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3 years ago

What is Join..?? Explain different types of joins.

Engineering

1 answer:

Anarel [89]3 years ago

5 0

Answer:

mainly used to combine two tables based on a specified common field between them. If we talk in terms of Relational algebra, it is the cartesian product of two tables followed by the selection operation. Thus, we can execute the product and selection process on two tables using a single join statement. We can use either 'on' or 'using' clause in MySQL to apply predicates to the join queries.

A Join can be broadly divided into two types:

Inner Join

Outer Join

Inner Join is a join that can be used to return all the values that have matching values in both the tables. Inner Join can be depicted using the below diagram.

The inner join can be further divided into the following types:

Equi Join

Natural Join

Now let us learn about these inner joins one-by-one.

You might be interested in

Which of the following have the capacity to display formatted data?

sukhopar [10]

Answer:

D. A and B

Explanation:

1. The method Console.Write() is an overloaded method in a language like C#.

One of its variations could be as follows;

Console.Write(String format, Object a, Object b).

This contains three parameters and will write the text representation of the specified objects to the standard output stream using the information specified by the format specifier. Parameter 1 is format which is a composite format string representing the format specifier. Parameter 2 is a, which is the first object to be written using format. Parameter 3 is b, which is the second object to be written using format.

2. The method Console.WriteLine() has the same characteristics as Console.Write() above, except that it writes the text representation of the specified objects, followed by current line terminator then to the standard output stream using the information specified by the format specifier.

3. Console.WriteFormat() does not exist, at least not in C# or .NET

Therefore, Console.Write() and Console.WriteLine() have the capacity to display formatted data.

Hope this helps!

6 0

3 years ago

Q.14

Kay [80]

Answer:

A. optical isolation

Explanation:

well I can't really give a good explanation because I also saw the same question in my exams and option A was the correct answer

6 0

3 years ago

Teachers

IRINA_888 [86]

Uh I’m just gonna say yes because I think this is just something random

5 0

3 years ago

a turbine operating at steady state at 500 kPa, 860 K and exists at 100 kPa. A temperature sensor indicates that the exit air te

noname [10]

Answer:

Given:

P₁ = 500 kPa

T₁ = 860 K

P₂= 100 kPa

T₂ = 460 K

Let's take entropy properties of T1 and T2 from ideal properties of air,

at T = 860K, s(T₁) = 2.79783 kJ/kg.K

at T = 460K, s(T₂) = 2.13407 kJ/kg.K

using entropy balance equation:

$\frac{\sigma _cv}{m} = s(T_2)- s(T_1) - R In [\frac{P_2}{P_1}]$

$\frac{\sigma _cv}{m} = 2.79783 - 2.13407 - 0.287 In [\frac{100}{500}]$

= - 0.2018 kJ/kg. K

In this case the entropy is negative, which means the value of exit temperature is not correct, beacause entropy should always be positive(>0).

8 0

3 years ago

An inverted tee lintel is made of two 8" x 1/2" steel plates. Calculate the maximum bending stress in tension and compression wh

Kamila [148]

Answer:

hello your question lacks some information attached is the complete question

A) (i)maximum bending stress in tension = 0.287 * 10^6 Ib-in

(ii) maximum bending stress in compression = 0.7413*10^6 Ib-in

B) (i) The average shear stress at the neutral axis = 0.7904 *10 ^5 psi

(ii) Average shear stress at the web = 18.289 * 10^5 psi

(iii) Average shear stress at the Flange = 1.143 *10^5 psi

Explanation:

First we calculate the centroid of the section,then we calculate the moment of inertia and maximum moment of the beam( find attached the calculation)

A) Calculate the maximum bending stress in tension and compression

lintel load = 10000 Ib

simple span = 6 ft

( (moment of inertia*Y)/ I ) = MAXIMUM BENDING STRESS

I = 53.54

i) The maximum bending stress (fb) in tension=

= $\frac{M_{mm}Y }{I}$ = $\frac{6.48 * 10^6 * 2.375}{53.54}$ = 0.287 * 10^6 Ib-in