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AfilCa [17]
3 years ago
6

*100 POINTS

Engineering
2 answers:
mylen [45]3 years ago
8 0

Answer:

D

Explanation:

lisov135 [29]3 years ago
7 0
I think it’s A i could be wrong honestly this is just a guess from my knowledge
You might be interested in
Create a Relational Schema for the following scenario. Include all primary and foreign keys and list any assumptions you make. 
NARA [144]

Answer:

See explaination

Explanation:

Some of the Important Observation used in Entity Relational Diagram.

1. A Chemist can be involved one project or many projects

2. A project can hire just one or many chemists.

3. A chemist can lead over one or many projects

4. A project can have only one project leader chemist.

5. A project can be done in 1 or many laboratories.

6. A laboratory can host 0 or 1 projects.

7. A laboratory can have 1 or many equipments.

8. An Equipment can be in 1 or 0 laboratories.

9. A chemist can have 0 or many types of equipments.

10. An Equipment can have 0 or 1 chemist at a time

Notations:

PK=Primary Key:

A primary key, also called a primary keyword, is a key in a relational database that is unique for each record. It is a unique identifier, such as a driver license number, telephone number

FK=Foreign Key:

A foreign key is a column or group of columns in a relational database table that provides a link between data in two tables. It acts as a cross-reference between tables because it references the primary key of another table, thereby establishing a link between them

Composite key:

A composite key is a combination of two or more columns in a table that can be used to uniquely identify each row in the table when the columns are combined uniqueness is guaranteed, but when it is taken individually it does not guarantee uniqueness

4 0
3 years ago
A well-insulated tank in a vapor power plant operates at steady state. Saturated liquid water enters at inlet 1 at a rate of 125
kompoz [17]

Answer:

a. The mass flow rate (in lbm/s) is 135lbm/s

b. The temperature (in o F) is 200.8°F

Explanation:

We assume that potential energy and kinetic energy are negligible and the control volume operates at a steady state.

Given

a. The mass flow rate (in lbm/s) is 135lbm/s

b.

m1 = Rate at inlet 1 = 125lbm/s

m2 = Rate at inlet 2 = 10lbm/s

The mass flow rate (in lbm/s) is calculated as m1 + m2

Mass flow rate = 125lbm/s + 10lbm/s

Mass flow rate = 135lbm/s

Hence, the mass flow rate (in lbm/s) is 135lbm/s

b. To calculate the temperature.

First we need to determine the enthalpy h1 at 14.7psia

Using table A-3E (thermodynamics)

h1 = 180.15 Btu/Ibm

h2 at 14.7psia and 60°F = 28.08 Btu/Ibm

Calculating h3 using the following formula

h3 = (h1m1 + h2m2) / M3

h3 = (180.15 * 125 + 28.08 * 10)/135

h3 = 168.8855555555555

h3 = 168.89 Btu/Ibm

To get the final temperature; we make use of table A-2E of thermodynamics.

Because h3 < h1, it means the liquid is at a compressed state.

The corresponding temperature at h3 = 168.89 is 200.8°F

The temperature (in o F) is 200.8°F

6 0
3 years ago
A homeowner consumes 260 kWh of energy in July when the family is on vacation most of the time. Determine the average cost per k
Llana [10]

Answer:

16.2 cents

Explanation:

Given that a homeowner consumes 260 kWh of energy in July when the family is on vacation most of the time.

Where Base monthly charge of $10.00. First 100 kWh per month at 16 cents/kWh. Next 200 kWh per month at 10 cents/kWh. Over 300 kWh per month at 6 cents/kWh.

For the first 100 kWh:

16 cent × 100 = 1600 cents = 16 dollars

Since 1 dollar = 100 cents

For the remaining energy:

260 - 100 = 160 kwh

10 cents × 160 = 1600 cents = 16 dollars

The total cost = 10 + 16 + 16 = 42 dollars

Note that the base monthly of 10 dollars is added.

The cost of 260 kWh of energy consumption in July is 42 dollars

To determine the average cost per kWh for the month of July, divide the total cost by the total energy consumed.

That is, 42 / 260 = 0.1615 dollars

Convert it to cents by multiplying the result by 100.

0.1615 × 100 = 16.15 cents

Approximately 16.2 cents

7 0
3 years ago
A double-pane insulated window consists of two 1 cm thick pieces of glass separated by a 1.8 cm layer of air. The window measure
Elanso [62]

Answer:

(b). T = 22.55 ⁰C

(c). q = 557.8 W

Explanation:

we take follow a step by step process to solving this problem.

from the question, we have that

The two glass pieces is separated by a 1.8 cm distance layer of air.

the thickness of glass piece is 1 cm

width = 4 m

the height = 3 m

(a). the sketch of the thermal circuit is uploaded in the picture below.

(b).  the thermal resistance due to the conduction in the first glass plane is given thus;

R₁ = Lg / Kg A ................(1)

given that Kg rep. the thermal conductivity of the glass plane

A = conduction surface area

Lg = Thickness of glass plane4

taking the thermal conductivity of glass plane as Kg = 0.78 w/mk

inputting values into equation (1) we have,

R₁ = [1 (cm) ˣ 1 (m)/100 (cm)] / [(0.78 w/mk)(4m ˣ 3m)]

R₁ = 1.068 ˣ 10 ⁻³ k/w

Being that we have same thermal resistance in the first and second plane,

therefore R₁ = R₃ = 1.068 ˣ 10 ⁻³ k/w

⇒ Also the thermal resistance between air and glass as a result of the conduction by the layer is given thus

R₂ = La/KaA .....................(2)

given Ka = thermal conductivity of air

A = surface area

La = thickness of air

substituting values into the equation we have

R₂ = [1.8 (cm) ˣ 1 (m)/100 (cm)] / [(0.0262 w/mk)(4m ˣ 3m)]

R₂ = 5.73 ˣ 10⁻² k/w

Given the thermal resistance on the outer surface due to convection, we have

R₄ = 1/hA

inputting value gives R₄ = 1 / (12 w/m² ˣ 12m) = 6.94 ˣ 10⁻³k/w

R₄ = 6.94 ˣ 10⁻³k/w

Finally the sum total of thermal resistance = R₁ + R₂ + R₃ + R₄

R-total = 0.0663 kw

From this we can calculate the rate of heat loss

using  q = Ti - To / R-total ..............(3)

given Ti and To is the inside and outside temperature i.e. 27⁰C and -10⁰C

from equation (3),

q = 27- (-10) / 0.0063 = 557.8 W

q = 557.8 W  

⇒ Applying the heat transfer formula for inside surface glass temperature gives;

q = Ti - T₂ / R₃ + R₄

T₂ = Ti - q (R₃ + R₄)

T₂ = 27 - 557.8 (1.068ˣ10⁻³ + 6.94ˣ10⁻³ ) = 22.55°C

T₂ = 22.55°C

cheers i hope this helps

8 0
3 years ago
It is the same as force. b. Stress that is created by plate collision is the same everywhere and reflects the total force produc
Vlada [557]

Answer:

Explanation:

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7 0
2 years ago
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