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Sergeeva-Olga [200]
4 years ago
5

Intrinsic semiconduction is a property of a pure material. (True , False )

Engineering
1 answer:
denis-greek [22]4 years ago
5 0

True.

An intrinsic semiconductor is a pure semiconductor. At room temperature it behaves as an insulator because it only has a few free and hollow electrons due to thermal energy.

In an intrinsic semiconductor there are also electron fluxes and gaps, although the total current resulting is zero. This is because the action of thermal energy produces free electrons and gaps in pairs, so there are as many free electrons as there are gaps with which the total current is zero.

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Reika [66]
50 is the answer!! aka B
7 0
3 years ago
In your opinion, should a marketing plan be created for each and every product under a single brand should the marketing be for
Kay [80]

Answer:

Both of them

Explanation:

When creating a marketing plan you need to think of the objective or goals that you have, you should probably always have a running campaign to help the brand itself to be constantly seen by the audience, while at the same time having specific campaigns for the products that you want to push or to promote the most. For example, not all of Nike's ads are aimed to sell more sneakers or sports clothing, some of them are just to keep them on the conversation and growing their brand, while the hype up campaign for the release of a new pair of sneakers or collection is done at the same time. So you should always go for both.

8 0
3 years ago
Forced air at T= 25ºC and V= 12 m/s is used to cool electronic elements on a circuit board. One such element is a chip, 4 mm ×4
tia_tia [17]

Answer:

The surface temperature of the chip 39.99°C

Explanation:

Assumptions

1. Steady state condition.

2. Power dissipated within the chip is lost by convection across upper surface only.

3. Chip surface is isothermal

4. The average heat transfer coefficient for the chip surface is equivalent to the local value x - L

Properties: from table. See attachment (4)

    See attachment for complete solution to the problem

i had problem submitting the answer in the answer box

8 0
3 years ago
For the same cross-sectional area, which column provides the higher buckling load: a circular bar or a circular tube?
juin [17]

Answer:

Circular tube

Explanation:

Now for better understanding lets take an example

Lets take

Diameter of solid bar= 4\sqrt{2} cm

Outer diameter of tube =6 cm

Inner diameter of tube=2 cm

So from we can say that both tubes have equal cross sectional area.

We know that buckling load is given as P = \dfrac{\pi ^2EI}{L_e^2}      

If area moment of inertia(I) is high then buckling load will be high.

We know that  area moment of inertia(I)

For circular tube I = \dfrac{\pi }{64}(D_o^4-D_i^4)

For circular bar I = \dfrac{\pi }{64}D^4  

Now by putting the values

    For circular tube I=62.83 cm^4

  For circular bar I=50.26 cm^4

So we can say that for same cross sectional area the  area moment of inertia(I) is high for tube as compare to bar.So buckling load  will be higher in tube as compare to bar.

3 0
3 years ago
without using the routh hurwitz criterion, determing if the followign systems are asymptotically stable, marginally stable, or u
lesantik [10]

Answer:

The system is marginally stable.

Explanation:

Transfer function, M(s) = [10(s+2)]/(s³ + 3s² + 5s)

In control the stability properties of a system can be obtained from just the characteristic equation of its closed loop transfer function.

- The condition for stability is that all the roots of the characteristic equation be negative and real.

- The condition for asymptotic stability is that all the real parts of the roots must all be negative, since there'll be complex roots.

- The condition for marginal stability is that the real part of all the complex roots are negative, the roots without real parts must have distinct imaginary parts.

- The condition for instability is for at least one of the roots to be positive. Or if there are complex roots, the real part of the roots being positive indicates instability.

The characteristic equation for this transfer function is (s³ + 3s² + 5s)

Solving this polynomial

s = 0

s = [-3 - √(11i)]/2

s = [-3 + √(11i)]/2

These roots have all their real parts to be negative, and the zero root has a distinct imaginary part, hence the system is marginally stable

4 0
4 years ago
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