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Korolek [52]
4 years ago
5

User Location: Inside Viewport Viewport Location: Fresh Viewport in downtown San Fransisco, California Query: [Burger King] Resu

lt: Burger King (located just outside the Fresh Viewport) What steps should you take to rate this result?
Engineering
1 answer:
8_murik_8 [283]4 years ago
4 0

Answer:

Steps should you take to rate this result are

  • Check the pin location.
  • Check the official website for the address .
  • Check if there are closer results that we are not returning.
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I have a Dutch oven that looks like this what do I do?
dangina [55]

Answer:

baking soda and vinegar dish soap

Explanation:

it will create a bubbles and let it sit for 3 hours and it will go away

3 0
3 years ago
Refer to Figure 9-18. A #_____ electrode lead and workpiece lead should be used to carry 150 amperes of electricity 75 feet to t
Anni [7]

Answer:

AC or

Explanation:

BC

8 0
2 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
Which two statements about professional technical jobs in the energy industry are correct?
Tanya [424]
The answer is both B and D
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3 years ago
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A gas at 1650C and 0.96 atm enters a plug flow reactor that converts 91% of the material. What is the amount of gas (in units of
yawa3891 [41]
I’d trying using photomath or tigermath
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4 years ago
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