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3 years ago

What separates the work of technology transfer research from implementation of the products of such research?

Engineering

1 answer:

tatuchka [14]3 years ago

4 0

Answer:

The thing that separates the work of technology transfer research from implementation of the products of such research is:

Technology transfer research looks at how technology may be transferred but does not actually make the transfer.

Explanation:

This suggests that technology transfer research is different from the technology transfer (implementation) itself. The first stops at making scientific investigations into technology transfer activities while the next step performs the actual transfer or implementation. In other words, technology transfer conveys the results of scientific and technological research to the marketplace and to the wider society. It is the bridge-builder between the research and the implementation.

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Explain the advantages of continuous insulation in envelope assemblies and why it is better to have the continuous insulation on

Luden [163]

Answer:

Continuous insulation helps in eliminating the necessity of applying extra materials to achieve moisture barrier demands, reducing labor and material cost.

Continuous insulation helps building last much longer over a period of time,without the need to upgrade and repair.

Inside the placement of continuous insulation in the envelope needed includes the following; the foundation wall or slab insulation, balcony interface, bond joist insulation, insulation against sub floor, on the interior of masonry wall.

Explanation:

Solution:

The analytical path to success contains a well planned and designed building exterior, which avoids the loss of energy, control cost and the maximization of technology advancement in materials.

Property installed continuous insulation on the exterior can also execute as an air barrier. the flashing of wall penetrations can form into a drainage plane. this plane can stop potentially damaging moisture from entering into the wall assembly.

By making use of continuous insulation it helps in removing necessity of applying extra materials to achieve moisture barrier demands, reducing labor and material cost.

Continuous insulation helps building to withstand the test of time without the need to upgrade and repair

Now inside the placement of continuous insulation in the envelope is needed as follows:

Foundation wall or slab insulation
Insulation against sub floor
Balcony interfaces
On the interior of masonry wall
Bond joist insulation

6 0

4 years ago

What are the two types of furnaces used in steel production?

Kobotan [32]

Explanation:

The two types of furnaces used in steel production are:

<u>Basic oxygen furnace </u>

In basic oxygen furnace, iron is combined with the varying amounts of the steel scrap and also small amounts of the flux in the Blast Furnace. Lance is introduced in vessel and blows about 99% of the pure oxygen causing rise in temperature to about 1700°C. This temperature melts scrap and the impurities are oxidized and results in the liquid steel.

<u>Electric arc furnace</u>

Electric arc furnace reuses existing steel. Furnace is charged with the steel scrap. It operates on basis of electrical charge between the two electrodes providing heat for process. Power is supplied through electrodes placed in furnace, which produce arc of the electricity through scrap steel which raises temperature to about 1600˚C. This temperature melts scrap and the impurities can be removed through use of the fluxes and results in the liquid steel.

3 0

4 years ago

An uncharged capacitor and a resistor are connected in series to a source of voltage. If the voltage = 7.41 Volts, C = 11.5 µFar

Musya8 [376]

Answer:

a) RC = 1.03 mseg.

b) Qmax = CV = 85.2 μC

c) Q = 53.9 μC

Explanation:

a) In a RC circuit, during the transient period, the capacitor charges exponentially (starting from 0 due to the voltage in the capacitor can´t change instantaneously) with time, being the exponent -t/RC.

This product RC, which defines the rate at which the capacitor charges, is called the time constant of the circuit.

In this case , it can be calculated as follows:

ζ = R C = 89.4 Ω . 11.5 μF = 1.03 mseg.

b) As the charge begins to build up the capacitor plates, a voltage establishes between plates, that opposes to the battery voltage. When this voltage is equal to the battery one, the capacitor reaches to the maximum charge, which is, by definition, as follows:

Q = C V = 11.5 μF . 7.41 V = 85.2 μC

c) During the charging process, the charge increases following this equation:

Q = CV (1 - e⁻t/RC)

When t = RC, the expression for Q is as follows:

Q = CV ( 1- e⁻¹) = 0.63 x CV = 53.9 μC

6 0

4 years ago

An AM radio transmitter radiates 550 kW at a frequency of 740 kHz. How many photons per second does the emitter emit?

kifflom [539]

Answer:

1121.7 × 10³⁰ photons per second

Explanation:

Data provided in the question:

Power transmitted by the AM radio,P = 550 kW = 550 × 10³ W

Frequency of AM radio, f = 740 kHz = 740 × 10³ Hz

Now,

P = $\frac{NE}{t}$

here,

N is the number of photons

t is the time

E = energy = hf

h = plank's constant = 6.626 × 10⁻³⁴ m² kg / s

Thus,

P = $\frac{NE}{t}$ = $\frac{N\times(6.626\times10^{-34}\times740\times10^{3})}{1}$ [t = 1 s for per second]

550 × 10³ = $\frac{N\times(6.626\times10^{-34}\times740\times10^{3})}{1}$

550 = N × 4903.24 × 10⁻³⁴

N = 0.11217 × 10³⁴ = 1121.7 × 10³⁰ photons per second

7 0

3 years ago

Steam enters a well-insulated turbine operating at steady state at 4 MPa with a specific enthalpy of 3015.4 kJ/kg and a velocity

Sedbober [7]

Answer:

attached below

Explanation:

6 0

4 years ago