what item did a pair of inmates use to dig their way out of a virginia jail?

The logarithmic mean temperature difference is a useful parameter in heat transfer analysis as it accounts for the non-uniformity of temperature distribution and provides an exact representation of the average temperature difference for the entire tube.

The logarithmic mean temperature difference (∆Tln) is a widely used parameter in heat transfer analysis. It represents the average temperature difference between the fluids and the surface of the entire tube. It is an exact representation of the temperature difference because it takes into account the non-uniformity of temperature distribution over the length of the tube. This non-uniformity is caused by the flow of fluids and the heat transfer rate, which vary with distance along the tube. The logarithmic mean temperature difference allows for a more accurate calculation of the heat transfer rate compared to a simple average temperature difference.

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Answer:

Se obtendrán 116.66 litros de jugo concentrado, y el agua evaporada será por un total de 883.33 litros.

Explanation:

Dado que para conseguir jugo de naranja concentrada, se parte de un extracto con 7% en peso de sólidos el cual se mete a un evaporador al vacío, y en el evaporador se elimina el agua necesaria para que el jugo salga con una concentración del 60% de peso de sólidos, si se introducen al proceso 1000 kg/h de jugo diluido, para calcular la cantidad de agua evaporada y de jugo concentrado saliente se debe realizar el siguiente cálculo;

1000 x 0.07 = 70

60 = 70

100 = X

100 x 70 / 60 = X

7000 / 60 = X

116.66 = X

Por lo tanto, se obtendrán 116.66 litros de jugo concentrado, y el agua evaporada será por un total de 883.33 litros.

Lo siento, como modelo de lenguaje, no tengo acceso a material fuera de línea o físico, como el libro "Principles of Foundation Engineering 9th Edition". Sin embargo, si puede proporcionarme los detalles del problema o la pregunta, puedo intentar ayudarlo a resolverlo o proporcionar una explicación sobre el tema en cuestión.

918.63517 is the answer.

The description you gave specifically seems to fit an Advanced Persistent Threat (APT), which is a knowledgeable and well-funded adversary that uses a variety of assault vectors to accomplish its goals.

An opponent with advanced levels of knowledge and substantial resources is known as an advanced persistent threat (APT), and it is able to seize opportunities to further its goals by employing a variety of attack vectors (e.g., cyber, physical, and deception).

In contrast to other (less sophisticated or determined) attackers, an APT has actors/sources with an appropriate level of sophistication, organisation, and resources to carry out the assault. The APT actor(s) continue to target the particular organisation or entity and adjust as necessary to further their objectives.

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Answer:

it involves the detection of light emitted by a sample after it has absorbed light of a specific wavelength.

Step-by-step:

Fluorescence spectroscopy is more sensitive than UV-Vis absorption spectroscopy because it involves the detection of light emitted by a sample after it has absorbed light of a specific wavelength. This emission of light is called fluorescence.

In UV-Vis absorption spectroscopy, the amount of light absorbed by a sample is measured as a function of wavelength. The amount of light absorbed is proportional to the concentration of the absorbing species in the sample, but the sensitivity of this technique is limited by the intensity of the incident light and the path length of the sample.

On the other hand, fluorescence spectroscopy measures the emission of light that occurs when the excited molecules return to their ground state. This emitted light is typically at a longer wavelength than the absorbed light, and it is much weaker than the incident light. However, the sensitivity of fluorescence spectroscopy is enhanced by the fact that the emitted light is measured at right angles to the excitation light, which reduces background noise from scattered light and improves the signal-to-noise ratio.

Additionally, fluorescence spectroscopy can be more selective than UV-Vis absorption spectroscopy because it can detect specific molecular species based on their unique fluorescence spectra. This selectivity is due to the fact that the fluorescence emission spectra of different molecules can be quite distinct, even for molecules with similar UV-Vis absorption spectra.

Overall, the increased sensitivity and selectivity of fluorescence spectroscopy make it a powerful technique for the detection and quantification of trace amounts of fluorescent molecules in complex samples.

Hope this helps!

Answer:

Fluorescence and UV-Vis absorption spectroscopy are two commonly used analytical techniques in the field of chemistry. Both techniques rely on the interaction of light with molecules to provide information about their electronic structure and chemical properties. However, fluorescence is generally considered to be more sensitive than UV-Vis absorption spectroscopy for several reasons.

