An asphalt binder is mixed with aggregate and compacted into a sample. The mass of the dry sample is 1173.5 g, the mass of the sample submerged and then surface-dried with a damp towel is 1175.5 g, and the mass of the sample completely submerged in water is 652.5 g. Find the bulk specific gravity of the compacted sample.

Answers

Answer 1

Answer:

[tex]\mathbf{G_m = 2.25}[/tex]

Explanation:

From the given information:

Let the weight of the mix in the air be = [tex]W_{ma}[/tex]

Let the weight of the mix in water be = [tex]W_{mw}[/tex]; &

the bulk specific gravity be = [tex]G_m[/tex]

SO;

[tex]W_{mw} = W_{ma} - v \delta _{w} --- (1)[/tex]

Also;

[tex]G_m = \dfrac{W_{mw}}{v \delta_w} --- (2)[/tex]

From (2), make[tex]v \delta_w[/tex] the subject:

[tex]v \delta_w = \dfrac{W_{ma}}{G_m}[/tex]

Now, equation (1) can be rewritten as:

[tex]W_{mw} = W_{ma} - \dfrac{W_{ma}}{G_m}[/tex]

[tex]G_m = \dfrac{W_{ma}}{W_{ma} - W_{mw}}[/tex]

Replacing the values;

[tex]G_m = \dfrac{1173.5}{1173.5 -652.5}[/tex]

[tex]G_m = \dfrac{1173.5}{521}[/tex]

[tex]\mathbf{G_m = 2.25}[/tex]


Related Questions

Which option identifies the step of the implementation phase represented in the following scenario?

A company in the gaming industry decides to integrate movement controls in its newest hardware release. It sends an online survey to interactive gamers to identify their highest priorities.

establishing a process and budget

using communication tools

building and assembling a team

setting up a change order process

Answers

Answer:

Which option identifies the step of the implementation phase represented in the following scenario?

A company in the gaming industry decides to integrate movement controls in its newest hardware release. It sends an online survey to interactive gamers to identify their highest priorities.

establishing a process and budget

using communication tools

building and assembling a team

setting up a change order process

Explanation:

#carryonml

Answer:

using communication tools

Explanation:

The correct answer is using communication tools. Communication tools such as online surveys help project teams identify customers’ wants and needs.

A cylindrical bar of metal having a diameter of 19.9 mm and a length of 186 mm is deformed elastically in tension with a force of 42600 N. Given that the elastic modulus and Poisson's ratio of the metal are 67.1 GPa and 0.34, respectively, determine the following: (a) The amount by which this specimen will elongate (in mm) in the direction of the applied stress. (b) The change in diameter of the specimen (in mm). Indicate an increase in diameter with a positive number and a decrease with a negative number.

Answers

Answer:

a) 0.347 mm

b)  - 0.0126 mm

Explanation:

Diameter of metal bar = 19.9 mm

length = 186 mm = 0.186 m

Tension force = 42600 N

elastic modulus = 67.1 GPa

Poisson's ratio = 0.34

a) calculate the amount by which the specimen will elongate

first calculate the area of metal bar

Area = πd^2 / 4 = π/4 ( 19.9 )^2  = 3.11 * 10^-4 m^2

Elongation ( E ) = б / e = P/A * L / ΔL

and ΔL = PL / AE

hence the elongation ( ΔL) = [ (42600 * 0.186 ) / ( 3.4*10^-4 *  67.1 * 10^9 ) ]

                                            = 3.47 * 10^-4 m ≈ 0.347 mm

b) Calculate the change in diameter of specimen

μ = -( Δd / d) / ( ΔL/L )

0.34 = - (Δd / 19.9 ) / ( 0.347 / 186 )

∴ Δd = 0.34 * 0.00186 * 19.9  = - 0.0126 mm

A tiger cub has a pattern of stripes on it for that is similar to that of his parents where are the instructions stored that provide information for a tigers for a pattern

Answers

probably in it's chromosomes

Q-) please give me a reference about Tack coat? Pleae i need it please??!!

Answers

Answer:

Tack coat is a sprayed application of an asphalt binder upon an existing asphalt or Portland cement concrete pavement prior to an overlay, or between layers of new asphalt concrete.

Explanation:

When an arbitrary substance undergoes an ideal throttling process through a valve at steady state, (SELECT ALL THAT APPLY) inlet and outlet mass flowrates will be equal. inlet and outlet specific enthalpies will be equal. inlet and outlet pressures will be equal. inlet and outlet mass flowrates will be equal. inlet and outlet specific enthalpies will be equal. inlet and outlet temperatures will be equal.

