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ch4aika [34]
3 years ago
8

Tech A says that full floating pistons use keepers to retain the piston pin. Tech B says that some piston pins are press fitted

in the rod. Who is correct?
Engineering
1 answer:
Dvinal [7]3 years ago
8 0
Gfffbbnjhhdb ghhfjgfjgbj hjvfhbfgvvfhh Gn hhvh hhchv high jump
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Multilane roads use what to divide lanes of traffic moving in the same direction.
Rina8888 [55]

Answer:

broken white lines

4 0
3 years ago
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37. In ______ combination of drugs, the effects of one drug cancel or diminish
Korvikt [17]

Answer:

In antagonistic combination of drugs, the effects of ine drug cancel or diminish the effects of another

6 0
3 years ago
Which option identifies the most likely scaling factor in the above drawing of the Mercury capsule and booster unit? 1:1,000 1:1
Elina [12.6K]

Answer:

I would say 10000;1 but thats just me

Explanation:

3 0
3 years ago
For each of the cases below, determine if the heat engine satisfies the first law (energy equation) and if it violates the secon
stepan [7]

Answer:

From first law of thermodynamics(energy conservation)

Qa= Qr+W

Qa=Heat added to the engine

Qr=heat rejected from the engine

W=work output from the engine

Second law:

It is impossible to construct a heat engine that will deliver the work with out rejecting heat.

In other word ,if engine take heat then it will reject some amount heat and will deliver some amount of work.

1.

QH=6 kW,

QL=4 kW,

W=2 kW

6 KW= 4 + 2  KW

It satisfy the first law.

Here heat is also rejected from the engine that is why it satisfy second law.

2.

QH=6 kW, QL=0 kW, W=6 kW

This satisfy first law but does not satisfy second law because heat rejection is zero.

3.

QH=6 kW   ,   QL=2 kW,      W=5 kW

This does not satisfy first as well as second law.Because summation of heat rejection and work can not be greater than heat addition or we can say that energy is not conserve.

4.

QH=6 kW,   QL=6 kW,   W=0 kW

This satisfy first law only and does not satisfy second law.

6 0
3 years ago
Show the bias polarities and depletion regions of an npn BJT in the normal active, saturation, and cutoff modes of operation. Dr
Troyanec [42]

Complete Question:

Show the bias polarities and depletion regions of an npn BJT in the normal active, saturation, and cutoff modes of operation. Draw the three sketches one below the other to (qualitatively) reflect the depletion widths for these biases, and the relative emitter, base, and collector doping.

Consider a BJT with a base transport factor of 1.0 and an emitter injection efficiency of 0.5.

Calculate roughly by what factor would doubling the base width of a BJT would increase, decrease, or leave unchanged the emitter injection efficiency and base transport factor? Repeat for the case of emitter doping increased 5 × =. Explain with key equations, and assume other BJT parameters remain unchanged!

Answer & Explanation:

[Find the attachments]

Step 1 :

Emitter and base, collector, and base are forward biased then BJT is in saturation region. Emitter and base is forward biased and base and collector in reverse biased then BJT is in active region.

Emitter and base, collector and base are reverse biased then BJT in cut off region.

Three sketches one below the other is shown in Figure 1.

[find the figure in attachment]

Step 2:

Value of base widths of saturation, active and cut off operated BJT are value of Base width of saturated region operated BJT is less than base width in active region operated BJT. Value of base width of active region operated BJT is less than base width in cut off region operated BJT.

Saturation region operated base width of BJT is < Active region operated base width of BJT is < Cut off region operated base width of BJT.

[For  Steps 3 4 5 6 and 7 find attachments]

8 0
4 years ago
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