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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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Answer:

The stress is calculated as $1.003\times 10^{9}\ Pa$

Solution:

As per the question:

Length of the wire, l = 75.2 cm = 0.752 m

Diameter of the circular cross-section, d = 0.560 mm = $0.560\times 10^{- 3}\ m$

Mass of the weight attached, m = 25.2 kg

Elongation in the wire, $\Delta l = 1.10\ mm = 1.10\times 10^{- 3}\ m$

Now,

The stress in the wire is given by:

$Stress,\ \sigma = \frac{Force,\ F}{Area,\ A}$ (1)

Now,

Force is due to the weight of the attached weight:

F = mg = $25.2\times 9.8 = 246.96\ N$

Cross sectional Area, A = $\pi (\frac{d}{2})^{2} = \pi (\frac{0.560\times 10^{- 3}}{2})^{2} = 2.46\times 10^{- 7}\ m^{2}$

Using these values in eqn (1):

$\sigma = \frac{246.96}{2.46\times 10^{- 7}} = 1.003\times 10^{9}\ Pa$

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Answer:

Explanation:

Given that,

Mass of star M(star) = 1.99×10^30kg

Gravitational constant G

G = 6.67×10^−11 N⋅m²/kg²

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d = 25,000m

R = d/2 = 25,000/2

R = 12,500m

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Then, the person mass which is constant can be determined using

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