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jarptica [38.1K]
3 years ago
8

A substance is round in one container and has a volume of one liter. In a much larger container, the substance is rectangular bu

t still has a volume of one liter.
What state of matter is the substance in?
Physics
1 answer:
UkoKoshka [18]3 years ago
8 0
Answer: liquid

explanation: 1 liter is a measurement of liquids, not solids, or gases.
Liquids also have a set volume, but can flow to take the shape of the bottom of their container.
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Answer:Questlon 20 of 20 Which change to an object would quadruple its kinetic energy?

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B. Increasing its velocitato twice its original value

C. Reducing its velocity to one-half of its original value O

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C. Reducing its velocity to one-half of its original value O

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A. Reducing its mass to one-half of its original value

B. Increasing its velocitato twice its original value

C. Reducing its velocity to one-half of its original value O

D. Increasing its mass to twice its original value ​Questlon 20 of 20 Which change to an object would quadruple its kinetic energy?

A. Reducing its mass to one-half of its original value

B. Increasing its velocitato twice its original value

C. Reducing its velocity to one-half of its original value O

D. Increasing its mass to twice its original value ​Questlon 20 of 20 Which change to an object would quadruple its kinetic energy?

A. Reducing its mass to one-half of its original value

B. Increasing its velocitato twice its original value

C. Reducing its velocity to one-half of its original value O

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A. Reducing its mass to one-half of its original value

B. Increasing its velocitato twice its original value

C. Reducing its velocity to one-half of its original value O

D. Increasing its mass to twice its original value ​Questlon 20 of 20 Which change to an object would quadruple its kinetic energy?

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A. Reducing its mass to one-half of its original value

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Explanation:

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2 years ago
Read 2 more answers
What type of energy is related to mass and the speed of an object?
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An energy that is related to mass and speed would most likely be kinetic energy.
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2 years ago
What is the force per unit area at this point acting normal to the surface with unit nor- Side View √√ mal vector n = (1/ 2)ex +
Mumz [18]

Complete Question:

Given \sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right] at a point. What is the force per unit area at this point acting normal to the surface with\b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z   ? Are there any shear stresses acting on this surface?

Answer:

Force per unit area, \sigma_n = 28 MPa

There are shear stresses acting on the surface since \tau \neq 0

Explanation:

\sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right]

equation of the normal, \b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z

\b n = \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right]

Traction vector on n, T_n = \sigma \b n

T_n =  \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right] \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right]

T_n = \left[\begin{array}{ccc}\frac{23}{\sqrt{2} }\\0\\\frac{27}{\sqrt{33} }\end{array}\right]

T_n = \frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z

To get the Force per unit area acting normal to the surface, find the dot product of the traction vector and the normal.

\sigma_n = T_n . \b n

\sigma \b n = (\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z) . ((1/ \sqrt{2} ) \b e_x + 0 \b  e_y +(1/ \sqrt{2}) \b e_z)\\\\\sigma \b n = 28 MPa

If the shear stress, \tau, is calculated and it is not equal to zero, this means there are shear stresses.

\tau = T_n  - \sigma_n \b n

\tau =  [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - 28( (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z)\\\\\tau =  [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - [ (28/ \sqrt{2} ) \b e_x + (28/ \sqrt{2}) \b e_z]\\\\\tau =  \frac{-5}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{5}{\sqrt{2} } \b e_z

\tau = \sqrt{(-5/\sqrt{2})^2  + (27/\sqrt{2})^2 + (5/\sqrt{2})^2} \\\\ \tau = 19.74 MPa

Since \tau \neq 0, there are shear stresses acting on the surface.

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