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mr Goodwill [35]
4 years ago
11

If you are a government authority what extend will you modify the existing policy

Engineering
1 answer:
mr_godi [17]4 years ago
4 0

Answer:

kk

Explanation:

dkdndidodd ndidkjeeiwonejeeidmdnddkdidfmndd

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Problem 3: Soil Classification using the AASHTO and USCS Systems
nataly862011 [7]

<u>Solution:</u>

Given\\                   \(\quad W=3000 Ib , \quad m=\frac{W}{g}=\frac{3000}{322} \ slug =93.1677 slug\)\\K_{e q}=2160 lbs / wp =2100 \frac{ lbs }{10} \frac{ x 12}{1 ft }=(2160 \times 12) lb / ft$$

a) The natural frequency

\begin{aligned}&\left(\omega_{n}\right)=\sqrt{\frac{K_{e q}}{m}}\\&=\sqrt{\frac{2160 \times 12}{93.1677}}\\&\omega_{n}=16.68 \text { rad } | s\\&\omega_{n}=\frac{2 \pi}{T}\\&16.68=\frac{2 \pi}{T}\\&T=0.3767 s\end{aligned}

b)

Given, \(t=10 s , \quad y(t)=6 in = A\)\\\(y(t)=A \cos \left(\omega_{n} t+\phi\right) \rightarrow 0\)\\\(6=6 \cos (16.68 \times 10+\phi)\)\\\(1=\cos (166.8+\phi)\)\\\(166.8+\phi=0\)\\\phi=-166.8\)\\At \(t=0, \quad y(0)=6 \cos (16.68 \times 0-166.8)\) {y(0)}=-5.74 in

5 0
3 years ago
How can any student outside apply for studying engineering at Cambridge University​
telo118 [61]
Admission to the Engineering course at Cambridge is highly competitive, both in terms of the numbers and quality of applicants. In considering applicants, Colleges look for evidence both of academic ability and of motivation towards Engineering. There are no absolute standards required of A Level achievement, but it should be noted that the average entrant to the Department has three A* grades. You need to get top marks in Maths and Physics.All Colleges strongly prefer applicants for Engineering to be taking a third subject that is relevant to Engineering.
Hope that helps and good luck if you are applying. Can you please mark this as brainliest and press the thank you button and if you have any further questions please let me know!!
3 0
3 years ago
What steps would you take to design an improved toothpaste container?
algol [13]

Answer:

A. Identify the need, recognize limitations of current toothpaste containers, and then brainstorm ideas on how to improve the existing

Explanation:

To design an improved toothpaste container, we must identify the needs of the customer, one of the major need is to make the container attractive to the sight. This is the first thing that will prompt a customer to wanting to buy the product (The reflectance/appearance).

Then recognize the limitation of the current design, what needed change. This will help in determining what is needed to be included and what should be removed based on identified customers need.

The last step is to brainstorm ideas on how to improve the existing designs. Get ideas from other colleagues because there is a saying that two heads are better than one. This will help in coming to a reasonable conclusion on the new design after taking careful consideration of people's opinion.

7 0
3 years ago
Five kilograms of air at 427°C and 600 kPa are contained in a piston–cylinder device. The air expands adiabatically until the pr
son4ous [18]

Answer:

The entropy change of the air is 0.240kJ/kgK

Explanation:

T_{1} =427+273K,T_{1} =700K\\P_{1} =600kPa\\P_{2} =100kPa

T_{2}  is unknown

we can apply the following expression to find T_{2}

-w_{out} =mc_{v} (T_{2} -T_{1} )

T_{2} =T_{1} -\frac{w_{out } }{mc_{v} }

now substitute

T_{2} =700K-\frac{600kJ}{5kg*0.718kJ/kgK} \\T_{2}=533K

To find entropy change of the air we can apply the ideal gas relationship

Δs_{air}=c_{p} ln\frac{T_{2} }{T_{1} } -Rln\frac{P_{2} }{P_{1} }

Δs_{air} =1.005*ln(\frac{533}{700})-0.287* in(\frac{100}{600} )

Δs_{air} =0.240kJ/kgK

4 0
3 years ago
A well insulated rigid tank contains 4 kg of argon gas at 450 kPa and 30 C. A valve is opened, allowing the argon to escape unti
natima [27]

Answer:

Final mass of Argon=  2.46 kg

Explanation:

Initial mass of Argon gas ( M1 ) = 4 kg

P1 = 450 kPa

T1 = 30°C = 303 K

P2 = 200 kPa

k ( specific heat ratio of Argon ) = 1.667

assuming a reversible adiabatic process

<u>Calculate the value of the M2 </u>

Applying ideal gas equation ( PV = mRT )

P₁V / P₂V = m₁ RT₁ / m₂ RT₂

hence : m2 = P₂T₁ / P₁T₂ * m₁

                   = (200 * 303 ) / (450 * 219 ) * 4

                   = 2.46 kg

<em>Note: Calculation for T2 is attached below</em>

5 0
3 years ago
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