Consider the circuit shown below. a. Determine the differential equation relating outputs \( y_{2}(t) \) to the input \( x(t) \).

Put your printer through a hard reset. Try turning off both the printer and your computer, then restarting both if that doesn't fix the problem. Then try to remove and reinstall your printer's driver.

A problem with the printer itself may exist if the status of your printer reads "Printer in error state." Check to see if the printer is on and wired or wirelessly linked to your computer. Make sure the cover is closed and the paper isn't jammed, and check if there isn't enough paper or ink.

Of all printer errors, paper jams are arguably the most dreadful. There are a few problems that might be present when it comes to smudges, faded lettering, and poor image quality.

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The best practice that Raven should follow is to use DC2 or DC3 as the Domain Naming Master of the following is a best practice that Raven should follow. The correct option is A.

The management of the addition or deletion of domains from the forest is the responsibility of the Domain Naming Master. For redundancy and fault tolerance, it is advised to split the FSMO roles among several domain controllers.

Since DC1 already has a copy of the global catalog, it is advantageous to choose a different domain controller (DC2 or DC3) as the Domain Naming Master to disperse the workload and guarantee high availability. This ensures that the forest's operations may continue even if one domain controller goes offline and prevents the creation of a single point of failure.

Thus, the ideal selection is option A.

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The orbital motion of water particles due to surface waves stops at a depth of "one-half the wavelength" (Option C).

Deep water waves (depth, D >λ /2) and shallow water waves (D < λ /20) both exhibit particle orbital motion. However, while deep water waves have a circular orbit, shallow water waves have an elliptical orbit with the primary axis in the horizontal plane [2, 3, 4, 5].

Instead, the water particles travel in circular orbits with a radius equal to wave height. This orbital motion arises because water waves include both longitudinally (side to side) and transverse (up and down) components, resulting in a circular motion.

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From the question above, :Width of the channel = 4 m

Depth of the flow = 2 m

Discharge = 20 m³/sec

Change in channel width = 1 mAs the discharge is constant, we have the equation of continuity as;

Q = A₁V₁ = A₂V₂

Where,Q = Discharge in m³/s

V₁ = Velocity of fluid in the original channel

A₁ = Area of the flow in the original channel

V₂ = Velocity of fluid in the new channel

A₂ = Area of the flow in the new channel

As the channel is rectangular in shape, we can write the equation as;

Q = W₁ * D₁ * V₁ = W₂ * D₂ * V₂

Where, D₁ = D₂ = D (Depth of flow)

W₁ = 4 m

W₂ = 4 + 1 = 5 m

V₁ = Velocity of fluid in the original channel

V₂ = Velocity of fluid in the new channel∴

4 * 2 * V₁ = 5 * D₂ * V₂∴ 8V₁ = 5D₂V₂∴ V₂ = 8/5 * V₁

The velocity of fluid in the new channel = 1.6V₁

Type of flow in the original channel can be determined using the Froude number as;

Fr = V₁/ √gD

Where g is the acceleration due to gravity

Fr₁ = V₁/ √gD = V₁/ √(9.81 * 2) = V₁/ 4.429

Fr₂ = V₂/ √ gD = (1.6V₁)/ √(9.81 * 2) = 1.6V₁/ 4.429

Fr₁ = V₁/ √gD = V₁/ 4.429

Fr₂ = 1.6V₁/ 4.429

Fr₁ < 1 → Subcritical flow

Fr₂ < 1 → Subcritical flow

As the value of Fr is less than 1, the type of flow is Subcritical flow

.Change in depth of the flow;

D₂ - D₁ = (W₁/W₂ - 1) * D₁D₂ - 2 = (4/5 - 1) * 2

D₂ - 2 = -0.4∴ D₂ = 1.6 m

Change in depth of the flow = D₂ - D₁ = 1.6 - 2 = -0.4 mA

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a. The height of a tree is the maximum number of edges in any path from the root to a leaf node. To find the height of the tree with root 70, we need to traverse the tree and determine the longest path from the root to a leaf. The height of the tree is the length of that path.

b. To find the height of the tree with root 70, we need to traverse the tree and determine the longest path from the root to a leaf. The height of the tree is the length of that path.

c. To list the path from the node with info 92 to the node with info 78, we start from the root and traverse the tree following the appropriate branches at each node until we reach the target node. We record the nodes visited during this traversal to obtain the path from node 92 to node 78.

d. To insert a node with info 30 in the binary search tree, we start at the root and compare the values until we find the appropriate position for insertion. The nodes visited during this process are the nodes that are traversed to insert the node with info 30. After inserting 30, we redraw the tree and determine its height.

e. To delete node 45 from the binary search tree, we need to find the node and rearrange the tree accordingly. After deleting 45, we redraw the tree and determine its height.

f. To delete nodes 63 and 82 from the binary search tree, we need to find the nodes and rearrange the tree accordingly. We perform the deletions in the specified order, redrawing the tree after each deletion. After deleting both nodes, we determine the height of the resulting tree.

g. To delete node 70 from the binary search tree, we need to find the node and rearrange the tree accordingly. After deleting 70, we redraw the tree and determine its height.

