For a steel alloy it has been determined that a carburizing heat treatment of 9-h duration will raise the carbon concentration to 0.38 wt% at a point 1.2 mm from the surface. Estimate the time (in h) necessary to achieve the same concentration at a 6.4 mm position for an identical steel and at the same carburizing temperature.

Answer:

True, That is correct. Soil removed from an excavation site is indeed called spoil.

Spoil definition: The waste material (such as soil) brought up during the course of an excavation.

Hope I helped! Sorry if not. Have a wonderful week and follow me for more help! Remember your worth and love yourself! Adíos! ;D

Yes, that is true. It is true that spoil refers to soil excavated from an excavation site.

Thus,  The debris that is dug up during an excavation, such as soil. Depending on the location and the type of excavation, the composition of the spoil material can change.

It could consist of dirt, gravel, boulders, debris, or other substances that were removed or moved during the excavation process.

Typically, the spoil is put aside and used for a variety of tasks, including backfilling, grading, or disposal, as necessary for the project and permitted by local laws.

Thus, Yes, that is true. It is true that spoil refers to soil excavated from an excavation site.

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Answer:

c   Waste stack

Explanation:

Answer:

Electrical engineering is an engineering discipline concerned with the study, design and application of equipment, devices and systems which use electricity, electronics, and electromagnetism.

Explanation:

Answer:

Attached below is the derived next state equations

Explanation:

Attached below is the derived next state equations

used for the solution of the given problem.

Answer:

Explanation:

Given that:

[tex]y = \int^t_og'(t-s) f(s) ds \ \text{is solution to } \ my"ky= f(t)[/tex]

where;

[tex]g'(0) = \dfrac{1}{m}[/tex]     and [tex]mg"+kg = 0[/tex]

[tex]\text{Using Leibniz Formula to prove the above equation:}[/tex]

[tex]\dfrac{d}{dt} \int ^{b(t)}_{a(t)} \ f (t,s) \ ds = f(t,b(t) ) * \dfrac{d}{dt}b(t) - f(t,a(t)) *\dfrac{d}{dt}a(t) + \int ^{b(t)}_{a(t)}\dfrac{\partial}{\partial t} f(t,s) \ dt[/tex]

So, [tex]y = \int ^t_0 g' (t-s) f(s) \ ds[/tex]

[tex]\text{By differentiation with respect to t;}[/tex]

[tex]y' = g'(o) f(t) \dfrac{d}{dt}t- 0 + \int^{t}_{0}g'' (t-s) f(s) ds \\ \\ y' = \dfrac{1}{m}f(t) + \int ^t_0 g'' (ts) f(s) \ ds[/tex]

[tex]y'' = \dfrac{1}{m} f'(t) + g"(0) f(t) + \int^t_o g"'(t-s) f(s)ds --- (1)[/tex]

[tex]Since \ \ mg" (t) +kg (t) = 0 \\ \\ \implies g" (t) = -\dfrac{k}{m} g(t) --- (111) \\ \\ put \ t \ =0 \ we \ get;\\g" (0) = - \dfrac{k}{m } g(0) \\ \\ g"(0) = 0 \ \ \ \ ( because \ g(0) =0) \\ \\[/tex]

[tex]Now \ differentiating \ equation (111) \ with \ respect \ to \ t \\ \\ g"'(t) = -\dfrac{k}{m}g(t) \\ \\ replacing \ it \ into \ equation \ (1) \\ \\ y" = \dfrac{1}{m}f' (t) + 0 + \int ^t_o \dfrac{-k}{m}g' (t-s) f(s) \ ds \\ \\ y" = \dfrac{1}{m}f' (t) - \dfrac{k}{m} \int ^t_o g' (t-s) \ f(s) \ ds \\ \\ y" = \dfrac{1}{m}f'(t) - \dfrac{k}{m}y \\ \\ my" = f'(t)-ky \\ \\ \implies \mathbf{ my" +ky = f'(t)}[/tex]

Answer:

C Electrical Connections

Explanation:

In reading says . However, electrical

connections aren’t shown in construction drawings.

Answer:

Magnitude of force P = 25715.1517 N

Explanation:

Given - The wires each have a diameter of 12 mm, length of 0.6 m, and are made from 304 stainless steel.

