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Ket [755]
3 years ago
9

Ben leads a team of a few engineers at a robotics firm. A couple of them would like to improve their skills by taking additional

courses in specialized areas. What are these courses in specialized areas called? A. technical certificate courses B. undergraduate certificate courses C. master’s degree courses D. online courses
Engineering
1 answer:
Anit [1.1K]3 years ago
4 0

Answer:

i dont know

Explanation:

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What is 94*738^389428394
Lady_Fox [76]

Answer:

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Explanation:

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3. Low-voltage conductors rarely cause<br> injuries.
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True
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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
3 years ago
How do you take a picture
Mrrafil [7]
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7 0
3 years ago
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There are four distinct events that occur in the Carnot cycle. Name the events and describe each one.
OLEGan [10]

Answer and Explanation:

There four events or stages of reversible nature in a Carnot cycle are:

1. Isothermal expansion of gas

2. Adiabatic expansion

3. Isothermal compression of gas

4.Adiabatic compression of gas

All these 4 processes are reversible processes.

1. A reversible Isothermal expansion of gas takes place in Carnot cycle where an ideal gas  absorbs or intake certain quantity of heat from a heat reservoir or source at an elevated temperature which results in the expansion of gas and work is done on the  surroundings.

2. A reversible expansion of gas adiabatically takes place in Carnot cycle in an environment with thermal insulation where the gas expand and work is done on the surrounding which results in the lowering of the temperature of the system.

3. A reversible compression of gas  isothermally takes place such that the work is done on the system by the surrounding resulting in heat loss.

4. A reversible compression of gas adiabatically, takes place in an environment with thermal insulation and the work is done continuously on the system by the surroundings resulting in the rise in temperature.

8 0
3 years ago
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