In conclusion, by following the instructions given and performing the required operations on the binary search tree, we can find the heights of the tree in different scenarios and draw the resulting trees.

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Answer:

busway

Explanation:

The required thickness of the cast iron column is approximately 16.8mm. we can use the formula for the buckling stress of a column:σ_critical = π^2 * E / (KL/r)^2

What is buckling Stress?

The abrupt change in shape (deformation) of a structural component under load, such as the bowing of a column under compression or the wrinkling of a plate under shear, is referred to as buckling in structural engineering.

To find the required thickness of the cast iron column, we can use the formula for the buckling stress of a column:

σ_critical = π^2 * E / (KL/r)^2

where σ_critical is the critical buckling stress, E is the modulus of elasticity of cast iron, K is the Rankine's constant (1/1600 for a fixed-fixed column), L = length of the column, and r = radius of the column.

The external diameter of the column is 250mm, so the radius is r = 250/2 = 125mm.

The modulus of elasticity for cast iron is around 200,000 N/mm^2.

We can then calculate the critical buckling stress:

σ_critical = π^2 * 200,000 / (1/1600 * 5000/125)^2

σ_critical = 80 N/mm^2

Since the working stress of the column is 80 N/mm^2, which is equal to the critical buckling stress, we can calculate the required thickness t using the formula for stress in a column:

σ = P / (π * r^2 / 4)

where P is the load on the column, and π * r^2 / 4 is the cross-sectional area of the column.

Given that the load on the column is 1000 KN, we can calculate the required thickness:

t = P * 4 / (π * r^2 * σ)

t = (1000 * 1000) * 4 / (π * 125^2 * 80)

t = 16.8 mm

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The critical crack length for a through crack in a thick plate of 7150-T651 aluminum alloy under uniaxial tension can be determined using fracture mechanics principles. For this particular alloy, the fracture toughness (Kic) is given as 25.5 MPa√m, and the ultimate tensile strength (σf) is 400 MPa.

By assuming a yield strength (Y) equal to the square root of pi, the critical crack length can be calculated.

Fracture mechanics is a field of study that deals with the behavior of materials when subjected to cracks or flaws. The critical crack length represents the maximum allowable size of a crack within a material before it fails catastrophically. In this case, we are considering a through crack in a thick plate of 7150-T651 aluminum alloy that is subjected to uniaxial tension.

To determine the critical crack length, we need to use the fracture toughness value (Kic) and the ultimate tensile strength (σf) of the material. The fracture toughness measures a material's resistance to crack propagation, and it is given as 25.5 MPa√m for this particular aluminum alloy. The ultimate tensile strength represents the maximum stress the material can withstand before failure, and it is specified as 400 MPa.

In fracture mechanics, the critical crack length can be calculated using the equation:

a = (Kic / (Y * σf))²

Here, Y represents the yield strength of the material. In this case, Y is assumed to be equal to the square root of pi (√π). By substituting the given values into the equation, we can determine the critical crack length for the 7150-T651 aluminum alloy under uniaxial tension.

The calculation of the critical crack length will provide a value in terms of the crack size (a). This value represents the maximum allowable crack length before failure occurs. It is crucial to consider the critical crack length in engineering applications to ensure the structural integrity and safety of components made from the aluminum alloy.

