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3 years ago

Engineering Questions

Engineering

2 answers:

valkas [14]3 years ago

6 0

5 is the correct one to choose for this

ankoles [38]3 years ago

6 0

1. the branch of science and technology concerned with the design, building, and use of engines, machines, and structures

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The _______ is a function that describes how the pinna, ear canal, head, and torso change the intensity of sounds with different

scoundrel [369]

Answer:

B: Directional Transfer Function

Explanation:

The function that describes how the pinna, ear canal, head, and torso change the intensity of sounds with different frequencies that arrive at each ear from different locations in space is called Directional Transfer Function.

8 0

2 years ago

A 650-kN column load is supported on a 1.5 m square, 0.5 m deep spread footing. The soil below is a well-graded, normally consol

insens350 [35]

<u>Explanation:</u>

Determine the weight of footing

$W_{f}=\gamma(L)(B)(D)$

Where $W_{f}$ is the weight of footing, γ is the unit weight of concrete, L is the length of footing is the width of footing, and D is the depth of footing

Substitute $2 m \text { for } L, 1.5 m \text { for } B, 0.5 m \text { for } D \text { and } 23.6 kN / m ^{3}$ for γ in the equation

$\begin{aligned}W_{f} &=\left(23.6 kN / m ^{3}\right)(2 m )(1.5 m )(0.5 m ) \\&=35.4 kN\end{aligned}$

Therefore, the weight of the footing is 35.4 kN

Determine the initial vertical effective stress.

$\sigma_{z p}^{\prime}=\gamma(D+B)-u$

Here, $\sigma_{z^{p}}^{\prime}$ is initial vertical stress at a depth below ground surface γ is the unit weight of soil, D is depth and u is pore water pressure.

Substitute $18 kN / m ^{3} \text { for } \gamma, 1.5 m \text { for } B, 0.5 m \text { for } D \text { and } 0$ for u in the equation

$\begin{aligned}\sigma_{z p}^{\prime} &=\left(18 kN / m ^{3}\right)(1.5+0.5) m -0 \\&=36 kPa\end{aligned}$

Therefore, the initial vertical stress is 36 kPa

Determine the vertical effective stress.

$\sigma_{z D}^{\prime}=\gamma D$

Here, $\sigma_{z^{p}}^{\prime}$ is initial vertical stress at a depth below ground surface γ is the unit weight of soil, D is depth and u is pore water pressure.

Substitute $18 kN / m ^{3}\) for $\gamma, 0.5 m$ for $D$ and 0 for \(u$ in the equation.

$\begin{aligned}\sigma_{z b}^{\prime} &=\left(18 kN / m ^{3}\right)(0.5 m )-0 \\&=9 kPa\end{aligned}$

Therefore, the vertical stress at a depth below the ground surface is

9 kPa

Determine the influence factor at the midpoint of soil layer,

$I_{e p}=0.5+0.1 \sqrt{\frac{q-\sigma_{s 0}^{\prime}}{\sigma_{z p}^{\prime}}}$

Here $I_{e p}$ is the influence factor at the midpoint of soil layer $\sigma_{z^{p}}^{\prime}$ is initial vertical stress, $\sigma_{z^{p}}^{\prime}$ is vertical effective stress, and $Q$ is bearing pressure

Substitute $36 kPa for $\sigma_{z p}^{\prime}, 228.47$ kPa for $q,$ and 9 kPa for $\sigma_{z D}^{\prime}$$ in the equation $\begin{aligned}I_{\epsilon P} &=0.5+0.1 \sqrt{\frac{228.47 kPa -9 kPa }{36 kPa }} \\&=0.75\end{aligned}$

Therefore the influence factor at the midpoint of the soil layer is 0.693

6 0

3 years ago

X-Ray Inspection

Softa [21]

The answer is the test is being tested towards the lungs the test is done by scanning your body the tools are called “the x rat visional lock space” and the rubber tool is called a deeldo it’s purple with a pencil looking shape perfect for the body.

5 0

3 years ago

You are recording a friend's 3-minute song with 24 bits per sample at 96 kHz sampling rate for a 5.1 surround sound system (6 ch

djyliett [7]

Answer:

save space = 1.9626 Gibibits

Explanation:

given data

recording song time = 3 minute = 180 seconds

song per sample = 24 bits per sample

frequency = 96 kHz

surround sound system = 5.1

solution

first we get here required space for 24 bits at 96 kHz that is

required space = 24 × 96 × 10³ × 6 × 180

required space = 2.48832 × $10^{9}$ bits

as 1 Gib bit = $2^{30}$ bits

so required space = 2.48832 × $10^{9}$ bits ÷ $2^{30}$ bits

required space = 2.3174 Gibibits ...............1

and

space required to save record 16 bit at 44.1 kHz

space required = 16 × 44.1 × 10³ × 3 × 180

space required = 0.381024 × $10^{9}$ bits

space required = 0.381024 × $10^{9}$ bits ÷ $2^{30}$ bits

space required = 0.3548 Gibibits ...........2

we get here that save space in 16 bit at 44.1 kHz

save space = 2.3174 Gibibit - 0.3548 Gibibits

save space = 1.9626 Gibibits

8 0

3 years ago

Technician A says that forma hardened steel may have different strenght áreas. Technician B says that aluminum collapses in a pr

stich3 [128]

Technician B is correct because the way aluminum collapses can be predicted.

Hardened steel and aluminum are two metals used for different purposes including:

Construction.
Appliances.
Small utensils.
Airplanes.
Vehicles.

These two materials have slightly different features in terms of resistance, flexibility, etc.

In the case of hardened steel, this is considered to be malleable but strong. This means it is possible to change its shape under some conditions but it can resist great forces and pressure. Moreover, if the hardening process is carried out properly all the areas should be equally strong.

On the other hand, aluminum is recognized due to its durability and for being lighter than other materials. Despite this, aluminum is more flexible than steel and collapses under weaker forces. This has been widely studied because aluminum collapse shows a predictable pattern.

Based on this, only technician B is correct.

Learn more in: brainly.com/question/24043240

5 0

2 years ago