If these components have weights WA = 50000 lb , WB=8000lb, and WC=6000lb, determine the normal reactions of the wheels D, E, and F on the ground.

Answer:

B. restricting the dislocation motion

Explanation:

Solid solution strengthening is a type of alloying that is carried out by the addition of the atoms of the element used for the alloying to the crystallized lattice structure of the base metal, which the metal that would be strengthened. The purpose of this act is to increase the strength of metals. It actually works by impeding or restricting the motion in the crystal lattice structure of metals thus making them more difficult to deform.

The solute atoms used for strengthening could be interstitial or substitutional. The interstitial solute atoms work by moving in between the space in the atoms of the base metal while the substitutional solute atoms make a replacement with the solvent atoms in the base metal.

It is to be noted that a hypertonic solution have the capacity to make cells to shrink.

In a hypertonic solution, the concentration of solutes (e.g., salts, sugars) outside the cell is higher than inside the cell.

As a result, water moves out of the cell through osmosis, trying to equalize the concentration, causing the cell to lose water and shrink.

This process is commonly observed in biology when examining the effect of different solutions on cells, such as in red blood cells or plant cells.

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Answer:

false

Explanation:

the changing of a prisoner sentence or another penalty to another less severe

Explanation:

(a) Aluminum alloys are generally not viable as lightweight structural materials in humid environments because they are highly susceptible to corrosion by water vapor.

False, aluminium is not susceptible to any corrosion by the presence of water vapor.

(b) Aluminum alloys are generally superior to pure aluminum, in terms of yield strength, because their micro structures often contain precipitate phases that strain the lattice, thereby hardening the alloy relative to pure aluminum.

True.

(c) Aluminum is not very workable at high temperatures in air, in terms of extrusion and rolling, because a non-protective oxide grows and consumes the metal, converting it to a hard and brittle ceramic.

False, aluminium is stable at high temperatures and does not oxidizes.

(d) Compared to most other metals, like steel, pure aluminum is very resistant to creep deformation.

False,pure aluminium is not resistant to the creep deformation.

(e) The relatively low melting point of aluminum is often considered a significant limitation for high-temperature structural applications.

False.

In this exercise, we have to analyze the statements that deal with aluminum and its properties, thus classifying it as true or false:

A) False

B) True

C) False

D) False

E) True

Analyzing the statements we can classify them as:

(a) For this statement we can say that it is False, aluminium is not susceptible to any corrosion by the presence of water vapor.

(b) For this statement we can say that it is True.

(c) For this statement we can say that it is False, aluminium is stable at high temperatures and does not oxidizes.

(d) For this statement we can say that it is False, pure aluminium is not resistant to the creep deformation.

(e) For this statement we can say that it is True.

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Answer:

c. stochastic.

Explanation:

A stochastic model is a tool in statistics, used to estimate the probability distributions of intended outcomes by the allowance of random variation in any number of the inputs over time. For a stochastic model, Inputs to a quantitative model are uncertain, and the value of the output from a stochastic model cannot be easily determined, even if the value of the input that can be determined is known. The distributions of the resulting outcomes of a stochastic model is usually due to the large number of simulations involved, and it is widely used as a statistical tool in the life sciences.

Answer:

The volume percentage of graphite is 10.197 per cent.

Explanation:

The volume percent of graphite is the ratio of the volume occupied by the graphite phase to the volume occupied by the graphite and ferrite phases. The weight percent in the cast iron is 3.2 wt% (graphite) and 96.8 wt% (ferrite). The volume percentage of graphite is:

[tex]\%V_{gr} = \frac{V_{gr}}{V_{gr}+V_{fe}} \times 100\,\%[/tex]

Where:

[tex]V_{gr}[/tex] - Volume occupied by the graphite phase, measured in cubic centimeters.

[tex]V_{fe}[/tex] - Volume occupied by the graphite phase, measured in cubic centimeters.

The expression is expanded by using the definition of density and subsequently simplified:

[tex]\%V_{gr} = \frac{\frac{m_{gr}}{\rho_{gr}} }{\frac{m_{gr}}{\rho_{gr}}+\frac{m_{fe}}{\rho_{fe}}}\times 100\,\%[/tex]

Where:

[tex]m_{fe}[/tex], [tex]m_{gr}[/tex] - Masses of the ferrite and graphite phases, measured in grams.

[tex]\rho_{fe}, \rho_{gr}[/tex] - Densities of the ferrite and graphite phases, measured in grams per cubic centimeter.