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Answer:

(a) L ≈ 0.397 H

(b) Z ≈ 52.5 kOhm

(c) θ ≈ 0.025 radians

Step-by-step explanation:

To calculate the values for an RL low pass filter with a cut-off frequency of 4 kHz and using an R-value of 10 kOhm, we can use the following formulas:

(a) L = R / (2 * π * f_c)

(b) Z = √(R^2 + (2 * π * f)^2)

(c) θ = atan((2 * π * f) / R)

Given:

R = 10 kOhm

f = 25 kHz

f_c = 4 kHz

(a) L = R / (2 * π * f_c)

L = 10 kOhm / (2 * π * 4 kHz)

L ≈ 0.397 H

(b) Z = √(R^2 + (2 * π * f)^2)

Z = √((10 kOhm)^2 + (2 * π * 25 kHz)^2)

Z ≈ 52.5 kOhm

(c) θ = atan((2 * π * f) / R)

θ = atan((2 * π * 25 kHz) / 10 kOhm)

θ ≈ 0.025 radians

Therefore, the calculated values are:

(a) L ≈ 0.397 H

(b) Z ≈ 52.5 kOhm

(c) θ ≈ 0.025 radians

Answer:

Beam of 25" depth and 12" width is sufficient.

I've attached a detailed section of the beam.

Explanation:

We are given;

Beam Span; L = 20 ft

Dead load; DL = 0.50 k/ft

Live load; LL = 0.65 k/ft.

Beam width; b = 12 inches

From ACI code, ultimate load is given as;

W_u = 1.2DL + 1.6LL

Thus;

W_u = 1.2(0.5) + 1.6(0.65)

W_u = 1.64 k/ft

Now, ultimate moment is given by the formula;

M_u = (W_u × L²)/8

M_u = (1.64 × 20²)/8

M_u = 82 k-ft

Since span is 20 ft, it's a bit larger than the average span beams, thus, let's try a depth of d = 25 inches.

Effective depth of a beam is given by the formula;

d_eff = d - clear cover - stirrup diameter - ½Main bar diameter

Now, let's adopt the following;

Clear cover = 1.5"

Stirrup diameter = 0.5"

Main bar diameter = 1"

Thus;

d_eff = 25" - 1.5" - 0.5" - ½(1")

d_eff = 22.5"

Now, let's find steel ratio(ρ) ;

ρ = Total A_s/(b × d_eff)

Now, A_s = ½ × area of main diameter bar

Thus, A_s = ½ × π × 1² = 0.785 in²

Let's use Nominal number of 3 bars as our main diameter bars.

Thus, total A_s = 3 × 0.785

Total A_s = 2.355 in²

Hence;

ρ = 2.355/(22.5 × 12)

ρ = 0.008722

Design moment Capacity is given;

M_n = Φ * ρ * Fy * b * d²[1 – (0.59ρfy/fc’)]/12

Φ is 0.9

f’c = 4,000 psi = 4 kpsi

fy = 60,000 psi = 60 kpsi

M_n = 0.9 × 0.008722 × 60 × 12 × 22.5²[1 - (0.59 × 0.008722 × 60/4)]/12

M_n = 220.03 k-ft

Thus: M_n > M_u

Thus, the beam of 25" depth and 12" width is sufficient.

Answer:

volunteer coordinator

Explanation:

because they are volunteering for that and in most of the cases they do not expect to be paid

Answer:

999999999999999999999

The decomposition of Liquid A follows a first-order kinetics. It means that the rate of reaction is proportional to the concentration of A present at any given time.

The rate constant for the reaction is k. The formula for the rate of a first-order reaction is given as follows:r = k[A]where, r is the rate of reaction, and [A] is the concentration of A at any given time.The time taken for the conversion of 50% of A is given as 5 minutes. The concentration of A remaining is 50% of the initial concentration. The rate of reaction at this point is:r = k[0.5 A0]where, A0 is the initial concentration of A.

Since the reaction follows a first-order kinetics, the rate constant k will remain constant throughout the reaction.To calculate the time taken for the conversion of 75% of A, we can use the following equation:ln ([A]t/[A]0) = -ktwhere, [A]t is the concentration of A remaining after time t, and [A]0 is the initial concentration of A. We know that [A]t = 0.25[A]0.Substituting these values, we get:ln (0.25) = -k(t2 - t1)where, t1 = 5 minutes (time taken for 50% conversion), and t2 is the time taken for 75% conversion.Solving for t2, we get:t2 = t1 + (1/k) ln(0.25)Substituting the value of k from the rate equation, we get:t2 = 5 + (1/k) ln(0.25 [A]0)Therefore, we need to know the value of the rate constant k to calculate the time taken for 75% conversion.

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Answer:

No. of transistors = \($4.1524 \times 10^{10}$\) transistors

Explanation:

Given that:

N = \($1920 \times 10^{0.163(Y-1980)}$\)

Y = 2025

N = \($1920 \times 10^{0.163(2025-1980)}$\)

N = \($4.1524 \times 10^{10}$\) transistors

Now at Y = 1980

Number of transistors N = 1920

Therefore,

\($1000 = 10^{0.163(Y-1980)}$\)

\($\log_{10} 1000=0.163(Y-1980)$\)

\($\frac{3}{0.163}=Y-1980$\)

18 ≅ 18.4 = Y - 1980

Y = 1980 + 18

   = 1998

So, to increase multiples of 1000 transistors. it takes 18 years.