Answers

Answer:

15x

Explanation:

Two ball bearings from different manufacturers are being considered for a certain application. Bearing A has a catalog rating of 2.12 kN based on a catalog rating system of 3000 hours at 500 rev/min. Bearing B has a catalog rating of 7.5 kN based on a catalog that rates at 106 cycles. For a given application, determine which bearing can carry the larger load.

Answers

Answer:

F[tex]_D[/tex] for A > F[tex]_D[/tex] for B

Hence, Bearing A can carry the larger load

Explanation:

Given the data in the question,

First lets consider an application which requires desired speed of n₀ and a desired life of L₀.

Lets start with Bearing A

so we write the relation between desired load and life catalog load and life;

[tex]F_R(L_Rn_R60)^{1/a}[/tex] = [tex]F_D(L_Dn_D60)^{1/a}[/tex]

where F[tex]_R[/tex] is the catalog rating( 2.12 kN)

L[tex]_R[/tex] is the rating life ( 3000 hours )

n[tex]_R[/tex] is the rating speed ( 500 rev/min )

F[tex]_D[/tex] is the desired load

L[tex]_D[/tex] is the desired life ( L₀ )

n[tex]_D[/tex]  is the the desired speed ( n₀ )

Now as we know, a = 3 for ball bearings

so we substitute

[tex]2.12( 3000 * 500 * 60 )^{1/3[/tex]  =  [tex]F_D( L_0n_060)^{1/3[/tex]    

950.0578 = [tex]F_D( L_0n_0)^{1/3} 3.914867[/tex]    

950.0578 / 3.914867 = [tex]F_D( L_0n_0)^{1/3}[/tex]

242.6794 =   [tex]F_D( L_0n_0)^{1/3}[/tex]

F[tex]_D[/tex] for A =  (242.6794 / [tex]( L_0n_0)^{1/3}[/tex] ) kN

Therefore the load that bearing A can carry is  (242.6794 / [tex]( L_0n_0)^{1/3}[/tex] ) kN

Next is Bearing B

[tex]F_R(L_Rn_R60)^{1/a}[/tex] = [tex]F_D(L_Dn_D60)^{1/a}[/tex]

F[tex]_R[/tex] = 7.5 kN, [tex](L_Rn_R60) = 10^6[/tex]

Also, for ball bearings, a = 3

so we substitute

[tex]7.5(10^6)^{1/3[/tex] = [tex]F_D(L_0n_060)^{1/3}[/tex]

750 =  [tex]F_D(L_0n_0)^{1/3} 3.914867[/tex]

750 / 3.914867  =  [tex]F_D(L_0n_0)^{1/3}[/tex]

191.5773 = [tex]F_D(L_0n_0)^{1/3}[/tex]

F[tex]_D[/tex] for B = ( 191.5773 / [tex](L_0n_0)^{1/3}[/tex] ) kN

Therefore, the load that bearing B can carry is  ( 191.5773 / [tex](L_0n_0)^{1/3}[/tex] ) kN

Now, comparing the Two results above,

we can say;

F[tex]_D[/tex] for A > F[tex]_D[/tex] for B

Hence, Bearing A can carry the larger load

Nate needs to replace the cable to his lamp. He is stripping it to connect it to the termils. What should he remember to do with the knife

Answers

Answer: i got you its d

Explanation:had the smae question as you

For maximum safety, an electrician should learn
A. all OSHA standards and requirements by heart.
B. to perform most tasks with one hand.
c. to perform tasks in low light situations.
D. every NEC code by heart.

Answers

Answer:A

Explanation:The minimum level of education required to become an electrician is a high school diploma or equivalency degree, like the General Education Diploma (GED). This educational step is important on the journey to becoming an electrician because the high school curriculum covers the basic principles used on the job.

Discuss in detail the manners of interaction with opposite gender

Answers

Answer:

8 Tips on Better Communication with the Opposite Sex

Put emotions away. Ladies, this one is more aimed at us, for the most part. ...

Forget your pride. In discussions, especially these days, people always want to be the one that prevails. .

Put yourself in their shoes. .

Listen. ...

Respond. ...

Actually communicate. ...

Be detailed. ...

Don't communicate too much.