Firstly, fluorescence is a more selective technique than UV-Vis absorption spectroscopy. When a molecule absorbs light, it undergoes a transition from its ground state to an excited state. This transition can occur via a number of different pathways, depending on the energy of the absorbed light and the electronic structure of the molecule. In contrast, fluorescence occurs when a molecule emits light after being excited by light of a specific wavelength. This means that fluorescence only occurs when certain conditions are met, such as the presence of specific functional groups or the correct excitation wavelength. As a result, fluorescence is generally more selective than UV-Vis absorption spectroscopy, which can detect any absorbing species in a sample.

Secondly, fluorescence is generally more sensitive than UV-Vis absorption spectroscopy because it produces a larger signal-to-noise ratio. In UV-Vis absorption spectroscopy, the signal is proportional to the concentration of absorbing species in the sample. However, the signal is also affected by other factors such as path length, instrument sensitivity, and background noise. In contrast, fluorescence produces a much stronger signal because it involves emission of light at a different wavelength than that used for excitation. This means that background noise and other interfering factors are less likely to affect the signal-to-noise ratio in fluorescence measurements.

Finally, fluorescence is more sensitive than UV-Vis absorption spectroscopy because it can detect lower concentrations of analyte in a sample. This is because fluorescence is an amplification process – each absorbed photon can result in multiple emitted photons if the molecule undergoes multiple cycles of excitation and emission. This amplification effect means that even low concentrations of fluorescent analyte can produce a measurable signal, whereas in UV-Vis absorption spectroscopy, the signal becomes weaker as the concentration of analyte decreases.

In summary, fluorescence is more sensitive than UV-Vis absorption spectroscopy because it is a more selective technique, produces a larger signal-to-noise ratio, and can detect lower concentrations of analyte in a sample.

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Answer:

  4.5 m/s = 16.2 km/h

Explanation:

The speed is the ratio of distance to time:

  speed = (180 m)/(40 s) = 4.5 m/s

The concentration of sulfur dioxide in ppm at the Atascadero Community Health Center HVAC air intakes is approximately 0.307 ppm.

Part A: Estimating the concentration in μg/m3 of SO2 at the Atascadero Community Health Center’s HVAC air intakes located on the roof (12.3 m above ground) as a result of emissions from the stack as shown below:The Gaussian plume equation with reflection is given as;Where;Q = emission rate in gm/sH = stack height, mU = wind speed, m/sσy, σz = standard deviation in the vertical and horizontal directions, mZ = height above ground, mx = distance from the stack, mΔH = effective stack height, mP = atmospheric pressure, N/m2g = gravitational acceleration, 9.81 m/s2Ts, Ta = stack gas temperature and ambient temperature respectively, KPaW = humidity of the flue gas.The following values have been given;Q = 1300 gm/sH = 13.6 mU = 5.44 m/sσy, σz = 1.63* (x/H)1/3 = 1.63 * (12.3/13.6)1/3 = 1.22 mZ = 12.3 mx = 0ΔH = 0.2 * H = 2.72 mP = 101.3 kPa (assume standard pressure)g = 9.81 m/s2Ts = 132+273 = 405 KTa = 22.1+273 = 295.1 KW = 0.012 kg water/kg dry air (given for a sunny day)The value of β, which is a constant given as;Beta, β = (2 x ΔH x g) / Ts = 0.0042/s.To convert g/s to μg/m3, we will use the following formula;1 gm/m3 = 1000 μg/m3.Quality check;Q = 1300 g/s = 1.3 kg/s which is similar to 1.3*3600 = 4680 kg/hr. Assuming a flow rate of 500 m3/hr, then the concentration should be around 10 mg/m3 as a maximum. Our answer should not be greater than this value.Calculation;Therefore, the concentration in μg/m3 of SO2 at the Atascadero Community Health Center’s HVAC air intakes located on the roof (12.3 m above ground) is approximately 189 μg/m3. The calculated value of 189 μg/m3 is less than the maximum value of 10 mg/m3, and therefore, it is reasonable.Part B:Sulfur dioxide, SO2, MW = 64.066 g/mol. The concentration of SO2 in ppm at the Atascadero Community Health Center HVAC air intakes can be determined as follows;The concentration in ppm is calculated using the following formula;1 ppm = (MW x Conc. in μg/m3) / (24.45 x temperature in K / P)Substituting the known values;T = 295.1 K (temperature)P = 101.3 KPa (pressure)MW = 64.066 g/moleConcentration in μg/m3 = 189 μg/m3Therefore, concentration in ppm is; Therefore, the concentration of sulfur dioxide in ppm at the Atascadero Community Health Center HVAC air intakes is approximately 0.307 ppm.Part C: If the stack gas velocity increases, the plume rise would most likely decrease. Assume all other factors do not change. Therefore, the answer is option 3, decrease.Part D: If the emission rate was doubled, the concentration at the receptor would double. Assume all other factors do not change. Therefore, the answer is option 1, double.

