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choli [55]
3 years ago
13

A fold is a _ in a rock , and a fault is a _in a rock?

Physics
2 answers:
777dan777 [17]3 years ago
3 0
A geological fold<span> occurs when one or a stack of originally flat and planar surfaces, such as sedimentary strata, are bent or curved as a result of permanent deformation.

So A fold is a Bend? in a rock. Maybe.

</span>A fault<span> is a planar fracture or discontinuity in a volume of </span>rock<span>, across which there has been significant displacement as a result of </span>rock<span>-mass movement.</span>
marissa [1.9K]3 years ago
3 0

Answer:

A bold is a bent in a rock.

A fault is a crack in a rock

Explanation:

A bold is a bent in a rock.

In earth's structure, when a flat plate of sedimentary rocks are pressed due to the plates movement into one another, the plate structure takes the shape of a bent structure or curved structure. The rocks buckle into folds. Fold are commonly seen in rock structure that is layered.

A fault is a crack in a rock.

Faults in a rock structure is formed due to stress. Faults are associated with the boundaries of the earth's tectonic plate. Fault is a zone of fracture between two blocks of mountain rocks and it causes relative motion in two sets of rock blocks.

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Answer:

The position of the spring in terms of g, m & k is x = \frac{m g}{k}

Explanation:

Stiffness of the spring = k

Mass = m

When a mass m is attached with the spring then spring stretched. in that case the force exerted on the spring is equal to weight of the mass attached.

⇒ Force exerted on the spring F = k x

⇒ m g = k x

⇒ x = \frac{m g}{k}

This is the position of the spring in terms of g, m & k.

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3 years ago
Two long, straight wires are separated by a distance of 9.15 cm . One wire carries a current of 2.79 A , the other carries a cur
Dafna1 [17]

Answer:

The force is the same

Explanation:

The force per meter exerted between two wires carrying a current is given by the formula

\frac{F}{L}=\frac{\mu_0 I_1 I_2}{2\pi r}

where

\mu_0 is the vacuum permeability

I_1 is the current in the 1st wire

I_2 is the current in the 2nd wire

r is the separation between the wires

In this problem

I_1=2.79 A\\I_2=4.36 A\\r = 9.15 cm = 0.0915 m

Substituting, we find the force per unit length on the two wires:

\frac{F}{L}=\frac{(4\pi \cdot 10^{-7})(2.79)(4.36)}{2\pi (0.0915)}=2.66\cdot 10^{-5}N

However, the formula is the same for the two wires: this means that the force per meter exerted on the two wires is the same.

The same conclusion comes out  from Newton's third law of motion, which states that when an object A exerts a force on an object B, then object B exerts an equal and opposite force on object A (action-reaction). If we apply the law to this situation, we see that the force exerted by wire 1 on wire 2 is the same as the force exerted by wire 2 on wire 1 (however the direction is opposite).

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Yosef applies an input force of 50 N to a crowbar. The crowbar applies a force of 750 N to the lid of a crate. What is the mecha
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Read 2 more answers
A muon has a rest mass energy of 105.7 MeV, and it decays into an electron and a massless particle. If all the lost mass is conv
sergeinik [125]

Answer:

The electron’s velocity is 0.9999 c m/s.

Explanation:

Given that,

Rest mass energy of muon = 105.7 MeV

We know the rest mass of electron = 0.511 Mev

We need to calculate the value of γ

Using formula of energy

K_{rel}=(\gamma-1)mc^2

\dfrac{K_{rel}}{mc^2}=\gamma-1

Put the value into the formula

\gamma=\dfrac{105.7}{0.511}+1

\gamma=208

We need to calculate the electron’s velocity

Using formula of velocity

\gamma=\dfrac{1}{\sqrt{1-(\dfrac{v}{c})^2}}

\gamma^2=\dfrac{1}{1-\dfrac{v^2}{c^2}}

\gamma^2-\gamma^2\times\dfrac{v^2}{c^2}=1

v^2=\dfrac{1-\gamma^2}{-\gamma^2}\times c^2

Put the value into the formula

v^2=\dfrac{1-(208)^2}{-208^2}\times c^2

v=c\sqrt{\dfrac{1-(208)^2}{-208^2}}

v=0.9999 c\ m/s

Hence, The electron’s velocity is 0.9999 c m/s.

6 0
2 years ago
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