To find - Determine the magnitude of force P so that the rigid beam tilts 0.015∘.

Proof -

Given that,

Diameter = 12 mm = 0.012 m

Length = 0.6 m

[tex]\theta[/tex] = 0.015°

Youngs modulus of elasticity of 34 stainless steel is 193 GPa

Now,

By applying the conditions of equilibrium, we have

∑fₓ = 0, ∑[tex]f_{y}[/tex] = 0, ∑M = 0

If ∑[tex]M_{A}[/tex] = 0

⇒[tex]F_{BC}[/tex]×0.9 - P × 0.6 = 0

⇒[tex]F_{BC}[/tex]×3 - P × 2 = 0

⇒[tex]F_{BC}[/tex] = [tex]\frac{2P}{3}[/tex]

If ∑[tex]M_{B}[/tex] = 0

⇒[tex]F_{AD}[/tex]×0.9 = P × 0.3

⇒[tex]F_{AD}[/tex] ×3 = P

⇒[tex]F_{AD}[/tex] = [tex]\frac{P}{3}[/tex]

Now,

Area, A = [tex]\frac{\pi }{4} X (0.012)^{2}[/tex] = 1.3097 × 10⁻⁴ m²

We know that,

Change in Length , [tex]\delta[/tex] = [tex]\frac{P l}{A E}[/tex]

Now,

[tex]\delta_{AD} = \frac{P(0.6)}{3(1.3097)(10^{-4}) (193)(10^{9} }[/tex] = 9.1626 × 10⁻⁹ P

[tex]\delta_{BC} = \frac{2P(0.6)}{3(1.3097)(10^{-4}) (193)(10^{9} }[/tex] = 1.83253 × 10⁻⁸ P

Given that,

[tex]\theta[/tex] = 0.015°

⇒[tex]\theta[/tex] = 2.618 × 10⁻⁴ rad

So,

[tex]\theta = \frac{\delta_{BC} - \delta_{AD}}{0.9}[/tex]

⇒2.618 × 10⁻⁴ = (  1.83253 × 10⁻⁸ P - 9.1626 × 10⁻⁹ P) / 0.9

⇒P = 25715.1517 N

∴ we get

Magnitude of force P = 25715.1517 N

The magnitude of Force P is; P = 25715.15 N

If we draw a free body diagram of the rigid beam system, then for beam AB we can take moments in the following manner;

Taking moments about point A, we have;

(F_bc * 0.9) - P(0.6) = 0

F_bc = ²/₃P

Taking moments about B gives;

P(0.3) - F_ad * 0.9 = 0

F_ad = ¹/₃P

Normal stress for BC is;

σ_bc = F_bc/A_bc

σ_bc = (²/₃P)/(π * 0.006²)

σ_bc = (²/₃P)/(1.131 × 10⁻⁴) N/m²

σ_ad = (¹/₃P)/(π * 0.006²)

σ_ad = (¹/₃P)/(1.131 × 10⁻⁴) N/m²

We know that;

Elongation is; ΔL = PL/AE = (P/A) * (L/E)

Where E for 304 stainless steel is 193 GPa = 193 × 10⁹ Pa

Thus;

ΔL_bc =  (²/₃P)/(1.131 × 10⁻⁴) * (0.6/(193 × 10⁹))

ΔL_bc = 1.83253P × 10⁻⁸

Likewise;

ΔL_ad = (¹/₃P)/(1.131 × 10⁻⁴) * (0.6/(193 × 10⁹))

ΔL_ad = 9.1626P × 10⁻⁹ m

Converting the beam tilt angle from degrees to radians gives;

θ = 0.015° = 0.00026179939 rads

Using small angle analysis, we can say that;

θ = (ΔL_bc - ΔL_ad)/36

θ = P((1.83253P × 10⁻⁸) - (9.1626P × 10⁻⁹))/36

Solving gives P = 25715.15 N

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Answer:

I would say Firewall security. Malware is pretty corrupting, but the firewall I think is good for protection if you have a good one.

Explanation:

Answer:

Explanation:

This is Answer....