Explanation:

A 10 cm thick slab (density 8530 kg/m3 , specific heat 380J/kg K, conductivity 110 W/m K) initially at 650C is being cooled by air at 15C (convection coefficient 220 W/m2 K) at left surface. The right surface is insulated. We want to estimate temperature in the slab. a. What will be the temperature of the slab after a very long time

Answers

Answer:

The temperature after a long time will return to 15°C

Explanation:

Determine the temperature of the slab after a very long time

First we calculate the heat flow for m^2 area normal to the surface

= q / A = 650°c - 15°C / ( 1 / h  + L / K )

           = 635°c  / ( 1 / 220 + 0.1 / 110 )  = 116.416 kw/m^2

Total heat content in the slab is calculated as

= m* c * ΔT

= 8530 * A * 0.1 * 380 * ( 650 - 15 )

= 205828.9  kJ/m^2

The temperature will return to 15°C after a long time

What regulates the car engines temperature

Answers

Answer:

car’s thermostat is used to regulate the temperature of the engine to help the engine stay cool.

Explanation:

Answer:

Coolant

Explanation:

For a bronze alloy, the stress at which plastic deformation begins is 284 MPa and the modulus of elasticity is 106 GPa. (a) What is the maximum load that can be applied to a specimen having a cross-sectional area of 310 mm2 without plastic deformation? (b) If the original specimen length is 120 mm, what is the maximum length to which it may be stretched without causing plastic deformation?

Answers

Answer:

a) the maximum load is 88,040 N

b)

the maximum length to which the specimen may be stretched is 0.12032148 mm

Explanation:

Given the data in the question;

the stress at which plastic deformation begins σ = 284 MPa = 2.84 × 10⁸ Pa

modulus of elasticity E = 106 GPa = 1.06 × 10¹¹ Pa

a)

Area A = 310 mm² = 310 × 10⁻⁶ m ( without plastic deformation )

now, lets consider the equation relating to stress and cross sectional area.

σ = F / A₀

hence, maximum load F = σA₀  

so we substitute

F = (2.84 × 10⁸) × (310 × 10⁻⁶)

F = 88,040 N

Therefore, the maximum load is 88,040 N

b)

Initial length specimen l₀ = 120 mm  = 120 × 10⁻³ m

using engineering strain, ε = (l₁ - l₀)/l₀

Also from Hooke's law, σ = Eε

so from the equation above;

l₁ = l₀( ε + 1 )

l₁ = l₀( σ/E + 1 )

so we substitute

l₁ = (120 × 10⁻³)( (2.84 × 10⁸)/(1.06 × 10¹¹)) + 1 )

l₁ = (120 × 10⁻³) ( 1.002679 )

l₁ = 0.12032148 mm

Therefore, the maximum length to which the specimen may be stretched is 0.12032148 mm

Which two technologies were combined to create product life cycle management (PLM) software?

CAD and a database

spreadsheets and graphics

a database and spreadsheets

CAD and spreadsheets

Answers

Answer:

CAD and a database

Explanation:

The correct answer is CAD and a database. When American Motors Corportation introduced the Jeep Cherokee, it implemented CAD to increase engineering productivity and combined that with a new communications system.

The answer is. CAD and a database

Some project managers prefer the PERT chart over the Gantt chart because it clearly illustrates task dependencies. A PERT chart, however, can be much more difficult to interpret, especially on complex projects. Alternatively, some project managers may choose to use both techniques. If you are the project manager of a residential construction project, will you prefer PERT chart to Gantt chart? Explain why?

Answers

Answer:

PERT Chart and GANTT Chart

As the project manager of a residential construction project, I will prefer the PERT chart to the GANTT chart because a PERT chart displays task dependencies unlike a Gantt chart.  With the PERT chart, the sequence of tasks is clearly mapped out.  Dependent tasks are carried out when other tasks that they depend on have been executed.

Explanation:

By definition, a Gantt chart is like a bar chart that lays out project tasks and timelines using bars.  On the other hand, a PERT chart follows a structure in the form of flow charts or network diagrams.  It displays all the project tasks in separate boxes.  The boxes are then connected with arrows which clearly show the task dependencies.

