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Pepsi [2]
3 years ago
9

We can process oil into a lot of useful fuels to run our cars, trucks, and even airplanes. Oil is used for making lots of other

valuable products that we use every day. Can you name five of those products?
Engineering
1 answer:
Ostrovityanka [42]3 years ago
4 0

Answer:

Explanation:

Products of oil in our everyday life:

(1) Petro-Chemical Feedstock: These are by product of Refining of Oil which it is used extensively to make PET bottles, Paints, Polyester Shirts, Pocket combs e.t.c

(2) Asphalt : Used extensively to make Motor Road, highways

(3) Plastics : we use plastics in our everyday life, this is also a product of Refining of crude oil e.g PVC, Telephone casing, Tapes e.t.c

(4) Lubricating Oil/Grease : This is another product from crude oil Fractional Distillation.

(5) Propane/ Cooking Gas: This is also a product from oil which is used in our everyday life for cooking, grilling etc.

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On a piece of paper, sketch the x-y stress state and the properly oriented principal stress state. Use the resulting sketch to a
stealth61 [152]

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step and very detailed solution of the given problem

6 0
3 years ago
Suppose there are 76 packets entering a queue at the same time. Each packet is of size 5 MiB. The link transmission rate is 2.1
tia_tia [17]

Answer:

938.7 milliseconds

Explanation:

Since the transmission rate is in bits, we will need to convert the packet size to Bits.

1 bytes = 8 bits

1 MiB = 2^20 bytes = 8 × 2^20 bits

5 MiB = 5 × 8 × 2^20 bits.

The formula for queueing delay of <em>n-th</em> packet is :  (n - 1) × L/R

where L :  packet size = 5 × 8 × 2^20 bits, n: packet number = 48 and R : transmission rate =  2.1 Gbps = 2.1 × 10^9 bits per second.

Therefore queueing delay for 48th packet = ( (48-1) ×5 × 8 × 2^20)/2.1 × 10^9

queueing delay for 48th packet = (47 ×40× 2^20)/2.1 × 10^9

queueing delay for 48th packet = 0.938725181 seconds

queueing delay for 48th packet = 938.725181 milliseconds = 938.7 milliseconds

4 0
3 years ago
You want to plate a steel part having a surface area of 160 with a 0.002--thick layer of lead. The atomic mass of lead is 207.19
Pepsi [2]

Answer:

<u><em>To answer this question we assumed that the area units and the thickness units are given in inches.</em></u>

The number of atoms of lead required is 1.73x10²³.    

Explanation:

To find the number of atoms of lead we need to find first the volume of the plate:

V = A*t

<u>Where</u>:

A: is the surface area = 160

t: is the thickness = 0.002

<u><em>Assuming that the units given above are in inches we proceed to calculate the volume: </em></u>

V = A*t = 160 in^{2}*0.002 in = 0.32 in^{3}*(\frac{2.54 cm}{1 in})^{3} = 5.24 cm^{3}    

Now, using the density we can find the mass:

m = d*V = 11.36 g/cm^{3}*5.24 cm^{3} = 59.5 g

Finally, with the Avogadros number (N_{A}) and with the atomic mass (A) we can find the number of atoms (N):

N = \frac{m*N_{A}}{A} = \frac{59.5 g*6.022 \cdot 10^{23} atoms/mol}{207.19 g/mol} = 1.73 \cdot 10^{23} atoms    

Hence, the number of atoms of lead required is 1.73x10²³.

I hope it helps you!

3 0
3 years ago
Koch traded Machine 1 for Machine 2 when the fair market value of both machines was $60,000. Koch originally purchased Machine 1
Mariana [72]

Answer:

Koch's adjusted basis in machine 2 after the exchange is $60,000

Explanation:

given data

fair market value = $60,000

originally purchased Machine 1 = $76,900

Machine 1 adjusted basis = $40,950

Machine 2 seller purchase = $64,050

Machine 2 adjusted basis = $55,950

solution

As he exchanged machine for another at $60,000

and this exchanged in fair market

so adjusted basis =  $50,000

Adjusted basis is the price of the item that affects the factors that are considered price. These factors usually include taxes, depreciation value, and other costs of acquiring and maintaining a given item. Adjusted basis is important so the right amount to sell

Adjusted basis increases when a person deducts expenses from factor taxes and operating statements

so Koch's adjusted basis in machine 2 after the exchange is $60,000

3 0
3 years ago
An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 130 lb and
k0ka [10]

Answer:

A.) 0.3088

B.) 0.0017

C.) part A

Explanation:

A.)

z1= \frac{\left(150-137\right)}{27.7}=0.4693

z2=\frac{\left(201-137\right)}{27.7}=2.3105

P(0.4693

B.)

z1=\frac{150-137}{27.7/ \sqrt{39}} =2.9309\\z2=\frac{201-137}{27.7/ \sqrt{39}}=14.4289

\\P(2.9309

C.) Since the seat performance for an individual pilot is more important than 39 different pilots.

3 0
3 years ago
Read 2 more answers
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