Answer:

Die of intoxication by water first

Explanation:

We assume that the weight of the man is 154.35 pounds which is 70 kg

LD50 water = 90g per kg

Maximum concentration = 90x70

= 6300grams

Convert grams to liters

6300/100

= 6.3 litres

From here we get amount of kool aid

6.3 x 1.97x10^-4

= 1.24x10^-3

= 1.24grams

1.24 grams is below 420 kool aid is lower than LD50 with about 6 grams for 1 kg (6x70kg = 420). So 420 is lethal dose. But 1.24 is less than this so the man has to die of water intoxication first.

Answer:

a) 75%

b) 82%

Explanation:

Assumptions:

[tex]\text{The mechanical energy for water at turbine exit is negligible.} \\ \\ \text{The elevation of the lake remains constant.}[/tex]

Properties: The density of water [tex]\delta = 1000 kg/m^3[/tex]

Conversions:

[tex]165 \ ft \ to \ meters = 50 m \\ \\7000 \ lbm/s \ to \ kilogram/sec = 3175 kg/s \\ \\1564 \ hp \ to \ kilowatt = 1166 kw \\ \\[/tex]

Analysis:

Note that the bottom of the lake is the reference level. The potential energy of water at the surface becomes gh. Consider that kinetic energy of water at the lake surface & the turbine exit is negligible and the pressure at both locations is the atmospheric pressure and change in the mechanical energy of water between lake surface & turbine exit are:

[tex]e_{mech_{in}} - e_{mech_{out}} = gh - 0[/tex]

Then;

[tex]gh = (9.8 m/s^2) (50 m) \times \dfrac{1 \ kJ/kg}{1000 m^2/s^2}[/tex]

gh = 0.491 kJ/kg

[tex]\Delta E_{mech \ fluid} = m(e_{mech_{in}} - e_{mech_{out}} ) \\ \\ = 3175 kg/s \times 0.491 kJ/kg[/tex]

= 1559 kW

Therefore; the overall efficiency is:

[tex]\eta _{overall} = \eta_{turbine- generator} = \dfrac{W_{elect\ out}}{\Delta E_{mech \fluid}}[/tex]

[tex]= \dfrac{1166 \ kW}{1559 \ kW}[/tex]

= 0.75

= 75%

b) mechanical efficiency of the turbine:

[tex]\eta_{turbine- generator} = \eta_{turbine}\times \eta_{generator}[/tex]

thus;

[tex]\eta_{turbine} = \dfrac{\eta_{[turbine- generator]} }{\eta_{generator}} \\ \\ \eta_{turbine} = \dfrac{0.75}{0.92} \\ \\ \eta_{turbine} = 0.82 \\ \\ \eta_{turbine} = 82\%[/tex]

Answer:

Explanation:

From the information given:

[tex]E_1 = 68 \ GPa \\ \\ E_2 = 201 \ GPa \\ \\ d = 25 \ mm \ \\ \\ D = 45 \ mm \ \\ \\ L = 761 \ mm \\ \\ P = -88 kN[/tex]

The total load is distributed across both the rod and tube:

[tex]P = P_1+P_2 --- (1)[/tex]

Since this is a composite column; the elongation of both aluminum rod & steel tube is equal.

[tex]\delta_1=\delta_2[/tex]

[tex]\dfrac{P_1L}{A_1E_1}= \dfrac{P_2L}{A_2E_2}[/tex]

[tex]\dfrac{P_1 \times 0.761}{(\dfrac{\pi}{4}\times .0025^2 ) \times 68\times 10^4}= \dfrac{P_2\times 0.761}{(\dfrac{\pi}{4}\times (0.045^2-0.025^2))\times 201 \times 10^9}[/tex]

[tex]P_1(2.27984775\times 10^{-8}) = P_2(3.44326686\times 10^{-9})[/tex]

[tex]P_2 = \dfrac{ (2.27984775\times 10^{-8}) P_1}{(3.44326686\times 10^{-9})}[/tex]

[tex]P_2 = 6.6212 \ P_1[/tex]

Replace [tex]P_2[/tex] into equation (1)

[tex]P= P_1 + 6.6212 \ P_1\\ \\ P= 7.6212\ P_1 \\ \\ -88 = 7.6212 \ P_1 \\ \\ P_1 = \dfrac{-88}{7.6212} \\ \\ P_1 = -11.547 \ kN[/tex]

Finally, to determine the normal stress in aluminum rod:

[tex]\sigma _1 = \dfrac{P_1}{A_1} \\ \\ \sigma _1 = \dfrac{-11.547 \times 10^3}{\dfrac{\pi}{4} \times 25^2}[/tex]

[tex]\sigma_1 = - 23.523 \ MPa}[/tex]

Thus, the normal stress = 23.523 MPa in compression.