[tex]\%V_{gr} = \frac{1}{1+\frac{\frac{m_{fe}}{\rho_{fe}} }{\frac{m_{gr}}{\rho_{gr}} } }\times 100\,\%[/tex]

[tex]\%V_{gr} = \frac{1}{1 + \left(\frac{\rho_{gr}}{\rho_{fe}} \right)\cdot\left(\frac{m_{fe}}{m_{gr}} \right)} \times 100\,\%[/tex]

If [tex]\rho_{gr} = 2.3\,\frac{g}{cm^{3}}[/tex], [tex]\rho_{fe} = 7.9\,\frac{g}{cm^{3}}[/tex], [tex]m_{gr} = 3.2\,g[/tex] and [tex]m_{fe} = 96.8\,g[/tex], the volume percentage of graphite is:

[tex]\%V_{gr} = \frac{1}{1+\left(\frac{2.3\,\frac{g}{cm^{3}} }{7.9\,\frac{g}{cm^{3}} } \right)\cdot \left(\frac{96.8\,g}{3.2\,g} \right)} \times 100\,\%[/tex]

[tex]\%V_{gr} = 10.197\,\%V[/tex]

The volume percentage of graphite is 10.197 per cent.

Following are the solution to the given points:

[tex]\to C_{Gr} = 100\\\\ \to C_{\alpha}= 0[/tex]From [tex]Fe-F_{\frac{e}{3}} c[/tex] diagram.  

[tex]\to W_{\alpha} =\frac{C_{Gr}-C_{o}}{C_{Gr}-C_{\alpha}}[/tex]

           [tex]= \frac{100-3.6}{100-0} \\\\= \frac{100-3.6}{100} \\\\= \frac{96.4}{100} \\\\=0.964[/tex]

Calculating the weight fraction of graphite:  

[tex]\to W_{Gr}=\frac{C_0 - c_d}{C_{Gr} -c_d}[/tex]

            [tex]= \frac{3.6-0}{100-0} \\\\ = \frac{3.6}{100} \\\\= 0.036[/tex]

Calculating the volume percent of graphite:

[tex]\to V_{Gr}=\frac{\frac{W_{Gr}}{P_{Gr}}}{\frac{w_{\alpha}}{P_{\alpha}}+ \frac{W_{Gr}}{P_{Gr}}}[/tex]

           [tex]=\frac{\frac{0.036}{2.3}}{\frac{0.964}{7.9}+\frac{0.036}{2.3}}\\\\=0.11368 \times 100\%\\\\=11.368\%[/tex]

Therefore, the final answer is "0.964, 0.036, and 11.368%"

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Answer:

point B where [tex]Q_B = 101.25 \ in^3[/tex]  has the largest Q value at section a–a

Explanation:

The missing diagram that is suppose to be attached to this question can be found in the attached file below.

So from the given information ;we are to determine the  point that  has the largest Q value at section a–a

In order to do that; we will work hand in hand with the image attached below.

From the image attached ; we will realize that there are 8 blocks aligned on top on another in the R.H.S of the image with the total of 12 in; meaning that each block contains 1.5 in each.

We also have block partitioned into different point segments . i,e A,B,C, D

For point A ;

Let Q be the moment of the Area A;

SO ; [tex]Q_A = Area \times y_1[/tex]

where ;

[tex]y_1 = (6 - \dfrac{1.5}{2})[/tex]

[tex]y_1 = (6- 0.75)[/tex]

[tex]y_1 = 5.25 \ in[/tex]

[tex]Q_A =(L \times B) \times y_1[/tex]

[tex]Q_A =(6 \times 1.5) \times 5.25[/tex]

[tex]Q_A =47.25 \ in^3[/tex]

For point B ;

Let Q be the moment of the Area B;

SO ; [tex]Q_B = Area \times y_2[/tex]

where ;

[tex]y_2 = (6 - \dfrac{1.5 \times 3}{2})[/tex]

[tex]y_2= (6 - \dfrac{4.5}{2}})[/tex]

[tex]y_2 = (6 -2.25})[/tex]

[tex]y_2 = 3.75 \ in[/tex]

[tex]Q_B =(L \times B) \times y_1[/tex]

[tex]Q_B=(6 \times 4.5) \times 3.75[/tex]

[tex]Q_B = 101.25 \ in^3[/tex]

For point C ;

Let Q be the moment of the Area C;

SO ; [tex]Q_C = Area \times y_3[/tex]

where ;

[tex]y_3 = (6 - \dfrac{1.5 \times 2}{2})[/tex]

[tex]y_3 = (6 - 1.5})[/tex]

[tex]y_3= 4.5 \ in[/tex]

[tex]Q_C =(L \times B) \times y_1[/tex]

[tex]Q_C =(6 \times 3) \times 4.5[/tex]

[tex]Q_C=81 \ in^3[/tex]

For point D ;

Let Q be the moment of the Area D;

SO ; [tex]Q_D = Area \times y_4[/tex]

since there is no area about point D

Area = 0

[tex]Q_D =0 \times y_4[/tex]

[tex]Q_D = 0[/tex]

Thus; from the foregoing ; point B where [tex]Q_B = 101.25 \ in^3[/tex]  has the largest Q value at section a–a

When scaffolds are now being construct or deconstruct, a competent person must supervise the work and train everybody who'll be assisting, and the further discussion can be defined as follows:

Therefore, the final answer is "Option B".

