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2 years ago

3.8 LAB - Select lesson schedule with multiple joins

Engineering

1 answer:

dem82 [27]2 years ago

4 0

Answer:

The database has three tables for tracking horse-riding lessons: Horse with columns: ID - primary key; RegisteredName; Breed; Height; BirthDate.

Explanation:

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Select the correct answer. Which existing technology did engineers use to enhance the speed of propeller-driven airplanes

Musya8 [376]

metallurgy:

Explanation:

7 0

3 years ago

Lab scale tests performed on a cell broth with a viscosity of 5cP gave a specific cake resistance of 1 x1011 cm/g and a negligib

insens350 [35]

Answer:

5.118 m^3/hr

Explanation:

Given data:

viscosity of cell broth = 5cP

cake resistance = 1*1011 cm/g

dry basis per volume of filtrate = 20 g/liter

Diameter = 8m , Length = 12m

vacuum pressure = 80 kpa

cake formation time = 20 s

cycle time = 60 s

<u>Determine the filtration rate in volumes/hr expected fir the rotary vacuum filter</u>

attached below is a detailed solution of the question

Hence The filtration rate in volumes/hr expected for the rotary vacuum filter

V' = ( $\frac{60}{20}$ ) * 1706.0670

= 5118.201 liters ≈ 5.118 m^3/hr

4 0

3 years ago

How much calcium chloride will react with100mg of soda ash?

mixer [17]

Answer:

105mg of calcium chloride $CaCl_{2}$

Explanation:

The molecular formula of calcium chloride is $CaCl_{2}$ and the molecular formula of soda ash is $Na_{2}CO_{3}$

First of all you should write the balanced reaction between both compounds, so:

$CaCl_{2}+Na_{2}CO_{3}=2NaCl+CaCO_{3}$

Then you should have the molar mass of the two compounds:

Molar mass of $CaCl_{2}=110.98\frac{g}{mol}$

Molar mass of $Na_{2}CO_{3}=105.98\frac{g}{mol}$

Now with stoichiometry you can find the mass of calcium chloride that reacts with 100mg of soda ash, so:

$100mgNa_{2}CO_{3}*\frac{1molNa_{2}CO_{3}}{105.98gNa_{2}CO_{3}}*\frac{1molCaCl_{2}}{1molNa_{2}CO_{3}}*\frac{110.98gCaCl_{2}}{1molCaCl_{2}}=105mgCaCl_{2}$

6 0

4 years ago

A rectangular car-top carrier of 1.7-ft height, 5.0-ft length (front to back), and 4.2-ft width is attached to the top of a car.

Nataliya [291]

Answer:

$\Delta P =1.2 \frac{1.3}{2}(26.822m/s)^2 (4.2*1.7*(0.3048)^2)=13.88 hp$

Explanation:

We can assume that the general formula for the drag force is given by:

$D= C_D \frac{\rho}{2}V^2 A$

And we can see that is proportional to the area. On this case we can calculate the area with the product of the width and the height. And we can express the grad force like this:

$D_1 = C_{D1} \frac{\rho}{2}V^2 (wh)$

Where w is the width and h the height.

The last formula is without consider the area of the carrier, but if we use the area for the carrier we got:

$D_2 = C_{D2} \frac{\rho}{2}V^2 (wh+ A_{carrier})$

If we want to find the additional power added with the carrier we just need to take the difference between the multiplication of drag force by the velocity (assuming equal velocities for both cases) of the two cases, and we got:

$\Delta P = C_{D2} \frac{\rho}{2}V^2 (wh+ A_{carrier}) V- C_{D1} \frac{\rho}{2}V^2 (wh) V$