Answer:

a) The maximum flowrate of the pump is approximately 13,305.22 cm³/s

b) The pressure difference across the pump is approximately 293.118 kPa

Explanation:

The efficiency of the pump = 78%

The power of the pump = 5 -kW

The height of the pool above the underground water, h = 30 m

The diameter of the pipe on the intake side = 7 cm

The diameter of the pipe on the discharge side = 5 cm

a) The maximum flowrate of the pump is given as follows;

[tex]P = \dfrac{Q \cdot \rho \cdot g\cdot h}{\eta_t}[/tex]

Where;

P = The power of the pump

Q = The flowrate of the pump

ρ = The density of the fluid = 997 kg/m³

h = The head of the pump = 30 m

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

[tex]\eta_t[/tex] = The efficiency of the pump = 78%

[tex]\therefore Q_{max} = \dfrac{P \cdot \eta_t}{\rho \cdot g\cdot h}[/tex]

[tex]Q_{max}[/tex] = 5,000 × 0.78/(997 × 9.8 × 30) ≈ 0.0133 m³/s

The maximum flowrate of the pump [tex]Q_{max}[/tex] ≈ 0.013305 m³/s = 13,305.22 cm³/s

b) The pressure difference across the pump, ΔP = ρ·g·h

∴ ΔP = 997 kg/m³ × 9.8 m/s² × 30 m = 293.118 kPa

The pressure difference across the pump, ΔP ≈ 293.118 kPa

Answer: hello attached below is the properly written chain reaction  to your question

answer :

d[NO] / dt = [tex]k_{1f} [O] [N_{2}] + K_{2f} [N][O_{2} ] + K_{3f}[N][OH][/tex]

d[N] / dt = [tex]k_{1f} [O] [N_{2}] + K_{2f} [N][O_{2} ] - K_{3f}[N][OH][/tex]

Explanation:

write out expressions for  d[NO] / dt and d[N] / dt

Given :

properly written chain reaction ( attached below)

Expression for d[NO] / dt can be written as

[tex]k_{1f} [O] [N_{2}] + K_{2f} [N][O_{2} ] + K_{3f}[N][OH][/tex]

Expression for d[N] / dt can be written as

[tex]k_{1f} [O] [N_{2}] + K_{2f} [N][O_{2} ] - K_{3f}[N][OH][/tex]

Answer:

Distribution (or its more sophisticated counterpart, supply chain management) can add value to goods and services by making them more easily and conveniently available to consumers. ... This means that you need good wholesalers and good transportation systems to get your products to the retailers.

Explanation:

Distribution (or its more sophisticated counterpart, supply chain management) can add value to goods and services by making them more easily and conveniently available to consumers. ... This means that you need good wholesalers and good transportation systems to get your products to the retailers.

Answer:

Distribution (or its more sophisticated counterpart, supply chain management) can add value to goods and services by making them more easily and conveniently available to consumers.

Explanation:

hope it helps <33

Answer:

a

Explanation:

Answer:

El =50 degre

Explanation:

Yan ang answer

Answer:

1.933 KN-M

Explanation:

Determine the largest permissible bending moment when the composite bar is bent  horizontally

Given data :

modulus of elasticity of steel = 200 GPa

modulus of elasticity of aluminum = 75 GPa

Allowable stress for steel = 220 MPa

Allowable stress for Aluminum = 100 MPa

a = 10 mm

First step

determine moment of resistance when steel reaches its max permissible stress

next : determine moment of resistance when Aluminum reaches its max permissible stress

Finally Largest permissible bending moment of the composite Bar = 1.933 KN-M

attached below is a detailed solution

Answer: hello your question lacks the required diagram attached below is the diagram

answer :  29528.1  N/m^2

Explanation:

Given data :

dimensions of tank :

