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Elena L [17]
3 years ago
5

Match the bones with their common name.

Physics
1 answer:
mrs_skeptik [129]3 years ago
7 0

1. clavicle = collarbone

2. vertebrae = backbone

3. scapula = shoulder blade

4. femur = thigh

5. humerus = upper arm

6. patella = kneecap

7. cranium = skull

8. tibia = lower leg

9. radius/ulna = forearm

10. phalanges = fingers/toes

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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
A 120-kg roller coaster cart is being tested on a new track, and a crash-test dummy is loaded into itThe roller coaster starts f
Black_prince [1.1K]

Answer:

a) variation of the energy is equal to the work of the friction force

b) W = Em_{f} -Em₀ ,  c) he conservation of mechanical energy

Explanation:

a) In an analysis of this problem we can use the energy law, where at the moment the mechanical energy is started it is totally potential, and at the lowest point it is totally kinetic, we can suppose two possibilities, that the friction is zero and therefore by equalizing the energy we set the velocity at the lowest point.

 Another case is if the friction is different from zero and in this case the variation of the energy is equal to the work of the friction force, in value it will be lower than in the calculations.

b) the calluses that he would use are to hinder the worker's friction force and energy

          W = Em_{f} -Em₀

          N d = ½ m v² - m g (y₂-y₁)

          y₂-y₁ = 35 -10 = 25m

c) if there is no friction, the physical principle is the conservation of mechanical energy

 If there is friction, the principle is that the non-conservative work is equal to the variation of the energy

7 0
3 years ago
A jet plane comes in for a downward dive as shown in the figure below.
xxTIMURxx [149]

The solution would be like this for this specific problem:

<span>5.5 g = g + v^2/r </span><span>
<span>4.5 g = v^2/r </span>
<span>v^2 = 4.5 g * r </span>
<span>v = sqrt ( 4.5 *9.81m/s^2 * 350 m) </span>
v = 124 m/s</span>

So the pilot will black out for this dive at 124 m/s. I am hoping that these answers have satisfied your query and it will be able to help you in your endeavors, and if you would like, feel free to ask another question.

8 0
3 years ago
Read 2 more answers
What is a type of pulley that increases the size and effort of force
12345 [234]
The answer is Movable
5 0
3 years ago
(PLEASE HELP) What area of the Earth's mantle is the densest?
Nady [450]
The area nearest to the earth's core is the densest.. 
So your answer would be letter choice ( D ) . . . 

Hope it Helped :) 
7 0
3 years ago
Read 2 more answers
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