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3 years ago

Along boundaries plates are brittle and the crust fractures (breaks and cracks) forming ___

Physics

1 answer:

docker41 [41]3 years ago

5 0

Answer:

Along boundaries plates are brittle and the crust fractures (breaks and cracks) forming Faults

Explanation:

Brittle material are substances that looks hard but can be easily broken by strain. Brittle deformation leads to crack and fracture of the material involve. Brittle crust lacks elastic ability , elasticity in this rocks are almost absent.

Fractures and cracks in rocks form fault. Fault is a discontinuity in rocks with an appreciable displacement. Fault is a displacement in rocks relative to one part. Generally, Fault occur when brittle rock fracture or crack, the fractured rocks move relative to one another.

Differential stress can cause rocks to fracture. If that part of the crust lacks ductility the stress on the rock will cause a brittle deformation. The brittle deformation causes fracture(breaks and cracks) leading to structure like Faults and Joints.

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3 years ago

Your Lead Teaching Assistant is initially seated on the top of a hemispherical ice mound of radius R = 30 m. Approximate the ice

saw5 [17]

Answer:

<em>Height = 5.65 km</em>

Explanation:

$3\pi r^2$ is the circumference or we can say measures the boundary of hemisphere of friction-less ice that he is sitting on.

So, the height will be = 2 x 3.14 x (30)^2 = 5654.7 m = 5.65 km

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3 years ago

Read 2 more answers

C. A helium atom has a diameter of approximately 9.8 • 10-11 meters. What is the diameter of a helium atom in nanometers? Given

Studentka2010 [4]

$1\ nm = 10^{-9} m\\\\1\ m =\frac{1\ nm}{10^{-9}}$

Since the diameter of helium atom is approximately $9.8\times 10^{-11}\ m$ , therefore the diameter of helium atom in nano meter,

$9.8\times 10^{-11}\ m=9.8\times 10^{-11} \times\frac{1\ nm}{10^{-9}}=0.098\ nm$

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3 years ago

A telephone line has a signal-to-noise ratio of 1000 and a bandwidth of 4 KHz. What is the maximum data rate supported by this l

ivolga24 [154]

Answer:

The maximum data rate supported by this line is 39900 bps

Explanation:

The maximum data rate supported by this line can be obtained using the formula below

c = W*log2(S/N+1)

where;

c is the maximum data rate supported by the line

W is the bandwidth = 4kHz

S/N+1 is the signal to noise ratio = 1001

c = 4*log2(1001)

c = 39868.9 ≅ 39900 bps

Therefore, the maximum data rate supported by this line is 39900 bps

5 0

3 years ago

Jill can use a force of 12.0 N to lift a single box up 5.00 m. How many of these boxes must she lift it in a minute to use 60.0

Mama L [17]

Answer:

60 boxes

Explanation:

The work done by lifting a single box is equal to the force applied (the weight of the box) times the displacement of the box:

$W_1 = Fd=(12.0 N)(5.00 m)=60 J$

Power is related to the work done by the equation:

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

where W is the work done and t is the time. In this problem, we are told that the power used is P=60.0 W, while the time taken is t = 1 min = 60 s, so the total work done must be

$W=Pt=(60.0 W)(60 s)=3600 J$

Therefore, the number of boxes that she has to lift in order to use this power is the total work divided by the work done in lifting each box:

$N=\frac{W}{W_1}=\frac{3,600 J}{60 J}=60$

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3 years ago