1)Saturated steam at 1.20bar (absolute)is condensed on the outside ofahorizontal steel pipe with an inside and outside diameter of 0.620 inches and 0.750 inches, respectively. Cooling water enters the tubes at 60.0°F and leaves at 75.0°F at a velocity of 6.00ft/s. (HINT: You may assume laminar condensate flow.You many also assume that the mean bulk temperature of the cooling water is equal to the wall temperature on the outside of the pipe, T".You may also neglect the viscosity correction in your calculations.)a)What are the inside

Answers

Answer:

hi = 7026.8  W/m^2.k

Explanation:

Given data :

pressure of saturated steam = 1.2 bar

Horizontal steel pipe : inside diameter = 0.620 , outside diameter = 0.750 inches

temperature of water at entry = 60°F

temperature of water at exit = 75°F

velocity of water = 6 ft/s

Calculate the Inside convective heat transfer coefficient ( hi )

mean temperature ( Tm ) = 60 + 75 / 2 = 67.5°F ≈ 292.877 K ≈ 19.727°C

next : find the properties of water at this temperature ( 19.727°C )

thermal conductivity = 0.598  w/m.k

density = 1000 kg/m^3

specific heat ( Cp ) = 4.18 KJ/kg.k

viscosity = 0.001 pa.s

velocity of water = 6 ft/s ≈ 1.8288 m/s

∴ Re ( Reynolds number ) = 28712.16

and Prandtl number ( Pr ) = (4180 * 0.001) / 0.598  = 6.989

finally to determine the inside convective heat transfer coefficient we will apply the Dittos - Bolter equation

hi = 7026.8 w/m^2.k

attached below is the remaining solution

what is the answer to life the universe and everything
(worth 95 points!)

Answers

i would say nothing, time is endless and the cycle of life is endless, not only on earth but almost anywhere, people try to find answers like what’s at the bottom of the ocean and stuff like that but there’s no point in finding out because it has no benefit, life is made for different reasons so there’s not one answer to it

Answer:

In the absence of dark energy, a flat universe expands forever but at a continually decelerating rate, with expansion asymptotically approaching zero; with dark energy, the expansion rate of the universe initially slows down, due to the effects of gravity, but eventually increases, and the ultimate fate of the universe ...

Explanation:

I think it goes on forever.

MAPP gas, natural gas, propane, and acetylene can be used with oxygen to cut metal.
True
False

Answers

The answer is true.

r.a.t.e this car from 1/10

Answers

Answer:

8.5 i guess

Explanation:

Steam in a heating system flows through tubes whose outer diameter is 5 cm and whose walls are maintained at a temperature of 198.06°C. Circular copper alloy fins (k =285 W/m · °C) of outer diameter 6 cm and constant thickness 1 mm are attached to the tube. The space between the fins is 3 mm, and thus there are 250 fins per meter length of the tube. Heat is transferred to the surrounding water at T= 43.06°C, with a heat transfer coefficient of 5300 W/m2 · °C. Determine the increase in heat transfer from the tube per meter of its length as a result of adding fins and fin effectiveness

Answers

Answer:

rf

Explanation:

attached to the tube. The space between the fins is 3 mm, and thus there are 250 fins per meter length of the tube. Heat is transferred to the surrounding water at T= 43.06°C, with a heat transfer coefficient of 5300 W/m2 · °C. Determine the increase in heat transfer from the tube per meter of its length as a resu.

Assume, X Company Limited (XCL) is one of the leading 4th generation Life Insurance
Companies in Bangladesh. The Company is fully customer focused. This Life insurance company are
experimenting with analysis of consumer profiles (to determine whether a person eats healthy food,
exercises, smokes or drinks too much, has high-risk hobbies, and so on) to estimate life expectancy.
Companies might use the analysis to find populations to market policies to. From the perspective of
privacy, what are some of the key ethical or social issues raised? Evaluate some of them.

Answers

Answer:

The issues related to the privacy are:

1. Informational privacy

2. Discrimination factors

3. Biased grouping on the basis of Data mining

4. Lack of consent

5. Morally wrong

6. Illegal distribution of information risks

7. Possibility of threat to life

Let's look at some major concerns:

1. Informational privacy : The concept of privacy of the personal information is totally nullified when the information is being used for a purpose other than the intended one for which it was given. This unethical use of information even for general purposes is not correct and is a matter of concern. It is more like using the sensitive data of others for personal benefit which is purely objectionable and raises security issues. Sometimes the data is also shared with the potential employers which might have certain impacts we are unaware of.

2. Data mining issues : The process of using a certain information to arrive and understand the trend and outcomes is called data mining. In this case, the consumer's data undergoes grouping and might get placed in the wrong group rather than the actual one. Also, there can be a case of biasing towards the groups which are not be focused on, or are not a part of the intended audience. This leads to the discrimination factors if we see it from a social point of view.

