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iragen [17]
3 years ago
13

A MOSFET differs from a JFET mainly because

Engineering
1 answer:
Solnce55 [7]3 years ago
5 0

Answer:

The answer is option

C . the JFET has a PN junction

Explanation:

Not only is option C in the question a dissimilarity between the MOSFET and the JFET we can go on with some more dissimilarities.

1.MOSFET stands for Metal Oxide Silicon Field Effect Transistor or Metal          Oxide Semiconductor Field Effect Transistor.

  (JFET) stands for junction gate field-effect transistor (JFET)  

2. JFET is a three-terminal semiconductor device, whereas  MOFET a four-terminal semiconductor device.  

3. In terms of areas of application of  JFETs are used in low noise applications while MOSFETs,  are used for high noise applications

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For the following circuit diagram, if A=010 , B= 101.
Fantom [35]

Answer:

cgghhhh chick jjkkkkkki

4 0
3 years ago
Stream Piracy – Kaaterskill, NY. Check and double-click the Problem 15 folder. The dark blue and orange streams highlight the pr
baherus [9]

Answer:

b. The pirating streams are eroding headwardly to intersect more of the other streams’ drainage basins, causing water to be diverted down their steeper gradients.

Explanation:

From the Kaaterskill NY 15 minute map (1906), this shows two classic examples of stream capture.

The Kaaterskill Creek flow down the east relatively steep slopes into the Hudson River Valley. While, the Gooseberry Creek is a low gradient stream flowing down the west direction which in turn drains the higher parts of the Catskills in this area.

However, there is Headward erosion of Kaaterskill Creek which resulted to the capture of part of the headwaters of Gooseberry Creek.

The evidence for this is the presence of "barbed" (enters at obtuse rather than acute angle) tributary which enters Kaaterskill Creek from South Lake which was once a part of the Gooseberry Creek drainage system.

It should be noted again, that there is drainage divide between the Gooseberry and Kaaterskill drainage systems (just to the left of the word Twilight) which is located in the center of the valley.

As it progresses, this divide will then move westward as Kaaterskill captures more and more of the Gooseberry system.

5 0
3 years ago
Given asphalt content test data:
VMariaS [17]

Answer:

hello your question is incomplete attached below is the complete question

A) overall mean = 5.535,  standard deviation ≈ 0.3239

B ) upper limit = 5.85, lower limit = 5.0

C) Not all the samples meet the contract specifications

D) fluctuation ( unstable Asphalt content )

Explanation:

B) The daily average asphalt content has to obtained in order to determine the upper and lower control limits using an average asphalt content of 5.5% +/- 0.5% everyday

The upper limit : 14 may = ( 5.8 + 5.1 ) / 2 = 5.85

The lower limit : 16 may = ( 5.2 + 4.8 ) / 2 = 5.0

attached below is the required plot

C ) Not all the samples meet the contract specifications and the samples that do not meet up are samples from :

15 may and 16 may . this is because their Asphalt contents are 6.2 and 4.8 respectively and sample number 18 and 20

D ) what can be observed is that the ASPHALT content fluctuates between the dates while the contract specification is fixed

5 0
3 years ago
As part of a chemistry experiment, Austin ends up with a beaker containing a mixture of two liquids, which he knows have boiling
asambeis [7]

Answer:

Distillation, heat

Explanation:

Here in this question, we simply want to look at the best options that could fit in the gaps.

We have a mixture of liquids having boiling points which is far from each other.

Whenever we have a mixture of liquids with boiling points far away from each other, the best technique to use in separating them is to use distillation. That is why we have that as the best fit for the first missing gap.

Now, to get the liquids to separate from each other, we shall be needing the heating mantle for the application of heat. This ensures that the mixture is vaporized. After vaporization, the condensing tube will help to condense the vapor of each of the liquids once we reach the boiling point of either of the two.

Kindly note that the liquid with the lower temperature will evaporate first and will be first obtained. In fact after reaching a little above the boiling point of the lower boiling liquid, we can be sure that what we have left in the mixture pot is the second other liquid with the higher boiling point.

7 0
3 years ago
Problem 4.079 SI A rigid tank whose volume is 3 m3, initially containing air at 1 bar, 295 K, is connected by a valve to a large
salantis [7]

Answer:

Q_{cv} = -1007.86kJ

Explanation:

Our values are,

State 1

V=3m^3\\P_1=1bar\\T_1 = 295K

We know moreover for the tables A-15 that

u_1 = 210.49kJ/kg\\h_i = 295.17kJkg

State 2

P_2 =6bar\\T_2 = 296K\\T_f = 320K

For tables we know at T=320K

u_2 = 228.42kJ/kg

We need to use the ideal gas equation to estimate the mass, so

m_1 = \frac{p_1V}{RT_1}

m_1 = \frac{1bar*100kPa/1bar(3m^3)}{0.287kJ/kg.K(295k)}

m_1 = 3.54kg

Using now for the final mass:

m_2 = \frac{p_2V}{RT_2}

m_2 = \frac{1bar*100kPa/6bar(3m^3)}{0.287kJ/kg.K(320k)}

m_2 = 19.59kg

We only need to apply a energy balance equation:

Q_{cv}+m_ih_i = m_2u_2-m_1u_1

Q_{cv}=m_2u_2-m1_u_1-(m_2-m_1)h_i

Q_{cv} = (19.59)(228.42)-(3.54)(210.49)-(19.59-3.54)(295.17)

Q_{cv} = -1007.86kJ

The negative value indidicates heat ransfer from the system

7 0
3 years ago
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