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Answer:

c) 1.75 g/cm³

Explanation:

Given that

Radii of the A ion, r(c) = 0.137 nm

Radii of the X ion, r(a) = 0.241 nm

Atomic weight of the A ion, A(c) = 22.7 g/mol

Atomic weight of the X ion, A(a) = 91.4 g/mol

Avogadro's number, N = 6.02*10^23 per mol

Solution is attached below

Answer:

a) Engineering controls.

Explanation:

Hazard can be defined as any agent or source that has the potential to constitute danger and cause harm, damage or adverse bodily injuries, health effects on a vulnerable individual, property or group of people.

Generally, can be classified into various categories such as mechanical, biological, chemical, physical, psychosocial and ergonomic hazard. These hazards are either human induced or natural.

Some examples of hazard are radiation, fire, flood, chemicals, drought, vapor or steam, exposed live wire, dust particles, electrical circuits and equipments etc.

The first choice for how to reduce or eliminate a hazard is engineering controls. Engineering controls of hazards involves the process of protecting, shielding or guarding individuals by eliminating the agent of hazards or through the use of barriers between the hazard and the vulnerable individual or group of people.

Basically, engineering controls when properly designed, maintained and used effectively would help to mitigate hazards and keep the work environment relatively safe for workers.

Examples of engineering controls are;

1. Ventilation systems.

2. Machine or equipment guards.

3. Radiation shields.

4. Safety interlocks.

5. Sound dampening equipments.

Answer:

a) In turbulent flow, the tubes with rough surfaces have much higher friction factors than the tubes with smooth surfaces.

Explanation:

Turbulent flow is a type of fluid flow in which fluid will undergo irregular fluctuations. The tubes with rough surfaces have higher friction factors than the tubes with smooth surfaces. In laminar flow the effect of effect of surface roughness is negligible on friction factors.

Answer:

The power of force F is 115.2 W

Explanation:

Use following formula

Power  = F x V

[tex]F_{H}[/tex] = F cos0

[tex]F_{H}[/tex] = (30) x 4/5

[tex]F_{H}[/tex] = 24N

Now Calculate V using following formula

V = [tex]V_{0}[/tex] + at

[tex]V_{0}[/tex] = 0

a = [tex]F_{H}[/tex] / m

a = 24N / 20 kg

a = 1.2m / [tex]S^{2}[/tex]

no place value in the formula of V

V = 0 + (1.2)(4)

V = 4.8 m/s

So,

Power = [tex]F_{H}[/tex] x V

Power = 24 x 4.8

Power = 115.2 W

Answer:

The heat flow from the composite wall is 1283.263 watts.

Explanation:

The conductive heat flow through a material, measured in watts, is represented by the following expression:

[tex]\dot Q = \frac{\Delta T}{R_{T}}[/tex]

Where:

[tex]R_{T}[/tex] - Equivalent thermal resistance, measured in Celsius degrees per watt.

[tex]\Delta T[/tex] - Temperature gradient, measured in Celsius degress.

First, the equivalent thermal resistance needs to be determined after considering the characteristics described below:

1) B and C are configurated in parallel and in series with A and D. (Section II)

2) A and D are configurated in series. (Sections I and III)

Section II

[tex]\frac{1}{R_{II}} = \frac{1}{R_{B}} + \frac{1}{R_{C}}[/tex]

[tex]\frac{1}{R_{II}} = \frac{R_{B}+R_{C}}{R_{B}\cdot R_{C}}[/tex]

[tex]R_{II} = \frac{R_{B}\cdot R_{C}}{R_{B}+R_{C}}[/tex]

Section I

[tex]R_{I} = R_{A}[/tex]

Section III

[tex]R_{III} = R_{D}[/tex]

The equivalent thermal resistance is:

[tex]R_{T} = R_{I} + R_{II}+R_{III}[/tex]

The thermal of each component is modelled by this:

[tex]R = \frac{L}{k\cdot A}[/tex]

Where:

[tex]L[/tex] - Thickness of the brick, measured in meters.

