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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.

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How do you calculate the rate of heat transfer between two bodies with a small area of physical contact.

gavmur [86]

Answer:

So the rate of heat transfer to an object is equal to the thermal conductivity of the material the object is made from, multiplied by the surface area in contact, multiplied by the difference in temperature between the two objects, divided by the thickness of the material.

3 0

3 years ago

The number of vacancies in some hypothetical metal increases by a factor of 3 when the temperature is increased from 1020 ˚C to

weeeeeb [17]

Answer:

first step here is to substitute the 3 of your two equations into the second;

3 Ne^(-Q_v/k(1293)) = Ne^(-Q_v/k(1566))

Since 'N' is a constant, we can remove it from both sides.

We also want to combine our two Q_v values, so we can solve for Q_v, so we should put them both on the same side:

3 = e^(-Q_v/k(1293)) / e^(-Q_v/k(1566))

3 = e^(-Q_v/k(1293) + Q_v/k(1566) ) (index laws)

ln (3) = -Q_v/k(1293) + Q_v/k(1566) (log laws)

ln (3) = -0.13Q_v / k(1566) (addition of fractions)

Q_v = ln (3)* k * 1566 / -0.13 (rearranging the equation)

Now, as long as you know Boltzmann's constant it's just a matter of substituting it for k and plugging everything into a calculator.

6 0

4 years ago

Who works alongside and assists the engineers?

nika2105 [10]

Answer:

Assistants works alongside and assists the engineers.

5 0

3 years ago

Consider a single bacterial cell as a discrete particle with a diameter of 1x10-6m and a specific gravity of 1.01. Assuming lami

Alinara [238K]

Answer:

Sedimentation is not a good method.

Explanation:

We need to apply Stoke laws and assume that is valid here.

So,

$V_s= 418(G-1)d^2(\frac{3T+70}{100})$

Replacing the values,

$V_s= (418)(1.01-1)(10^-3)^3*(\frac{3*20+70}{100})\\V_s=5.434*10^-6mm/s$

Here then we calculate the time,

$t_{req}=\frac{x}{v}$

Where x= Distance, v= velocity

$t_{req}=\frac{1foot}{5.434*10^{-6}} = \frac{304.8}{5.434*10^{-6}}\\t_{req}=649.2days$

To calculate the surface required we need first to calculate the volume through the volume,

So,

$Q=4MGD=4*10^6*0.135 (ft^3/day)$

Then,

$V_{req} = Q*t\\V_{req}=4*10^6*0.134*649.2\\V_{req}=34.79*10^7ft^3$

Here we can calculate the surface

$S_{req}= \frac{Volume}{Distance}=\frac{34.79*10^7}{1}\\S_{req}=7987.8 Acres$

So, the requeriment of Area of tank and settlement time is huge, it's not a practical method.

6 0

3 years ago

A motor vehicle has a mass of 1.8 tonnes and its wheelbase is 3 m. The centre of gravity of the vehicle is situated in the centr

seropon [69]

Answer:

1) The normal reactions at the front wheel is 9909.375 N

The normal reactions at the rear wheel is 8090.625 N

2) The least coefficient of friction required between the tyres and the road is 0.625

Explanation:

1) The parameters given are as follows;

Speed, u = 90 km/h = 25 m/s

Distance, s it takes to come to rest = 50 m

Mass, m = 1.8 tonnes = 1,800 kg

From the equation of motion, we have;

v² - u² = 2·a·s

Where:

v = Final velocity = 0 m/s

a = acceleration

∴ 0² - 25² = 2 × a × 50

a = -6.25 m/s²

Force, F = mass, m × a = 1,800 × (-6.25) = -11,250 N

The coefficient of friction, μ, is given as follows;

$\mu =\dfrac{u^2}{2 \times g \times s} = \dfrac{25^2}{2 \times 10 \times 50} = 0.625$

Weight transfer is given as follows;

$W_{t}=\dfrac{0.625 \times 0.9}{3}\times \dfrac{6.25}{10}\times 18000 = 2109.375 \, N$

Therefore, we have for the car at rest;

Taking moment about the Center of Gravity CG;

$F_R$ × 1.3 = 1.7 × $F_F$

$F_R$ + $F_F$ = 18000

$F_R + \dfrac{1.3 }{1.7} \times F_R = 18000$

$F_R$ = 18000*17/30 = 10200 N

$F_F$ = 18000 N - 10200 N = 7800 N

Hence with the weight transfer, we have;

The normal reactions at the rear wheel $F_R$ = 10200 N - 2109.375 N = 8090.625 N

The normal reactions at the front wheel $F_F$ = 7800 N + 2109.375 N = 9909.375 N

2) The least coefficient of friction, μ, is given as follows;

$\mu = \dfrac{F}{R} = \dfrac{11250}{18000} = 0.625$

The least coefficient of friction, μ = 0.625.

3 0

4 years ago