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Answer:

Fast Fourier ( FFT ) algorithms speed up computation of Fourier coefficients by simply reducing the the computing time of a traditional direct form  Fourier series. it  achieves this by breaking complex DFTS into smaller DFTS to reduce its complexity and in turn reduce its computing time

Explanation:

Fast Fourier ( FFT ) algorithms speed up computation of Fourier coefficients by simply reducing the the computing time of a traditional direct form  Fourier series. it  achieves this by breaking complex DFTS into smaller DFTS to reduce its complexity and in turn reduce its computing time. an example of such FFT is Cooley-Tukey algorithm

The force acting on the aortic arch is -1907.26 N as per the given data.

The relationship between pressure, kinetic energy, and gravitational potential energy of a fluid in a container is known as Bernoulli's equation.

To determine the force acting on the aortic arch, we need to convert the units of velocity, pressure, and diameter to SI units (m/s, Pa, and m respectively).

Then, we can use the principle of continuity of flow and the equation of energy conservation (Bernoulli's equation) to calculate the force.

First, we'll convert the velocity, pressure, and diameter:

V1 = 90 cm/s = 0.9 m/s

P1g = 120 mmHg = 120 x 133.322 Pa = 16039.64 Pa

P2g = 115 mmHg = 115 x 133.322 Pa = 15345.83 Pa

d1 = 27 mm = 0.027 m

d2 = 22 mm = 0.022 m

Next, we'll use the principle of continuity of flow to calculate the velocity at the outlet (V2) and cross-sectional area (A2):

A1 x V1 = A2 x V2

\(V2 = V1 X (A1 / A2) = 0.9 x (\pi X (0.027^2) / 4) / (\pi X (0.022^2) / 4) = 1.33 m/s\)

Finally, we'll use Bernoulli's equation to calculate the pressure at the outlet (P2) and then calculate the force acting on the aortic arch:

\(P1g + (1/2) X \rho X V_1^2 = P2g + (1/2) \rho X V_2^2\)

\(P2 = P1g + (1/2) X \rho X V_1^2 - (1/2) X \rho X V_2^2 = 16039.64 + (1/2) X 1.057 X 0.9^2 - (1/2) X 1.057 X 1.33^2 = 14159.56 Pa\)

\(F = (P_2 - P_1g) X A1 = (14159.56 - 16039.64) X \pi X (0.027^2) / 4 = -1907.26 N\)

Thus, the force acting on the aortic arch is -1907.26 N. The negative sign indicates that the force is acting in the opposite direction to the flow, which is normal in fluid mechanics.

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Answer:

dddddddddddddddddddddddddddddddddddddd

Explanation:

Answer:

The correct answer is "\(O (n\times Log n)\)". A further explanation is given below.

Explanation:

Answer:

Concentric circles

Explanation:

Concentric circles are two or more circles which have the same center point. The region between two concentric circles is called an annulus.

The term "drill down" refers to a user's desire to delve deeper into the details or obtain more specific information about a particular topic or data set.

When a user wants to "drill down," it means they want to explore a topic in more depth or examine data at a more granular level. This often involves navigating through hierarchical structures or layers of information to access more specific or detailed data points.

For example, in data analysis, a user may start with an overview or summary of the data and then drill down to explore specific subsets or categories of the data. They can progressively narrow their focus by selecting specific dimensions, filters, or variables to analyze. By drilling down, users can gain insights, identify patterns, and uncover underlying trends or anomalies that may not be apparent at a higher level of aggregation.

Drilling down is a common practice in various domains, including business intelligence, data visualization, financial analysis, and web analytics. It allows users to explore data in a more interactive and exploratory manner, enabling better decision-making and a deeper understanding of the information at hand. By digging into the details, users can uncover hidden insights and make more informed conclusions based on the specific data subsets they are examining.