[tex]A[/tex] - Cross-section area, measured in square meters.

[tex]k[/tex] - Thermal conductivity, measured in watts per meter-Celsius degree.

If [tex]L_{A} = 0.03\,m[/tex], [tex]L_{B} = 0.08\,m[/tex], [tex]L_{C} = 0.08\,m[/tex], [tex]L_{D} = 0.05\,m[/tex], [tex]A_{A} = 0.01\,m^{2}[/tex], [tex]A_{B} = 3\times 10^{-3}\,m^{2}[/tex], [tex]A_{C} = 7\times 10^{-3}\,m^{2}[/tex], [tex]A_{D} = 0.01\,m^{2}[/tex], [tex]k_{A} = 150\,\frac{W}{m\cdot ^{\circ}C}[/tex], [tex]k_{B} = 25\,\frac{W}{m\cdot ^{\circ}C}[/tex], [tex]k_{C} = 60\,\frac{W}{m\cdot ^{\circ}C}[/tex] and [tex]k_{D} = 60\,\frac{W}{m\cdot ^{\circ}C}[/tex], then:

[tex]R_{A} = \frac{0.03\,m}{\left(150\,\frac{W}{m\cdot ^{\circ}C} \right)\cdot (0.01\,m^{2})}[/tex]

[tex]R_{A} = \frac{1}{50}\,\frac{^{\circ}C}{W}[/tex]

[tex]R_{B} = \frac{0.08\,m}{\left(25\,\frac{W}{m\cdot ^{\circ}C} \right)\cdot (3\times 10^{-3}\,m^{2})}[/tex]

[tex]R_{B} = \frac{16}{15}\,\frac{^{\circ}C}{W}[/tex]

[tex]R_{C} = \frac{0.08\,m}{\left(60\,\frac{W}{m\cdot ^{\circ}C} \right)\cdot (7\times 10^{-3}\,m^{2})}[/tex]

[tex]R_{C} = \frac{4}{21}\,\frac{^{\circ}C}{W}[/tex]

[tex]R_{D} = \frac{0.05\,m}{\left(60\,\frac{W}{m\cdot ^{\circ}C} \right)\cdot (0.01\,m^{2})}[/tex]

[tex]R_{D} = \frac{1}{12}\,\frac{^{\circ}C}{W}[/tex]

[tex]R_{I} = \frac{1}{50} \,\frac{^{\circ}C}{W}[/tex]

[tex]R_{III} = \frac{1}{12}\,\frac{^{\circ}C}{W}[/tex]

[tex]R_{II} = \frac{\left(\frac{16}{15}\,\frac{^{\circ}C}{W} \right)\cdot \left(\frac{4}{21}\,\frac{^{\circ}C}{W}\right)}{\frac{16}{15}\,\frac{^{\circ}C}{W} + \frac{4}{21}\,\frac{^{\circ}C}{W}}[/tex]

[tex]R_{II} = \frac{16}{99}\,\frac{^{\circ}C}{W}[/tex]

[tex]R_{T} = \frac{1}{50}\,\frac{^{\circ}C}{W} + \frac{16}{99}\,\frac{^{\circ}C}{W} + \frac{1}{12}\,\frac{^{\circ}C}{W}[/tex]

[tex]R_{T} = \frac{2623}{9900}\,\frac{^{\circ}C}{W}[/tex]

Now, if [tex]\Delta T = 400\,^{\circ}C - 60\,^{\circ}C = 340\,^{\circ}C[/tex] and [tex]R_{T} = \frac{2623}{9900}\,\frac{^{\circ}C}{W}[/tex], the heat flow is:

[tex]\dot Q = \frac{340\,^{\circ}C}{\frac{2623}{9900}\,\frac{^{\circ}C}{W} }[/tex]

[tex]\dot Q = 1283.263\,W[/tex]

The heat flow from the composite wall is 1283.263 watts.

Answer:

The temperature will be greater than 25°C

Explanation:

In an adiabatic process, heat is not transferred to or from the boundary of the system. The gain or loss of internal heat energy is solely from the work done on the system, or work done by the system. The work done on the system by the environment adds heat to the system, and work done by the system on its environment takes away heat from the system.

mathematically

Change in the internal energy of a system ΔU = ΔQ + ΔW

in an adiabatic process, ΔQ = 0

therefore

ΔU = ΔW

where ΔQ is the change in heat into the system

ΔW is the work done by or done on the system

when work is done on the system, it is conventionally negative, and vice versa.

also W = pΔv

where p is the pressure, and

Δv = change in volume of the system.

