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astraxan [27]
3 years ago
13

In the idealized Otto cycle, heat is added during: a. Isentropic Compression b. Constant (minimum) volume c. Constant (maximum)

volume d. Isentropic Expansion
Engineering
1 answer:
docker41 [41]3 years ago
3 0

Answer:

(b) Constant (minimum) volume

Explanation:

In the idealized Otto cycle there are 4 process that are

  • Reversible adiabatic compression  
  • Addition of heat at constant volume
  • Reversible adiabatic expansion
  • Rejection of constant volume

So from above discussion we can see that heat is added when there is constant (minimum) volume which is given in option (b) so option (b) will be the correct answer

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Assume the impedance of a circuit element is Z = (3 + j4) Ω. Determine the circuit element’s conductance and susceptance.
djyliett [7]

Answer:

B. G = 333 mS, B = j250 mS

Explanation:

impedance of a circuit element is Z = (3 + j4) Ω

The general equation for impedance

Z = (R + jX) Ω

where

R = resistance in ohm

X = reactance

R = 3Ω  X = 4Ω

Conductance = 1/R while Susceptance = 1/X

Conductance = 1/3 = 0.333S

= 333 mS

Susceptance = 1/4 = 0.25S

= 250mS

The right option is B. G = 333 mS, B = j250 mS

8 0
3 years ago
What is the shape of the output signal on a rigexpert analyzer?
Gekata [30.6K]

Answer:

Output signal shape: square, from 0.1 to 230 MHz. Output power: -10 dBm (at a load of 50 Ohms).

Explanation:

8 0
2 years ago
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Consider a multiprocessor system and a multithreaded program written using the many-to-many threading model. Let the number of u
Montano1993 [528]

Answer:

At the point when the quantity of bit strings is not exactly the quantity of processors, at that point a portion of the processors would stay inert since the scheduler maps just part strings to processors and not client level strings to processors. At the point when the quantity of part strings is actually equivalent to the quantity of processors, at that point it is conceivable that the entirety of the processors may be used all the while. Be that as it may, when a part string obstructs inside the portion (because of a page flaw or while summoning framework calls), the comparing processor would stay inert. When there are more portion strings than processors, a blocked piece string could be swapped out for another bit string that is prepared to execute, in this way expanding the use of the multiprocessor system.When the quantity of part strings is not exactly the quantity of processors, at that point a portion of the processors would stay inert since the scheduler maps just bit strings to processors and not client level strings to processors. At the point when the quantity of bit strings is actually equivalent to the quantity of processors, at that point it is conceivable that the entirety of the processors may be used at the same time. Be that as it may, when a part string hinders inside the piece (because of a page flaw or while summoning framework calls), the relating processor would stay inert. When there are more portion strings than processors, a blocked piece string could be swapped out for another bit string that is prepared to execute, along these lines expanding the usage of the multiprocessor framework.

4 0
3 years ago
Can you prove that the two bleu areas are the same without numbers please?
Svet_ta [14]

Answer:

\small{\boxed{\tt{\colorbox{green}{✓Verified\:answer}}}}\:

Just draw a line from point D join to point E

The triangle formed DME will be congruent to AMC

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3 years ago
30 points and brainiest if correct please help A, B, C, D
tatuchka [14]

Answer:

B. to lock the tape into place

Explanation:

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