Length = 5-m

Width = 4-m

Depth = 2.5-m

acceleration of tank = 2m/s^2

Determine the maximum gage pressure in the tank

Pa ( pressure at point A )  = s*g*h1

    = 10^3 * 9.81 * 3.01

    = 29528.1  N/m^2

attached below is the remaining part of the solution

Answer:

a) the inductance of the coil is 6 mH

b) the emf generated in the coil is 18 mV  

Explanation:

Given the data in the question;

N = 570 turns

diameter of tube d = 8.10 cm = 0.081 m

length of the wire-wrapped portion l =  35.0 cm = 0.35 m

a) the inductance of the coil (in mH)

inductance of solenoid

L = N²μA / l

A = πd²/4  

so

L = N²μ(πd²/4) / l

L = N²μ(πd²) / 4l

we know that μ = 4π × 10⁻⁷ TmA⁻¹

we substitute

L = [(570)² × 4π × 10⁻⁷× ( π × (0.081)² )] / 4(0.35)

L =  0.00841549 / 1.4

L = 6 × 10⁻³ H    

L = 6 × 10⁻³ × 1000 mH

L = 6 mH

Therefore, the inductance of the coil is 6 mH

b)

Emf ( ∈ ) = L di/dt

given that; di/dt = 3.00 A/sec

{∴ di = 3 - 0 = 3 and dt = 1 sec}

Emf ( ∈ ) = L di/dt

we substitute

⇒ 6 × 10⁻³ ( 3/1 )

= 18 × 10⁻³ V

= 18 × 10⁻³ × 1000

= 18 mV  

Therefore, the emf generated in the coil is 18 mV  

Answer:

Explanation:

The missing part of the question is attached in the diagram below, the second diagram shows the schematic view of the stress-strain curve and the plane stress.

From the given information:

The elastic modulus is:

[tex]E = \dfrac{\sigma}{\varepsilon} \\ \\ E = \dfrac{150 \ MPa}{0.0217} \\ \\ E = 69.124 \ GPa[/tex]

Hence, suppose 0.2% offset cuts the stress-strain curve at a designated point A from the image attached below, then the yield strength relating to the stress axis from the curve will be [tex]\sigma_y[/tex] = 270 MPa.

The shear yield strength by using von Mises criteria is estimated as;

[tex]\tau_1 = \dfrac{\sqrt{2}}{3}\sigma_y \\ \\ \tau_1 = \dfrac{\sqrt{2}}{3}*270 \\ \\ \tau_1 = 127.28 \ MPa[/tex]

The shear yield strength by using Tresca criteria is:

[tex]\tau_2 = \dfrac{1}{2}\sigma_y \\ \\ \tau_2= \dfrac{1}{2}*270 \\ \\ \tau_2 = 135 \ MPa[/tex]

Answer:

Explanation:

Crystalline melting temperature (Tm) is termed as the temperature required for a  crystalline polymer to change to a fluid or glasslike crystalline spaces of a semi-crystalline polymer liquefy (expanded sub-atomic movement).  

Crystallization of polymers is an interaction related with incomplete arrangement of their atomic and molecular chains. These chains crease together and structure requested districts called lamellae, which form bigger spheroidal designs named spherulites. Polymers can solidify after cooling from melting, mechanical extending, or dissolvable dissipation. Crystallization influences the optical, mechanical, and synthetic chemical properties of the polymer.  

For a crystalline polymer, a required polymer chain is present in or goes along a few crystalline and amorphous zones. The crystalline zones are comprised of intermolecular & intramolecular arrangements or deliberate and thus firmly stuffed plan of atoms or chain fragments, and an absence of it brings about the development of amorphous zones.  

The mechanical property boundary, for example, shear modulus expansions in the temperature of perception for polymer material framework.  

The temperature reaction of direct linear polymers might be seen as partitioned into three particularly separate fragments:  

1. Above Tm: The polymer stays as fluid whose consistency & viscosity would rely upon atomic molecular weight and temperature.  

2. Between Tm and Tg: This area may go between close to 100% crystalline & 100% amorphous chain atomic bunches relying upon the polymer underlying consistency. The amorphous part carries on similar to supercooled fluid in this section. The generally actual conduct of the polymer in this moderate portion is similar to an elastic rubber.  