In this case, work is done on the gas by compressing it from an initial volume to the new volume of the cylinder. The result is that the temperature of the gas will rise above the initial temperature of 25°C

Answer:

Option A - fail/ not fail

Explanation:

For this given problem, if the yield strength is now 45 ksi, using Distortion Energy Theory the material will _fail______ and using the Maximum Shear Stress Theory the material will ___not fail_______

Answer:

1) The three possible assumptions are

a) All processes are reversible internally

b) Air, which is the working fluid circulates continuously in a closed loop

cycle

c) The process of combustion is depicted as a heat addition process

2) The diagrams are attached

5) The net work per cycle is 845.88 kJ/kg

The power developed in horsepower ≈ 45374 hP

Explanation:

1) The three possible assumptions are

a) All processes are reversible internally

b) Air, which is the working fluid circulates continuously in a closed loop

cycle

c) The process of combustion is depicted as a heat addition process

2) The diagrams are attached

5) The dimension of the cylinder bore diameter = 3.7 in. = 0.09398 m

Stroke length = 3.4 in. = 0.08636 m.

The volume of the cylinder v₁= 0.08636 ×(0.09398²)/4 = 5.99×10⁻⁴ m³

The clearance volume = 16% of cylinder volume = 0.16×5.99×10⁻⁴ m³

The clearance volume, v₂  = 9.59 × 10⁻⁵ m³

p₁ = 14.5 lbf/in.² = 99973.981 Pa

T₁ = 60 F = 288.706 K

[tex]\dfrac{T_{2}}{T_{1}} = \left (\dfrac{v_{1}}{v_{2}} \right )^{K-1}[/tex]

Otto cycle T-S diagram

T₂ = 288.706*[tex]6.25^{0.393}[/tex] = 592.984 K

The maximum temperature = T₃ = 5200 R = 2888.89 K

[tex]\dfrac{T_{3}}{T_{4}} = \left (\dfrac{v_{4}}{v_{3}} \right )^{K-1}[/tex]

T₄ = 2888.89 / [tex]6.25^{0.393}[/tex] = 1406.5 K

Work done, W = [tex]c_v[/tex]×(T₃ - T₂) - [tex]c_v[/tex]×(T₄ - T₁)

0.718×(2888.89  - 592.984) - 0.718×(1406.5 - 288.706) = 845.88 kJ/kg

The power developed in an Otto cycle = W×Cycle per second

= 845.88 × 2400 / 60  = 33,835.377 kW = 45373.99 ≈ 45374 hP.

Answer:

help me why are you an enginer

Explanation:

because lives

Answer:

False

Explanation:

Because

The effectiveness (ϵ) of a heat exchanger is defined as the ratio of the actual heat transfer to the maximum possible heat transfer.

Answer:

Domain 1: Biological (includes neuroscience, consciousness, and sensation) Domain 2: Cognitive (includes the study of perception, cognition, memory, and intelligence) Domain 3: Development (includes learning and conditioning, lifespan development, and language) i hope this helps you.

Answer:

a) [tex]h_c = 0.1599 W/m^2-K[/tex]

b) [tex]H_{loss} = 5.02 W[/tex]

c) [tex]T_s = 302 K[/tex]

d) [tex]\dot{Q} = 25.125 W[/tex]

Explanation:

Non horizontal pipe diameter, d = 25 cm = 0.25 m

Radius, r = 0.25/2 = 0.125 m

Entry temperature, T₁ = 304 + 273 = 577 K

Exit temperature, T₂ = 284 + 273 = 557 K

Ambient temperature, [tex]T_a = 25^0 C = 298 K[/tex]

Pipe length, L = 10 m

Area, A = 2πrL

A = 2π * 0.125 * 10

A = 7.855 m²

Mass flow rate,

[tex]\dot{ m} = 4.5 tons/hr\\\dot{m} = \frac{4.5*1000}{3600} = 1.25 kg/sec[/tex]

Rate of heat transfer,

[tex]\dot{Q} = \dot{m} c_p ( T_1 - T_2)\\\dot{Q} = 1.25 * 1.005 * (577 - 557)\\\dot{Q} = 25.125 W[/tex]

a) To calculate the convection coefficient relationship for heat transfer by convection:

[tex]\dot{Q} = h_c A (T_1 - T_2)\\25.125 = h_c * 7.855 * (577 - 557)\\h_c = 0.1599 W/m^2 - K[/tex]

Note that we cannot calculate the heat loss by the pipe to the environment without first calculating the surface temperature of the pipe.

c) The surface temperature of the pipe:

Smear coefficient of the pipe, [tex]k_c = 0.8[/tex]

[tex]\dot{Q} = k_c A (T_s - T_a)\\25.125 = 0.8 * 7.855 * (T_s - 298)\\T_s = 302 K[/tex]

b) Heat loss from the pipe to the environment:

[tex]H_{loss} = h_c A(T_s - T_a)\\H_{loss} = 0.1599 * 7.855( 302 - 298)\\H_{loss} = 5.02 W[/tex]

d) The required fan control power is 25.125 W as calculated earlier above

Answer:

The answer is given below

Explanation:

A 60 Hz three-phase, three-wire overhead line has solid cylindrical conductors  arranged in the form of an equilateral triangle with 4 ft conductor spacing. Conductor  diameter is 0.5 in.

