Ben İngiliz oldum düzelte bilirmiyim

Which type of Artificial Intelligence (AI) can repeatedly perform tasks of limited scope?

seraphim [82]

Narrow AI is the type of Artificial Intelligence (AI) can repeatedly perform tasks of limited scope.

Natural Language Understanding (NLU) is used to describe extracting information from unstructured text using algorithms.

<h3>What is Narrow AI ?</h3>

This is known to be a form or a type of AI that is said to be able to carry out a dedicated task as a result of its intelligence.

Note that Narrow AI is the type of Artificial Intelligence (AI) can repeatedly perform tasks of limited scope.

Natural Language Understanding (NLU) is used to describe extracting information from unstructured text using algorithms.

Learn more about Artificial Intelligence from

brainly.com/question/25523571

#SPJ1

7 0

1 year ago

An ideal Rankine cycle with reheat uses water as the working fluid. The conditions at the inlet to the first-stage turbine are p

gogolik [260]

Answer:

myjmjkkyjm

Explanation:

Tyjjhmjtymtyyjmjymjy

4 0

2 years ago

'Energy' has the potential to:

Kazeer [188]

answer:

Energy' has the potential to:<u>do work</u>

b.do work

6 0

1 year ago

A weighted, frictionless piston-cylinder device initially contains 5.25 kg of R134a as saturated vapor at 500 kPa. The container

kykrilka [37]

Answer:

-6.326 KJ/K

Explanation:

A) the entropy change is defined as:

$delta S_{12}=\int\limits^2_1 \, \frac{dQ}{T}$

In an isobaric process heat (Q) is defined as:

$Q= m*Cp*dT$

Replacing in the equation for entropy

$delta S_{12}=\int\limits^2_1 \frac{m*Cp*dT}{T}$

m is the mass and Cp is the specific heat of R134a. We can considerer these values as constants so the expression for entropy would be:

$delta S_{12}= m*Cp*\int\limits^2_1 \frac{ dT }{T}$

Solving the integral we get the expression to estimate the entropy change in the system

$delta S_{12}= m*Cp *ln(\frac{T_{2}}{T_{1}})$

The mass is 5.25 Kg and Cp for R134a vapor can be consulted in tables, this value is $0.85\frac{kJ}{Kg*K}$

We can get the temperature at the beginning knowing that is saturated vapor at 500 KPa. Consulting the thermodynamic tables, we get that temperature of saturation at this pressure is: 288.86 K

The temperature in the final state we can get it from the heat expression, since we know how much heat was lost in the process (-976.71 kJ). By convention when heat is released by the system a negative sign is used to express it.

$Q= m*Cp*dT$

With $dt=T_{2}-T_{1}$ clearing for T2 we get:

$T_{2}=\frac{Q}{m*Cp}+T1= \frac{-976.71kJ}{5.25Kg*0.85\frac{kJ}{Kg*K}}+288.86 K =69.98 K$

Now we can estimate the entropy change in the system

$delta S_{12}= m*Cp*ln(\frac{T_{2}}{T_{1}})= 5.25Kg*0.85\frac{kJ}{Kg*K}*ln(\frac{69.98}{288.861})= -6.326\frac{kJ}{K}$

The entropy change in the system is negative because we are going from a state with a lot of disorder (high temperature) to one more organize (less temperature. This was done increasing the entropy of the surroundings.

b) see picture.

3 0

3 years ago

¿Cómo nos podría ayudar una hoja de cálculo en nuestro estudio?

Ghella [55]

Las hojas de cálculo en Excel facilitan los cálculos numéricos a través del uso de fórmulas; de manera fácil y rápida se pueden hacer operaciones aritméticas sobre cientos de miles de datos numéricos; por lo que se puede actualizar o corregir cualquiera de los datos numéricos y las operaciones se recalculan

3 0

2 years ago