3. Lack of consent : Use of information without the consent or awareness of the consumers raises concern over the business ethics followed by the company. No one deserves the right to misuse information for his personal benefits without any of its information to the consumer. It is morally wrong and againt the work ethics. Moreover, it raises trust issues between the two involved, and hence is socially unacceptable.Explanation:

The criminal and traffic code requires that a driver must have a valid driver's license in his/her
immediate possession at any time when operating a motor vehicle.
True
False

Answers

Answer:

true

Explanation:

the answer is true because if u don't have a valid license when operating a vehicle and you get pulled over you will get in trouble i know this because my parents got in trouble for it once

A cylindrical metal specimen having an original diameter of 10.33 mm and gauge length of 52.8 mm is pulled in tension until fracture occurs. The diameter at the point of fracture is 6.38 mm, and the fractured gauge length is 73.9 mm. Calculate the ductility in terms of (a) percent reduction in area (percent RA), and (b) percent elongation (percent EL).

Answers

Answer

a) 62 percent

b) 40 percent

Explanation:

Original diameter ( d[tex]_{i}[/tex] ) = 10.33 mm

Original Gauge length ( L[tex]_{i}[/tex] ) = 52.8 mm

diameter at point of fracture ( d[tex]_{f}[/tex] ) = 6.38 mm

New gauge length ( L[tex]_{f}[/tex] ) = 73.9 mm

Calculate ductility in terms of

a) percent reduction in area

percentage reduction = [ (A[tex]_{i}[/tex] - A[tex]_{f}[/tex] ) / A[tex]_{i}[/tex] ] * 100

A[tex]_{i}[/tex] ( initial area ) = π /4 di^2

= π /4 * ( 10.33 )^2 = 83.81 mm^2

A[tex]_{f}[/tex] ( final area ) = π /4 df^2

= π /4 ( 6.38)^2 = 31.97 mm^2

hence : %reduction = ( 83.81 - 31.97 ) / 83.81

= 0.62 = 62 percent

b ) percent elongation

percentage elongation = ( L[tex]_{f}[/tex] - L[tex]_{i}[/tex] ) / L[tex]_{i}[/tex]

= ( 73.9 - 52.8 ) / 52.8 = 0.40 = 40 percent

A 4-L pressure cooker has an operating pressure of 175 kPa. Initially, one-half of the volume is filled with liquid and the other half with vapor. If it is desired that the pressure cooker not run out of liquid water for 75 min, determine the highest rate of heat transfer allowed.

Answers

Answer:

the highest rate of heat transfer allowed is 0.9306 kW

Explanation:

Given the data in the question;

Volume = 4L = 0.004 m³

V[tex]_f[/tex] = V[tex]_g[/tex] = 0.002 m³

Using Table ( saturated water - pressure table);

at pressure p = 175 kPa;

v[tex]_f[/tex] = 0.001057 m³/kg

v[tex]_g[/tex] = 1.0037 m³/kg

u[tex]_f[/tex] = 486.82 kJ/kg

u[tex]_g[/tex] 2524.5 kJ/kg

h[tex]_g[/tex] = 2700.2 kJ/kg

So the initial mass of the water;

m₁ = V[tex]_f[/tex]/v[tex]_f[/tex] + V[tex]_g[/tex]/v[tex]_g[/tex]

we substitute

m₁ = 0.002/0.001057  + 0.002/1.0037

m₁ = 1.89414 kg

Now, the final mass will be;

m₂ = V/v[tex]_g[/tex]

m₂ = 0.004 / 1.0037

m₂ = 0.003985 kg

Now, mass leaving the pressure cooker is;

m[tex]_{out[/tex] = m₁ - m₂

m[tex]_{out[/tex] = 1.89414  - 0.003985

m[tex]_{out[/tex] = 1.890155 kg

so, Initial internal energy will be;

U₁ = m[tex]_f[/tex]u[tex]_f[/tex] + m[tex]_g[/tex]u[tex]_g[/tex]

U₁ = (V[tex]_f[/tex]/v[tex]_f[/tex])u[tex]_f[/tex]  + (V[tex]_g[/tex]/v[tex]_g[/tex])u[tex]_g[/tex]

we substitute

U₁ = (0.002/0.001057)(486.82)  + (0.002/1.0037)(2524.5)

U₁ = 921.135288 + 5.030387

U₁ = 926.165675 kJ

Now, using Energy balance;

E[tex]_{in[/tex] -  E[tex]_{out[/tex] = ΔE[tex]_{sys[/tex]