Given that:

The spacing between the conductors (D) = 4 ft

1 ft = 0.3048 m

D = 4 ft = 4 × 0.3048 m = 1.2192 m

The conductor diameter = 0.5 in

Radius of conductor (r) = 0.5/2 = 0.25 in = 0.00635 m

Frequency (f) = 60 Hz

The capacitance-to-neutral is given by:

[tex]C_n=\frac{2\pi \epsilon_0}{ln(\frac{D}{r} )} =\frac{2\pi *8.854*10^{-12}}{ln(1.2192/0.00635)}=1.058*10^{-11}\ F/m[/tex]

The admittance-to-neutral is given by:

[tex]Y_n=j2\pi fC_n=j*2\pi *60*1.058*10^{-11}*\frac{1000\ m}{1\ km}=j3.989*10^{-6}\ S/km[/tex]

Answer:

R = V / I ,   R = V² / P,     R = P / I²

Explanation:

For this exercise let's use ohm's law

      V = I R

      R = V / I

Electric power is defined by

      P = V I

ohm's law

      I = V / R

we substitute

      P = V (V / R)

      P = V² / R

      R = V² / P

the third way of calculation

      P = (i R) I

      P = R I²

      R = P / I²

Answer:

Explanation:

coarse aggregate (ca) = 65%,   SG = 2.701

Fine aggregate = 35%,    SG = 2.625

A) Bulk specific gravity of aggregate

   = [tex]\frac{65*2.701 + 35*2.625}{100} = 2.674[/tex]

B) Wm = 1257.9 g { weight in air }

    Ww = 740 g { submerged weight }

   therefore Bulk specific gravity of compacted specimen

   = [tex]\frac{Wm}{Wm-Ww}[/tex]  =  [tex]\frac{1257.9}{1257.9 - 740 }[/tex]  =  2.428

   Theoretical specific gravity = 2.511

Percent stone

= 100 - asphalt content - Vv

= 100 - 5 - 3.305 = 91.695%

c) percent of void

= [tex]\frac{9.511-2.428}{2.511} * 100[/tex]    Vv = 3.305%

d) let effective specific gravity in stone

     = [tex]\frac{91.695*unstone+ 5 *1.030 }{96.695} = 2.511[/tex]

    = Instone = 2.592 effective specific gravity of stone

e) Vv = 3.305%

f ) volume filled with asphalt (Vb) = [tex]\frac{\frac{Wb}{lnb} }{\frac{Wm}{Inm} } * 100[/tex]

           Vb = [tex]\frac{5 * 2.428}{1.030 * 100} * 100[/tex]

          Vb = 11.786 %

Volume of mineral aggregate = Vb + Vv

              VMA = 11.786 + 3.305 = 15.091

g) percent void filled with alphalt

     = Vb / VMA * 100

    VMA = 11.786 + 3.305 = 15.091

   percent void filled with alphalt

     = Vb / VMA * 100 = (11.786 / 15.091) * 100 = 78.1%

Answer:

The correct answer would be a) Engineering Controls.

Explanation:

If the controls are handled correctly, you can reduce and eliminate hazards so no one gets hurt. Engineering controls are absolutely necessary to prevent hazards.

Hope this helped! :)

Personal  protective equipment (PPE) is appropriate for controlling hazards

PPE are used for exposure to hazards when safe work practices and other forms of administrative controls cannot provide sufficient additional protection, a supplementary method of control is the use of protective clothing or equipment. PPE may also be appropriate for controlling hazards  while engineering and work practice controls are being installed.