It is subjective to determine the "most happy" and "most sad" song as it varies from person to person. Different songs can evoke different emotions in different people.

However, some songs that are commonly considered to be happy include "Happy" by Pharrell Williams, "Don't Stop Believin'" by Journey, and "I Will Always Love You" by Whitney Houston. Some songs that are commonly considered to be sad include "Everybody Hurts" by R.E.M., "Tears in Heaven" by Eric Clapton, and "My Heart Will Go On" by Celine Dion.

The emotions that a song evokes in a listener are subjective, meaning that they can vary from person to person. For example, a song that one person finds to be happy and uplifting, another person may find to be sad or melancholy. This is because emotions are personal experiences that are shaped by a person's individual life experiences, beliefs, and attitudes.

Therefore, Regarding the songs mentioned, "Happy" by Pharrell Williams, "Don't Stop Believin'" by Journey, and "I Will Always Love You" by Whitney Houston are considered happy songs because they have upbeat tempos, positive lyrics, and a cheerful sound that can make people feel good.

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Answer:

The velocity at section is approximately 42.2 m/s

Explanation:

For the water flowing through the pipe, we have;

The pressure at section (1), P₁ = 300 kPa

The pressure at section (2), P₂ = 100 kPa

The diameter at section (1), D₁ = 0.1 m

The height of section (1) above section (2), D₂ = 50 m

The velocity at section (1), v₁ = 20 m/s

Let 'v₂' represent the velocity at section (2)

According to Bernoulli's equation, we have;

\(z_1 + \dfrac{P_1}{\rho \cdot g} + \dfrac{v^2_1}{2 \cdot g} = z_2 + \dfrac{P_2}{\rho \cdot g} + \dfrac{v^2_2}{2 \cdot g}\)

Where;

ρ = The density of water = 997 kg/m³

g = The acceleration due to gravity = 9.8 m/s²

z₁ = 50 m

z₂ = The reference = 0 m

By plugging in the values, we have;

\(50 \, m + \dfrac{300 \ kPa}{997 \, kg/m^3 \times 9.8 \, m/s^2} + \dfrac{(20 \, m/s)^2}{2 \times 9.8 \, m/s^2} = \dfrac{100 \ kPa}{997 \, kg/m^3 \times 9.8 \, m/s^2} + \dfrac{v_2^2}{2 \times 9.8 \, m/s^2}\)50 m + 30.704358 m + 20.4081633 m = 10.234786 m + \(\dfrac{v_2^2}{2 \times 9.8 \, m/s^2}\)

50 m + 30.704358 m + 20.4081633 m - 10.234786 m = \(\dfrac{v_2^2}{2 \times 9.8 \, m/s^2}\)

90.8777353 m = \(\dfrac{v_2^2}{2 \times 9.8 \, m/s^2}\)

v₂² = 2 × 9.8 m/s² × 90.8777353 m

v₂² = 1,781.20361 m²/s²

v₂ = √(1,781.20361 m²/s²) ≈ 42.204308 m/s

The velocity at section (2), v₂ ≈ 42.2 m/s

When air undergoes adiabatic expansion, the process is governed by the equation PV^γ = constant, where γ is the ratio of specific heats. In this case, the value of γ for air is 1.4. We are given that one pound of air undergoes an adiabatic expansion from 200 psia to 50 psia, with an initial volume of 4 ft3/lbm and PV^1.4 = constant.

