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garri49 [273]
3 years ago
5

When water precipitates from the sky, runs off downhill along the ground, or infiltrates down into the soil, its gravitational p

otential energy is transforming into what other form of energy? (two words) *
Engineering
1 answer:
agasfer [191]3 years ago
8 0

Answer:

Geothermal energy.

Explanation:

Geothermal energy is called a renewable energy source because the water is replenished by rainfall, and the heat is continuously produced by the earth.

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What is the purpose of O-ring and valve seals in a cylinder head?
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its to show the shape is flat and only flat at the botom and top and you can set it up ther way and it wlll still look the same.

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Tech A says that speed density systems use vehicle speed and fuel density to determine injector pulse width. Tech B says that ma
bogdanovich [222]

The person that is correct based on the 2 statements from Tech A and Tech B is; Tech B

A mass flow sensor is defined as a sensor that is used to measure the mass flow rate of air entering a fuel-injected internal combustion engine and then sends a voltage that represents the airflow to the electronic control circuit.

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3 years ago
Does a thicker core make an electromagnet stronger?
mel-nik [20]

Answer:

Yes

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6 0
3 years ago
Read 2 more answers
A device that transforms electrical energy to mechanical:
xxTIMURxx [149]

Answer:

electric motor

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Explanation:

3 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
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