Find out more on Personal  protective equipment at: https://brainly.com/question/13720623

Answer:

hello your question lacks some information attached is the complete question

A) (i)maximum bending stress in tension = 0.287 * 10^6 Ib-in

    (ii) maximum bending stress in compression =  0.7413*10^6 Ib-in

B) (i)  The average shear stress at the neutral axis = 0.7904 *10 ^5 psi

    (ii)  Average shear stress at the web = 18.289 * 10^5 psi

    (iii) Average shear stress at the Flange = 1.143 *10^5 psi

Explanation:

First we calculate the centroid of the section,then we calculate the moment of inertia and maximum moment of the beam( find attached the calculation)

A) Calculate the maximum bending stress in tension and compression

lintel load = 10000 Ib

simple span = 6 ft

( (moment of inertia*Y)/ I ) = MAXIMUM BENDING STRESS

I = 53.54

i) The maximum bending stress (fb) in tension=

= [tex]\frac{M_{mm}Y }{I}[/tex]  = [tex]\frac{6.48 * 10^6 * 2.375}{53.54}[/tex] =  0.287 * 10^6 Ib-in

ii) The maximum bending stress (fb) in compression

= [tex]\frac{M_{mm}Y }{I}[/tex] = [tex]\frac{6.48 *10^6*(8.5-2.375)}{53.54}[/tex] = 0.7413*10^6 Ib-in

B) calculate the average shear stress at the neutral axis and the average shear stresses at the web and the flange

i) The average shear stress at the neutral axis

V = [tex]\frac{wL}{2}[/tex] = [tex]\frac{1000*6*12}{2}[/tex] = 3.6*10^5 Ib

Ay = 8 * 0.5 * (2.375 - 0.5 ) + 0.5 * (2.375 - [tex]\frac{0.5}{2}[/tex] ) * [tex]\frac{(2.375 - (\frac{0.5}{2} ))}{2}[/tex]

= 5.878 in^3

t = VQ / Ib  = ( 3.6*10^5 * 5.878 ) / (53.54 8 0.5) = 0.7904 *10 ^5 psi

ii) Average shear stress at the web ( value gotten from the shear stress at the flange )

t = 1.143 * 10^5 * (8 / 0.5 )  psi

  = 18.289 * 10^5 psi

iii) Average shear stress at the Flange

t = VQ / Ib = [tex]\frac{3.6*10^5 * 8*0.5*(2.375*(0.5/2))}{53.54 *0.5}[/tex]

= 1.143 *10^5

Answer:

minimum required diameter of the steel linkage is 3.57 mm

Explanation:

original length of linkage l = 10 m

force to be transmitted  f = 2 kN = 2000 N

extension e = 5 mm= 0.005 m

maximum stress σ = 200 N/mm^2 = [tex]2*10^{8} N/m^{2}[/tex]

maximum stress allowed on material σ = force/area

imputing values,

200 = 2000/area

area = 2000/([tex]2*10^{8}[/tex]) = [tex]10^{-5}[/tex] m^2

recall that area = [tex]\pi d^{2} /4[/tex]

[tex]10^{-5}[/tex] = [tex]\frac{3.142*d^{2} }{4}[/tex] = [tex]0.7855d^{2}[/tex]

[tex]d^{2} = \frac{10^{-5} }{0.7855}[/tex] = [tex]1.273*10^{-5}[/tex]

[tex]d = \sqrt{1.273*10^{-5} }[/tex] = [tex]3.57*10^{-3}[/tex] m = 3.57 mm

maximum diameter of  the steel linkage d = 3.57 mm

Answer:

32000 bits/seconds

Explanation:

Given that :

there are 16  signal combinations (states) = 2⁴

bits  n = 4

and a baud rate (number of signals/second) = 8000/second

Therefore; the number of bits per seconds can be calculated as follows:

Number of bits per seconds = bits  n × number of signal per seconds

Number of bits per seconds =  4 × 8000/second

Number of bits per seconds = 32000 bits/seconds

Answer:

115.2 W

Explanation:

The computation is shown below:

As we know that

Power = F . v

[tex]F_H = F cos \theta[/tex]

[tex]F_H = 30 \frac{4}{5}[/tex]

[tex]F_H = 24N[/tex]

Now we solve for V

[tex]V = V_0 + at[/tex]            a = 24N ÷ 20Kg

But V_0 = 0          a = 1.2 m/s^2

F_H = ma             V = 0 + (1.2) (4)

a = F_H ÷ m        V = 4.8 m/s

Therefore

Power = F_Hv

= (24) (4.8)

= 115.2 W

By applying the above formuals we can get the power

Answer:

The power produced by the turbine is 23309.1856 kW

Explanation:

h₁ = 3755.39

s₁ = 7.0955

s₂ = sf + x₂sfg  =

Interpolating fot the pressure at 3.25 bar gives;

570.935 +(3.25 - 3.2)/(3.3 - 3.2)*(575.500 - 570.935) = 573.2175

2156.92 +(3.25 - 3.2)/(3.3 - 3.2)*(2153.77- 2156.92) = 2155.345

h₂ = 573.2175 + 0.94*2155.345 = 2599.2418 kJ/kg

Power output of the turbine formula =

[tex]Q - \dot{W } = \dot{m}\left [ \left (h_{2}-h_{1} \right )+\dfrac{v_{2}^{2}- v_{1}^{2}}{2} + g(z_{2}-z_{1})\right ][/tex]