Let's calculate the final volume of the air using the initial and final pressures and the initial volume. Using the formula P1V1^γ = P2V2^γ and substituting the given values, we have: 200(4)^1.4 = 50(V2)^1.4V2 = (200(4)^1.4 / 50)^(1/1.4)V2 = 11.14 ft^3/lbmThe work done by the air is given by the equation W = ∆E + Q, where ∆E is the change in internal energy and Q is the heat added to or removed from the system. Since the process is adiabatic (Q = 0), the work done is equal to the change in internal energy. Let's calculate the work done:W = ∆E = C_v (T2 - T1)where C_v is the specific heat at constant volume, and T1 and T2 are the initial and final temperatures, respectively. The specific heat at constant volume for air is 0.1715 Btu/lbm·R. Let's calculate the final temperature of the air using the initial and final pressures and volumes and the equation P1V1^γ/T1 = P2V2^γ/T2.200(4)^1.4/T1 = 50(11.14)^1.4/T2T2 = T1 * (P2V2^γ / P1V1^γ)T2 = 1183.3 RLet's substitute the values into the equation for work done to get:W = C_v (T2 - T1)W = 0.1715 Btu/lbm·R (1183.3 R - 527.7 R)W = 0.1715 Btu/lbm·R (655.6 R)W = 112.3 Btu/lbmThe change in internal energy is also 112.3 Btu/lbm, since Q = 0. The change in temperature is T2 - T1 = 1183.3 R - 527.7 R = 655.6 R.Answer: The work done by the air is 112.3 Btu/lbm, and the change in internal energy and temperature of the gas are also 112.3 Btu/lbm and 655.6 R, respectively.

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Convenience receptacles that are installed in bedrooms in a house are required to be protected by a arc-fault circuit interrupter. (Option A)

Arc-fault circuit interrupters (AFCIs) are designed to detect dangerous electrical arcs in wiring and shut off power before they can cause a fire. AFCIs are required by the National Electrical Code (NEC) to be installed in certain areas of the home, including bedrooms, to provide increased electrical safety.


Ground-fault circuit interrupters (GFCIs) are also required in areas where water and electricity may come into contact, such as bathrooms, kitchens, and outdoor outlets, to protect against electric shock. Surge suppressors and surface-mounted incandescent luminaires are not required for bedroom receptacles.

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A true statement about the seal between the respirator and the wearer's face is: D. facial hair can prevent a respirator from properly sealing to the wearer’s face.

A respirator can be defined as a personal protective equipment that is typically worn over the face to protect the mouth and nose, while protecting the wearer from inhaling smoke, dust, or other toxic chemical substances.

In this context, a true statement about the seal between the respirator and the wearer's face is that facial hair has the ability to prevent a respirator from properly sealing to the wearer’s face.

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Answer:

D. the service technician

Explanation:

hope it's help

Answer:

c) the owner

Explanation:

this is because when there is no owner there is no manager

1. Scalability: Cloud computing offers the advantage of scalability, allowing businesses to easily scale their computing resources up or down based on their needs.

With cloud services, businesses can quickly adapt to changing demands, whether it's handling high traffic during peak periods or scaling down during periods of low activity. This flexibility eliminates the need for businesses to invest in expensive hardware or infrastructure that may remain underutilized.

2. Cost Efficiency: Cloud computing offers cost efficiency by reducing the need for upfront investments in hardware, software, and IT infrastructure. Instead of purchasing and maintaining physical servers, businesses can leverage cloud services and pay for the resources they consume on a subscription or pay-per-use basis.

This pay-as-you-go model allows businesses to optimize their costs, as they only pay for the resources they need when they need them. Additionally, cloud services often provide automatic updates, maintenance, and support, reducing the burden on internal IT teams function and further lowering costs.

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Answer:

False

Explanation:

It is not okay to download information without asking a lab instructor, even if it is for an online class. Academic integrity policies and copyright laws still apply to online courses and students are expected to follow ethical and legal guidelines when accessing and using information. If you are unsure about whether it is appropriate to download certain information or materials, it is always best to check with your instructor.

Answer:

\(\frac{e'_z}{e_z} = 0.87142\)

Explanation:

Given:-

- The diameter of the cylinder, d = 50 mm.

- The compressive load, F = 80 KN.

Solution:-

- We will form a 3-dimensional coordinate system. The z-direction is along the axial load, and x-y plane is categorized by lateral direction.

- Next we will write down principal strains ( εx, εy, εz ) in all three directions in terms of corresponding stresses ( σx, σy, σz ). The stress-strain relationships will be used for anisotropic material with poisson ratio ( ν ).

                          εx = - [ σx - ν( σy + σz ) ] / E

                          εy = - [ σy - ν( σx + σz ) ] / E

                          εz = - [ σz - ν( σy + σx ) ] / E

- First we will investigate the "no-restraint" case. That is cylinder to expand in lateral direction as usual and contract in compressive load direction. The stresses in the x-y plane are zero because there is " no-restraint" and the lateral expansion occurs only due to compressive load in axial direction. So σy= σx = 0, the 3-D stress - strain relationships can be simplified to:

                          εx =  [ ν*σz ] / E

                          εy = [ ν*σz ] / E

                          εz = - [ σz ] / E   .... Eq 1

- The "restraint" case is a bit tricky in the sense, that first: There is a restriction in the lateral expansion. Second: The restriction is partial in nature, such, that lateral expansion is not completely restrained but reduced to half.

