\frac{k.dq}{\sqrt{R^2+z^2} } \\\\= \frac{k(\alpha (2\pi R)dR}{\sqrt{R^2+z^2} }$ ----- ( 1 )

Integrating equation ( 1 ) over for full radius of a

∫dv = $\int\limits^a_o {\frac{k(\alpha (2\pi R)dR)}{\sqrt{R^2+z^2} } } \,$

V = $\pi k\alpha [ (a^2+z^2)^\frac{1}{2} -z ]$

= $\pi k (\frac{Q}{\pi \alpha ^2})[(a^2 +z^2)^{\frac{1}{2} } -z ]$

Therefore the electric potential V(z) = $(\frac{Q}{a^2} ) [ (a^2 + z^2)^{\frac{1}{2} } -z$

Also

The magnitude of the electric field on the z axis is : E = kб 2 $\pi$ ( 1 - [z / √(z² + a² ) ] )

Hence we can conclude that the answers to your question are as listed above.

Learn more about electric potential : brainly.com/question/25923373

7 0

2 years ago