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mrs_skeptik [129]
2 years ago
6

A vertebra is subjected to a shearing force of 535 N. Find the shear deformation, taking the vertebra to be a cylinder 3.00 cm h

igh and 4.00 cm in diameter. The shear modulus for vertebrae is 8.00 × 1010 N/m2.
Physics
1 answer:
Ganezh [65]2 years ago
5 0
4.00 is the answer i think im not sure either 4.00 or 8.00
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What is 62.125 kg rounded to two significant figures?
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Which BEST describes the physical properties of the Earth’s core?
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C) a solid lower part and a liquid upper part

Explanation:

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The energy required to completely remove the covalent bond between two
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Answer:

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7 0
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A circuit component with a high resistance will have a low electric current.
Zepler [3.9K]
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6 0
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The angle θ is slowly increased. Write an expression for the angle at which the block begins to move in terms of μs.
Reika [66]

Answer:

tan \theta = \mu_s

Explanation:

An object is at rest along a slope if the net force acting on it is zero. The equation of the forces along the direction parallel to the slope is:

mg sin \theta - \mu_s R =0 (1)

where

mg sin \theta is the component of the weight parallel to the slope, with m being the mass of the object, g the acceleration of gravity, \theta the angle of the slope

\mu_s R is the frictional force, with \mu_s being the coefficient of friction and R the normal reaction of the incline

The equation of the forces along the direction perpendicular to the slope is

R-mg cos \theta = 0

where

R is the normal reaction

mg cos \theta is the component of the weight perpendicular to the slope

Solving for R,

R=mg cos \theta

And substituting into (1)

mg sin \theta - \mu_s mg cos \theta = 0

Re-arranging the equation,

sin \theta = \mu_s cos \theta\\\rightarrow tan \theta = \mu_s

This the condition at which the equilibrium holds: when the tangent of the angle becomes larger than the value of \mu_s, the force of friction is no longer able to balance the component of the weight parallel to the slope, and so the object starts sliding down.

4 0
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