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3 years ago

What invention would you argue most impacted mechanical engineering

Engineering

1 answer:

Law Incorporation [45]3 years ago

3 0

Answer:

The wheel

Explanation:

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A scientist makes observations of a forest environment. Describe an example of a pattern that the scientist might’ve observe

krek1111 [17]

Answer:

The answer is different types of animals and plants, and the forest soil.

Explanation:

A forest Ecosystems consist of different animals and plants found there as well as the soil that in-habitat several microorganisms. While explaining the details of forest ecosystem is it important to describe that what kind of animals are found there and what is the role of forest soil in the growth of plantation found there.

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3 years ago

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Bars A and B have cross-sectional areas of 400 m2 and a modulus of elasticity of 200 GPa. A gap exists between bar A and the rig

NISA [10]

Answer:

axial stress in bar B = 25Mpa.

Deformation of bar A = 0.4mm.

Explanation:

PS: Kindly check the attached picture for the diagram showing the two bars that is to say the bar A and the bar B.

So, we are given the following data or information or parameters which we are going to use in solving this particular question or problem. Here they are;

The cross-sectional areas of Bars A and B = 400 mm2, the modulus of elasticity of bar A and bar B = 200 GPa, applied force = 10kN.

STEP ONE: The first step is to determine or calculate the axial stress in bar B. Therefore,

Axial stress in bar B = 10 × 10³ ÷ 400 × 10⁻⁶ = 25 Mpa.

STEP TWO: The second step here is to determine or calculate the deformation of bar A. Therefore,

The deformation of bar A = 20 × 10³ ×1.5 ÷ 400 × 10⁻⁶ × 200 × 10³ = 0.375 mm.

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3 years ago

Suppose you want to buy a new car. The car will be used mainly for going to work, shopping, running errands, and visiting friend

Sergeu [11.5K]

1. So you can complete the things you NEED to do faster. (I.e work, groceries)

2. So it’s easier for you to do leisure activities. (I.e visiting friends, shopping)

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3 years ago

Consider liquid n-hexane in a 50-mm diameter graduated cylinder. Air blows across the top of the cylinder. The distance from the

ra1l [238]

The evaporation rate of the n-Hexane is $7.85 \times 10^{-6} \mathrm{mol} / \mathrm{s}$

<u>Explanation</u>:

This is a situation regarding diffusing A through non-diffusing B.

A = n-Hexane B=Air

Where the molar flux is provided by,

$N_{A}=D_{A B} P_{T}\left(P_{A 1}-P_{A 2}\right) / R T z P_{b m}$

$\mathrm{D}_{\mathrm{AB}}=8.8 \times 10^{-6} \mathrm{m}^{2} / \mathrm{s}$

$P_{t}=1 a t m=101325 P a\\$

$\text { so, } P_{A 1}=$ the vapor pressure at hexane $25 \mathrm{C}$ $=20158.2 \mathrm{Pa}$

For wind, assume negligible hexane is present, hence $P_{A 2}=0$

Now,

$\mathrm{P}_{\mathrm{B} 1}=\mathrm{P}_{\mathrm{T}}-\mathrm{P}_{\mathrm{A} 1}=101325-20158.2 \mathrm{P}_{\mathrm{a}}$

$\mathrm{P}_{\mathrm{B} 2}=\mathrm{P}_{\mathrm{T}}-\mathrm{P}_{\mathrm{A} 2}=\mathrm{P}_{\mathrm{T}}=101325 \mathrm{Pa}$

$P_{B M}=\frac{\left(P_{B 2}-P_{B 1}\right)}{\log _{e}\left(P_{B 2} / P_{B 1}\right)}\\$

$=\frac{101325-81166.8}{\ln \left(\frac{101325}{81166.8}\right) \mathrm{Pa}}$

$=90873.57 \mathrm{Pa}$

$R=8.314 \mathrm{J} / \mathrm{mol}-\mathrm{K}$

$z=\text { distance }=20 \mathrm{cm}=0.2 \mathrm{m}\\$

where T = 298 K

substituting all in the equation, we get

$\begin{aligned}&\mathrm{N}_{\mathrm{A}}=\\&\left(8.8 \times 10^{-6} \mathrm{m}^{2} / \mathrm{s}\right) \times 101325 \mathrm{Pa} \times(20158.2 \mathrm{Pa}) /(8.314 \mathrm{J} / \mathrm{mol}-\mathrm{K} \times 0.2 \mathrm{m} \times 298 \mathrm{K}\\&\times 90873.57 \mathrm{Pa})\end{aligned}$

$=0.004 \mathrm{mol} / \mathrm{m}^{2} \mathrm{s}\\$

Now,Flux $\times$ area = Molar rate of evaporation

Evaporation rate = $0.004 \mathrm{mol} / \mathrm{m}^{2}-5 \mathrm{x}\left(\pi \mathrm{d}^{2} / 4 \mathrm{m}^{2}\right)=0.004 \times(3.14 \times 0.05 \times 0.05 / 4)$