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galben [10]
4 years ago
8

In no less than one hundred fifty words, explain how microstructures reduce friction within fluid power systems. Mention some of

the benefits of designing with microstructures.
Write your answer in the textbox below, not in an uploaded file.
Engineering
1 answer:
Cerrena [4.2K]4 years ago
6 0

Micro structure can be used to reduce friction and slipping of a smooth surface becoming possible to hold on.

<u>Explanation:</u>

Micro structure is a structure which is very fine and small. It can be made visible if it is seen with the help if a micro scope. It can not be seen other wise directly through the eyes because it is very small is size not becoming visible to the eyes.

These even though are very small is size can be helped to increase the friction on a smooth surface and reduce the slipping of the object off the smooth surface. This will lead to no hurting of the person off a smooth surface.

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In which of the following states would homes most likely have the deepest foundation?
Evgen [1.6K]

Answer:

D) Louisiana, to prevent homes from blowing away in hurricanes.

Explanation:

Louisiana is famous for hurricanes i think that if they were well prepared and had deeper foundation the damage may lower by a good lot.

5 0
3 years ago
A slug travels 3 centimeters in 3 seconds. A snail travels 6 centimeters in 6 seconds. Both travel at constant speeds. Mai says,
Irina-Kira [14]

Answer:

i dont agree with mai because they were both going 1cm per second

Explanation:

3÷3=1

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8 0
3 years ago
20 POINTS
Kobotan [32]

Answer:

True ...................

5 0
3 years ago
A plane wall of thickness 0.1 m and thermal conductivity 25 W/m·K having uniform volumetric heat generation of 0.3 MW/m3 is insu
Contact [7]

Answer:

T = 167 ° C

Explanation:

To solve the question we have the following known variables

Type of surface = plane wall ,

Thermal conductivity k = 25.0 W/m·K,  

Thickness L = 0.1 m,

Heat generation rate q' = 0.300 MW/m³,

Heat transfer coefficient hc = 400 W/m² ·K,

Ambient temperature T∞ = 32.0 °C

We are to determine the maximum temperature in the wall

Assumptions for the calculation are as follows

  • Negligible heat loss through the insulation
  • Steady state system
  • One dimensional conduction across the wall

Therefore by the one dimensional conduction equation we have

k\frac{d^{2}T }{dx^{2} } +q'_{G} = \rho c\frac{dT}{dt}

During steady state

\frac{dT}{dt} = 0 which gives k\frac{d^{2}T }{dx^{2} } +q'_{G} = 0

From which we have \frac{d^{2}T }{dx^{2} }  = -\frac{q'_{G}}{k}

Considering the boundary condition at x =0 where there is no heat loss

 \frac{dT}{dt} = 0 also at the other end of the plane wall we have

-k\frac{dT }{dx } = hc (T - T∞) at point x = L

Integrating the equation we have

\frac{dT }{dx }  = \frac{q'_{G}}{k} x+ C_{1} from which C₁ is evaluated from the first boundary condition thus

0 = \frac{q'_{G}}{k} (0)+ C_{1}  from which C₁ = 0

From the second integration we have

T  = -\frac{q'_{G}}{2k} x^{2} + C_{2}

From which we can solve for C₂ by substituting the T and the first derivative into the second boundary condition s follows

-k\frac{q'_{G}L}{k} = h_{c}( -\frac{q'_{G}L^{2} }{k}  + C_{2}-T∞) → C₂ = q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞

T(x) = \frac{q'_{G}}{2k} x^{2} + q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞ and T(x) = T∞ + \frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} )-x^{2} )

∴ Tmax → when x = 0 = T∞ + \frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} ))

Substituting the values we get

T = 167 ° C

4 0
4 years ago
30 points and brainiest if correct please help A, B, C, D
tatuchka [14]

Answer:

B. to lock the tape into place

Explanation:

the button on the front of the housing locks the tape into place when pressed, preventing the tape from being pulled out further it retracting

4 0
3 years ago
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