3.Below Tg: The polymer material saw as glass is hard and inflexible, showing and emanating a predetermined coefficient of thermal extension. The glass is more like a crystalline strong than the fluid in personal conduct standard regarding mechanical property boundaries. In regard to the molecular atomic request, in any case, the glass all the more intently takes after the fluid. There is little contrast between the direct linear and cross-connected polymer beneath Tg.

Where are your answer options?

Answer:

Implement a hazard communication program

Explanation: i took the quiz

Answer:

3.52×10⁶ atoms of Si and 7.05×10⁶ atoms of O

Explanation:

It is all about unit conversions.

The area of our MOS is 1 mm². So, we know that the thickness is 160 nm. This data can give us the volume. We convert nm to mm.

160 nm . 1×10⁻⁶ mm /1nm = 1.6×10⁻⁴ mm

By the way, now we can determine the volume of MOS, in order to work with density.

1.6×10⁻⁴ mm . 1 mm² = 1.6×10⁻⁴ mm³

But density is mg/m³, so we convert mm³ to m³

1.6×10⁻⁴ mm³ . 1×10⁻⁹ m³/mm³ = 1.6×10⁻¹³ m³

Now, we apply density to determine the mass of MOS

Density = mass /volume → Density . volume = mass

1.6×10⁻¹³ m³ . 2.20mg/m³ = 3.52×10⁻¹³ mg

To make more easier the calculate, we convert mg to g.

3.52×10⁻¹³ mg . 1g /1000mg = 3.52×10⁻¹⁶ g

To count the atoms, we determine molar mass of SiO₂ → 60.08 g/mol

We need to know moles of Si and O₂ in the MOS

Firstly, we determine amount of MOS: 3.52×10⁻¹⁶ g / 60.08 g/mol = 5.86×10⁻¹⁸ moles

1 mol of SiO₂ has 1 mol of Si and 2 mol of O so:

5.86×10⁻¹⁸ mol of SiO₂ may have:

(5.86×10⁻¹⁸ . 1) /1 = 5.86×10⁻¹⁸ moles of Si

(5.86×10⁻¹⁸ .2) /1 =  1.17×10⁻¹⁷ moles of O₂

Let's count the atoms (1 mol of anything contain NA particles)

5.86×10⁻¹⁸ mol of Si . 6.02×10²³ atoms/ mol = 3.52×10⁶ atoms of Si

1.17×10⁻¹⁷ mol of O₂  . 6.02×10²³ atoms/ mol = 7.05×10⁶ atoms of O

The number of Si and O atoms present per mm² of the oxide layer are respectively; 3.52 × 10⁶ atoms of Si and 7.05×10⁶ atoms of O

We are given;

Area of MOS device; A = 1 mm²

Thickness of layer; t = 160 nm = 1.6 × 10⁻⁴ mm

Formula for volume is;

V = Area * thickness

V = 1.6 × 10⁻⁴ mm × 1 mm² = 1.6 × 10⁻⁴ mm³

Converting volume to m³ gives;

V = 1.6 × 10⁻¹³ m³

Now, to get the mass of MOS, we will use the formula;

Mass = Density * volume

we are given density = 2.20mg/m³

Thus;

Mass = 1.6 × 10⁻¹³ m³ × 2.20mg/m³

Mass = 3.52 × 10⁻¹³ mg = 3.52×10⁻¹⁶ g

From periodic table, the molar mass of SiO₂ = 60.08 g/mol

Then;

Number of moles of MOS = (3.52 × 10⁻¹⁶ g)/60.08 g/mol

Number of moles of MOS = 5.86 × 10⁻¹⁸ moles

Now, 1 mol of SiO₂  is formed from 1 mol of Si and 2 mol of O. Thus;

5.86×10⁻¹⁸ mol of SiO₂ will have;

5.86×10⁻¹⁸ moles of Si

and (5.86×10⁻¹⁸ .2) =  11.72 × 10⁻¹⁸ moles of O₂

From Avogadro's number that 1 mol equals 6.02×10²³ atoms , we can say that;

5.86 × 10⁻¹⁸ mol of Si × 6.02 × 10²³ atoms/mol = 3.52 × 10⁶ atoms of Si

11.72 × 10⁻¹⁸ mol of O₂ × 6.02 × 10²³ atoms/mol = 7.05×10⁶ atoms of O

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Answer:

it is indeed C

Explanation:

Answer:

c

Explanation: