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mars1129 [50]
3 years ago
11

El movimiento de los cuerpos y la formación de imágenes son ejemplos de esta clase de fenómeno ​

Physics
1 answer:
77julia77 [94]3 years ago
7 0

Answer:

depende de que fenómenos nos referimos de acuerdo al los cuerpos de formación puede aver movimiento contante

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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.

3 0
3 years ago
The hydrocarbon used to manufacture foam plastics is called styrene. Analysis shows composition to be 92.25% C and 7.75% H. Anal
zhannawk [14.2K]

Answer:

molecular formula = C_{8} H_{8}

Explanation:

Given data

c = 92.25%

H = 7.75%

molar mass = 104 g/mol

to find out

the empirical and molecular formula for styrene

solution

we know that

styrene 1 g contain = 0.9225 g C and 0.0775 g H

so

C = 104 × 0.9225 g / 12 g/mol

C = 7.995 mol = approx 8 mol

and

H = 104 × 0.0775 g / 1 g/mol

H = 8.06 mol = approx 8 mol

so we say that 1 mole of styrene  have 8 mole of C and H

so

molecular formula = C_{8} H_{8}

8 0
4 years ago
OMG PLZ PLZ HELP!!
horsena [70]
1,3,4 I’m pretty sur e
4 0
3 years ago
Read 2 more answers
a lamp hangs from the ceiling at the height of 2.9 m. if the lamp breaks and falls to the floor what is its impact speed
nydimaria [60]
This is a problem that can be solved using free-fall motion analysis. Since the displacement (2.9m) is given, we can use the following equation to solve for the impact speed:

V^2 = 2gh

V = sqrt (2*9.8*2.9)
V = 7.54 m/s
5 0
4 years ago
Determine the centripetal force on a vehicle rounding a circular curve with a radius of 80 m at a constant speed of 90 km/h if t
quester [9]
Mass of the vehicle = 2000 kg
Velocity of the vehicle = 90 km/hr
                                   = 25 m/s
Radius of the curve = 80 m
Then
Centripetal force = [2000 * (25)^2]/80
                            = (2000 * 625)/80
                            = 15625 kg m/s
                            = 15625 Newton second
I hope that this is the answer that you were looking for and the answer has come to your desired help. 
3 0
3 years ago
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