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3 years ago

What is modulus of elasticity for shearing stress and strain called?

1 answer:

katrin2010 [14]3 years ago

8 0

The shear modulus, or the modulus of rigidity, is derived from the torsion of a cylindrical test piece. It describes the material's response to shear stress. The shear modulus is one of several quantities for measuring the stiffness of materials and it arises in the generalized Hooke's law. ...

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3 years ago

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3 years ago

if two point charges are separated by 1.5 cm and have charge values of 2.0 and -4.0, respectively, what is the value of the mutu

RUDIKE [14]

Complete question:

if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.

Answer:

The mutual force between the two point charges is 319.64 N

Explanation:

Given;

distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m

value of the charges, q₁ and q₂ = 2 μC and - μ4 C

Apply Coulomb's law;

$F = \frac{k|q_1||q_2|}{r^2}$

where;

F is the force of attraction between the two charges

|q₁| and |q₂| are the magnitude of the two charges

r is the distance between the two charges

k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²

$F = \frac{k|q_1||q_2|}{r^2} \\\\F = \frac{8.99*10^9 *4*10^{-6}*2*10^{-6}}{(1.5*10^{-2})^2} \\\\F = 319.64 \ N$

Therefore, the mutual force between the two point charges is 319.64 N

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3 years ago

When a 0.350-kg package is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The package is now di

AleksandrR [38]

To solve the problem it is necessary to use the concepts related to the calculation of periods by means of a spring constant.

We know that by Hooke's law

$F=kx$

Where,

k = Spring constant

x = Displacement

Re-arrange to find k,

$k= \frac{F}{x}$

$k= \frac{mg}{x}$

$k= \frac{(0.35)(9.8)}{12*10^{-2}}$

$k = 28.58N/m$

Perioricity in an elastic body is defined by

$T = 2\pi \sqrt{\frac{m}{k}}$

Where,

m = Mass

k = Spring constant

$T = 2\pi \sqrt{\frac{0.35}{28.58}}$

$T = 0.685s$

Therefore the period of the oscillations is 0.685s

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3 years ago

A particle moves at a constant speed in a circular path with a radius of r=2.06 cm. If the particle makes four revolutions each

nataly862011 [7]

The centripetal acceleration is $13.0 m/s^2$

Explanation:

For an object in uniform circular motion, the centripetal acceleration is given by

$a=\frac{v^2}{r}$

where

v is the speed of the object

r is the radius of the circle

The speed of the object is equal to the ratio between the length of the circumference ( $2\pi r$ ) and the period of revolution (T), so it can be rewritten as