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Answer:

  0°

Explanation:

The resistor voltage has the same phase as the circuit current. There is no phase difference.

Answer:

Explanation:

Therefore, the range of k that will make the transfer function g(s) stable is k < 7.2. Any value of k within this range will ensure that all the coefficients in the first column of the Routh array are positive, and the system will be stable.

To determine the stability of the transfer function g(s) = 4s^5 + 4s^3 + 8s^2 + 5ks + 9, we can use the Routh-Hurwitz stability criterion. First, we will create a Routh array using the coefficients of the polynomial.

| 4 | 8 | 9 |
| --- | --- | --- |
| 4 | 5k | 0 |
| 1.25k | 9 | 0 |
| 9 - 1.25k | 0 | 0 |
For the system to be stable, all the coefficients in the first column of the Routh array must be greater than zero. So, we can set the inequality 9 - 1.25k > 0 and solve for k to find the range of values that will make the system stable.
9 - 1.25k > 0
1.25k < 9
k < 7.2
Therefore, the range of k that will make the transfer function g(s) stable is k < 7.2. Any value of k within this range will ensure that all the coefficients in the first column of the Routh array are positive, and the system will be stable.

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The statement "Employees are not responsible for their own safety while at work" is false because Employees most certainly are responsible.

A multidisciplinary discipline dealing with the safety, health, and welfare of individuals at work is known as occupational safety and health, often known as occupational health and safety, occupational health, or occupational safety.

An environment that is safe and healthy for workers may minimize injury and sickness expenses, lower levels of absenteeism, boost output and quality, and improve employee morale. In other words, safety benefits the business.

Thus, the statement "Employees are not responsible for their own safety while at work" is false because Employees most certainly are responsible.

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Answer:

I dont think you can blur text in google docs

Answer:

somewhere around 34.2223 meters thick but that's what I am estimating.

Known :

Pg = 51 psi

Patm = 28 inHg = 13.7 psi

Solution :

Absolute pressure is gauge pressure plus the atmospheric pressure :

Pabs = Pg + Patm

Pabs = 51 + 13.7

Pabs = 64.8 psi

Answer:

OPTION A,Larger

HOPE IT HELPS

Answer:

Recycling prevents the emissions of many greenhouse gases and water pollutants.

Explanation:

Answer:

lower costs

Explanation:

sorry for adding an answer after a little over a year.

Answer:

r = 1.922 mm

Explanation:

We are given;

Yield stress; σ = 250 MPa = 250 N/mm²

Force; F = 29 KN = 29000 N

Now, formula for yield stress is;

σ = F/A

A = F/σ

Where A is area = πr²

Thus;

r² = 2900/250π

r² = 3.6924

r = √3.6924

r = 1.922 mm

Answer:

whatsthe question choices

Answer:

<E1, E2>.

Explanation:

So, in the question above we are given that the Two Electric field vectors E1 and E2 are perpendicular to each other. Thus, we are going to have the i and the j components for the two Electric Field that is E1 and E2 respectively. That is to say the addition we give us a resultant E which is an arbitrary vector;

E = |E| cos θi + |E| sin θj. -------------------(1).

Therefore, if we make use of the components division rule we will have something like what we have below;

x = |E2|/ |E| cos θ and y = |E1|/|E| sin θ

​

Therefore, we will now have;

E = x |E2| i + y |E1| j.

The base vectors is then Given as <E1, E2>.

At 1.2mpa pressure and 50c

What is pressure?
By pressing a knife against some fruit, one can see a straightforward illustration of pressure. The surface won't be cut if you press the flat part of the knife against the fruit. The force is dispersed over a wide area (low pressure).

a)Mass flow rate of the refrigerant
Therefore h1= condenser inlet enthalpy =278.28KJ/Kg
saturation temperature at 1.2mpa is 46.29C
Therefore the temperature of the condenser

T2 = 46.29C - 5
T2 = 41.29C

Now,
d)power consumed by compressor W = 3.3KW
Q4 = QL + w = Q4
QL = mR(h1-h2)-W
= 0.0498 x (278.26 - 110.19)-3.3
=5.074KW
Hence refrigerator load is 5.74Kg

(COP)r = 238/53
(Cop) = 4.490

Therefore the above values are the (a) mass flow rate of the refrigerant, b) the refrigerant load, c) the cop, and d) the minimum power input to the compressor for the same refrigeration load.

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