Which gives;

[tex]560 - \dot{W } = 8\left [ \left (2599.2418-3755.39 \right )+\dfrac{15^{2}- 60^{2}}{2} \right ][/tex]

= -8*((2599.2418 - 3755.39)+(15^2 - 60^2)/2 ) = -22749.1856

[tex]- \dot{W }[/tex] = -22749.1856 - 560 = -23309.1856 kJ

[tex]\dot{W }[/tex] = 23309.1856 kJ

Power produced by the turbine = Work done per second = 23309.1856 kW.

Answer:

a) The mass moves a distance of 0.625 m up the slope before coming to rest

b) The distance moved by the mass when it is connected to the spring is 0.6 m

c) [tex]\mu = 0.206[/tex]

Explanation:

Spring constant, k = 70 N/m

Compression, x = 0.50 m

Mass placed at the free end, m = 2.2 kg

angle, θ = 41°

Potential Energy stored in the spring, [tex]PE= 0.5 kx^2[/tex]

[tex]PE = 0.5 * 70 * 0.5^2\\PE = 8.75 J[/tex]

According to the principle of energy conservation

PE = mgh

8.75 = 2.2 * 9.8 * h

h = 0.41

If the mass moves a distance d from the spring

sin 41 = h/d

sin 41 = 0.41/d

d = 0.41/(sin 41)

d = 0.625 m

The mass moves a distance of 0.625 m up the slope before coming to rest

b) If the mass is attached to the spring

According to energy conservation principle:

Initial PE of spring = Final PE of spring + PE of block

[tex]0.5kx_1^2 = 0.5kx_2^2 + mgh\\x_2 = d - x_1 = d - 0.5\\h = d sin 41\\0.5*70*0.5^2 = 0.5*70*(d-0.5)^2 + 2.2*9.8*d*sin41\\8.75 = 35(d^2 - d + 0.25) + 14.15d\\8.75 = 35d^2 - 35d + 8.75 + 14.15d\\35d^2 = 20.85d\\d = 0.6 m[/tex]

The distance moved by the mass when it is connected to the spring is 0.6 m

3) The spring potential is converted to increased PE and work within the system.

mgh = Fd + 0.5kx²...........(1)

d = x , h = dsinθ

kinetic friction force , F = μmgcosθ

mgdsinθ + μmg(cosθ)d = 0.5kd²

mgsinθ + μmgcosθ = 0.5kd

sinθ + μcosθ = kd/(2mg)

[tex]\mu = \frac{\frac{kd}{2mg} - sin\theta}{cos\theta} \\\\\mu = \frac{\frac{70*0.5}{2*2.2*9.8} - sin41}{cos41} \\\\\mu = 0.206[/tex]

Answer:

A) l = 0.619m

B) l = 0.596m

C) μ = 0.314

Explanation:

The data given is:

k = 70 N/m

x = 0.5 m

m = 2.2 kg

θ = 41°

(FIGURES FOR EACH PART ARE ATTACHED AT THE BOTTOM. CONSULT THEM FOR BETTER UNDERSTANDING)

Gain in Gravitational Potential Energy = Loss in Elastic Potential Energy

mgh = (1/2)kx²

(2.2)(9.8)h = (1/2)(70)(0.5)²

h = 0.406 m

sinθ = h/l

l = h / sinθ

l = 0.406/sin41

l = 0.619m

Loss in Elastic Potential Energy in compressed spring = Gain in Gravitational Potential Energy + Gain in Elastic Potential Energy in stretched spring

(1/2)kx² = mgh + (1/2)k(l - 0.5)²

(1/2)(70)(0.5)² = (2.2)(9.8)(l·sin41)) + (1/2)(70)(l² - l + 1/4)

8.75 = 14.15(l) + 35(l²) - 35(l) + 8.75

35(l²) -20.85(l) = 0

l = 0.596m

Loss in Elastic Potential Energy = Gain in Gravitational Potential Energy + Work done against friction

(1/2)kx² = mgh + Fd

(1/2)kx² = mg(dsinθ) + μRd

(1/2)kx² = mg(dsinθ) + μ(mg · cosθ)d

(1/2)kx² = mgd (sinθ + μ(cosθ))

(1/2)(70)(0.5)² = (2)(9.8)(0.5) (sin41 + μcos41)

8.75 = 6.43 + 7.4μ

μ = 0.314