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sesenic [268]
4 years ago
14

Mr.Haussman has 17 students in his class

Engineering
2 answers:
FrozenT [24]4 years ago
5 0

Answer:

this question is incomplete

Tanya [424]4 years ago
3 0

Answer:

And? what we supposed to answer

Explanation:

Brainlest me

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Which term describes erosion?
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transports solid materials

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3 years ago
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explain how a roller coaster functions using gravity and momentum. Why does it not need an engine or a motor to speed it up and
vichka [17]

Answer:

A roller coaster does not have an engine to generate energy. This builds up a supply of potential energy that will be used to go down the hill as the train is pulled by gravity. Then all of that stored energy is released as kinetic energy which is what will get the train to go up the next hill.

Explanation:

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3 years ago
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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
Buildings designed to partially heat rooms by passive solar heating rely on
Paha777 [63]

Explanation:Image result for buildings designed to partially heat rooms by passive solar heating rely on

A strictly passive design will use the three natural heat transfer modes exclusively—conduction, convection, and radiation. In some applications, however, fans, ducts, and blowers may help with the distribution of heat through the building

4 0
3 years ago
For the reactions of ketone body metabolism, _______.
Artemon [7]

Answer:

Is

Explanation:

7 0
3 years ago
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