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2 years ago

What is the engine circulator for in a cooler for a reactor plsss help me

Engineering

2 answers:

strojnjashka [21]2 years ago

8 0

Answer:

I am COOL nobody can mess with me cuz they DROOL They are loungers for foulers they need to get beet to get the beat to mam

Explanation:

shutvik [7]2 years ago

7 0

Answer:

KEHSJAHSJAJAJ this is me I am not stealing ajjshe jwjsbsh I am cool. not you

I am gangster here. not you

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A gas flows through a one-inlet, one-exit control volume operating at steady state. Considering an adiabatic control volume with

Hunter-Best [27]

Answer:

b. equal to the specific entropy of the gas at the inlet.

Explanation:

Isentropic process is the process in which the entropy of the system remains unchanged. The word isentropic is formed from the combination of the prefix "iso" which means "equal" and the word entropy.

If a process is completely reversible, without the need to provide energy in the form of heat, then the process is isentropic.

3 0

3 years ago

Two substances, A and B, initially at different temperatures, come into contact and reach thermal equilibrium. The mass of subst

Kaylis [27]

Answer:

The specific heat capacity of substance A is 1.16 J/g

Explanation:

The substances A and B come to a thermal equilibrium, therefore, the heat given by the hotter substance B is absorbed by the colder substance A.

The equation becomes:

Heat release by Substance B = Heat Gained by Substance A

The heat can be calculated by the formula:

Heat = mCΔT

where,

m = mass of substance

C = specific heat capacity of substance

ΔT = difference in temperature of substance

Therefore, the equation becomes:

(mCΔT) of A = (mCΔT) of B

FOR SUBSTANCE A:

m = 6.01 g

ΔT = Final Temperature - Initial Temperature

ΔT = 46.1°C - 20°C = 26.1°C

C = ?

FOR SUBSTANCE B:

m = 25.6 g

ΔT = Initial Temperature - Final Temperature

ΔT = 52.2°C - 46.1°C = 6.1°C

C = 1.17 J/g

Therefore, eqn becomes:

(6.01 g)(C)(26.1°C) = (25.6 g)(1.17 J/g)(6.1°C)

C = (182.7072 J °C)/(156.861 g °C)

C = 1.16 J/g

5 0

3 years ago

What are two factors that determine the thermal energy of a substance

Sav [38]

Mass and chemical composition

6 0

3 years ago

Which of these cars traveled faster during time interval please show solution

Mazyrski [523]

Answer:

I think D is correct

Explanation:

C is decreasing function, probably worst

A is arctan -> in radian, the rate of increasing is very slow-> second worst

B(14) = ln(9*14) = 4.8

D(14) = sqrt(8+14^2)=14.2

3 0

3 years ago

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