- We will use the strains ( simplified expressions ) evaluated in " no-restraint case " and half them. So the new lateral strains ( εx', εy' ) would be:

                         εx' = - [ σx' - ν( σy' + σz ) ] / E = 0.5*εx

                         εx' = - [ σx' - ν( σy' + σz ) ] / E =  [ ν*σz ] / 2E

                         εy' = - [ σy' - ν( σx' + σz ) ] / E = 0.5*εy

                         εx' = - [ σy' - ν( σx' + σz ) ] / E =  [ ν*σz ] / 2E

- Now, we need to visualize the "enclosure". We see that the entire x-y plane and family of planes parallel to ( z = 0 - plane ) are enclosed by the well-fitted casing. However, the axial direction is free! So, in other words the reduction in lateral expansion has to be compensated by the axial direction. And that compensatory effect is governed by induced compressive stresses ( σx', σy' ) by the fitting on the cylinderical surface.

- We will use the relationhsips developed above and determine the induced compressive stresses ( σx', σy' ).

Note:  σx' = σy', The cylinder is radially enclosed around the entire surface.

Therefore,

                        - [ σx' - ν( σx'+ σz ) ] =  [ ν*σz ] / 2

                          σx' ( 1 - v ) = [ ν*σz ] / 2

                          σx' = σy' = [ ν*σz ] / [ 2*( 1 - v ) ]

- Now use the induced stresses in ( x-y ) plane and determine the new axial strain ( εz' ):

                           εz' = - [ σz - ν( σy' + σx' ) ] / E

                           εz' = - { σz - [ ν^2*σz ] / [ 1 - v ] } / E

                          εz' = - σz*{ 1 - [ ν^2 ] / [ 1 - v ] } / E  ... Eq2

- Now take the ratio of the axial strains determined in the second case ( Eq2 ) to the first case ( Eq1 ) as follows:

                            \(\frac{e'_z}{e_z} = \frac{- \frac{s_z}{E} * [ 1 - \frac{v^2}{1 - v} ] }{-\frac{s_z}{E}} \\\\\frac{e'_z}{e_z} = [ 1 - \frac{v^2}{1 - v} ] = [ 1 - \frac{0.3^2}{1 - 0.3} ] \\\\\frac{e'_z}{e_z} = 0.87142\)... Answer

Answer:

2.5=1500/Whp=> Whp=600 kWh

delWgain=1500-600=900 kWh

Money saved= 900* 6tk*=5400 tk

Ew Wakefield Hospital has only one portable X-ray machine. The emergency room staff claim to have the greatest need for the machine, but the surgeons in the operating room demand ready access to the machine. The conflict between these two groups is a result of scarcity.

Science and technology are the driving forces behind today's modern medicine, allowing us to identify and treat a range of medical problems. X-ray technology has had a significant impact on medicine, and it is commonly used to diagnose various diseases, making it an essential tool for hospitals.

Despite the advantages, the scarcity of portable X-ray machines creates difficulties for hospital employees, including a dispute between the emergency room and the operating room staff at Ew Wakefield Hospital. There is only one portable X-ray machine available at Ew Wakefield Hospital.

The conflict between the emergency room and the operating room staff is a result of scarcity. Both departments require ready access to the X-ray machine. Scarcity can generate rivalry, competition, and conflict when people and organizations compete for the same resource.

The staff's conflict is a direct result of the lack of accessibility to the X-ray machine. Thus, scarcity can be a significant factor contributing to interpersonal conflict.

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Answer:

the % recovery of aluminum product is 80.5%

the % purity of the aluminum product is 54.7%

Explanation:

feed rate to separator = 2500 kg/hr

in one hour, there will be 2500 kg/hr x 1 hr = 2500 kg of material is fed into the  machine

of this 2500 kg, the feed is known to contain 174 kg of aluminium and 2326 kg of rejects.

After the separation, 256 kg  is collected in the product stream.

of this 256 kg, 140 kg is aluminium.

% recovery of aluminium will be = mass of aluminium in material collected in the product stream ÷ mass of aluminium contained in the feed material

% recovery of aluminium = 140kg/174kg x 100% = 80.5%

% purity of the aluminium product = mass of aluminium in final product ÷ total mass of product collected in product stream

% purity of the aluminium product = 140kg/256kg

x 100% = 54.7%