QΔt - m[tex]_{out[/tex]h[tex]_{out[/tex] = m₂u₂ - U₁

QΔt - m[tex]_{out[/tex]h[tex]_g[/tex] = m₂u[tex]_g[/tex] - U₁

given that time = 75 min = 75 × 60s = 4500 sec

so we substitute

Q(4500) - ( 1.890155 × 2700.2 ) = ( 0.003985 × 2524.5 ) - 926.165675

Q(4500) - 5103.7965 = 10.06013 - 926.165675

Q(4500) = 10.06013 - 926.165675 + 5103.7965

Q(4500) = 4187.690955

Q = 4187.690955 / 4500

Q = 0.9306 kW

Therefore, the highest rate of heat transfer allowed is 0.9306 kW

How should backing plates, struts, levers, and other metal brake parts be cleaned?

Answers

Answer: Cleaning of mechanical parts is necessary to remove contaminants, and to avoid clogging of wastes which could restrict the functioning of the machine.

Explanation:

There are different agents used for cleaning different machine instruments to prevent their corrosion and experience proper cleaning.

Backing plates must be dry cleaned using a cotton cloth to remove the dirt, dust or any other dry contaminant.

Struts can be wet cleaned by applying alcoholic solvent.

Levers can be cleaned using a mineral spirit.

Metallic plates can be cleaned using water based solution or water.

Help meeeeeeeee plzzzzz need explanation

Answers

the picture is blank for me what does it say i can comment the answer plz mark brainlyist

As part of a heat treatment process, cylindrical, 304 stainless steel rods of 100-mm diameter are cooled from an initial temperature of 500 C by suspending them in an oil bath at 30 C. If a convection coefficient of 500 W/m2 K is maintained by circulation of the oil, how long does it take for the centerline of a rod to reach a temperature of 50 C, at which point it is withdrawn from the bath

Answers

Answer:

Explanation:

Given that:

diameter = 100 mm

initial temperature = 500 ° C

Conventional coefficient = 500 W/m^2 K

length  = 1 m

We obtain the following data from the tables A-1;

For the stainless steel of the rod [tex]\overline T = 548 \ K[/tex]

[tex]\rho = 7900 \ kg/m^3[/tex]

[tex]K = 19.0 \ W/mk \\ \\ C_p = 545 \ J/kg.K[/tex]

[tex]\alpha = 4.40 \times 10^{-6} \ m^2/s \\ \\ B_i = \dfrac{h(\rho/4)}{K} \\ \\ =0.657[/tex]

Here, we can't apply the lumped capacitance method, since Bi > 0.1

[tex]\theta_o = \dfrac{T_o-T_{\infty}}{T_i -T_\infty}} \\ \\ \theta_o = \dfrac{50-30}{500 -30}} \\ \\ \theta_o = 0.0426\\[/tex]

[tex]0.0426 = c_1 \ exp (- E^2_1 F_o_)\\ \\ \\ 0.0426 = 1.1382 \ exp (-10.9287)^2 \ f_o \\ \\ = f_o = \dfrac{In(0.0374)}{0.863} \\ \\ f_o = 3.81[/tex]

[tex]t_f = \dfrac{f_o r^2}{\alpha} \\ \\ t_f = \dfrac{3.81 \times (0.05)^2}{4.40 \times 10^{-6}} \\ \\ t_f= 2162.5 \\ \\ t_f = 36 mins[/tex]

However, on a single rod, the energy extracted is:

[tex]\theta = pcv (T_i - T_{\infty} )(1 - \dfrac{2 \theta}{c} J_1 (\zeta) ) \\ \\ = 7900 \\times 546 \times 0.007854 \times (500 -300) (1 - \dfrac{2 \times 0.0426}{1.3643}) \\ \\ \theta = 1.54 \times 10^7 \ J[/tex]

Hence, for centerline temperature at 50 °C;

The surface temperature is:

[tex]T(r_o,t) = T_{\infty} +(T_1 -T_{\infty}) \theta_o \ J_o(\zeta_1) \\ \\ = 30 + (500-30) \times 0.0426 \times 0.5386 \\ \\ \mathbf{T(r_o,t) = 41.69 ^0 \ C}[/tex]

In the production of soybean oil, dried and flaked soybeans are brought in contact with a solvent (often hexane) that extracts the oil and leaves behind the residual solids and a small amount of oil.

a. Draw flow diagram of the process, labeling the two feed streams (beans and solvent) and the leaving streams (solids and extract).
b. The soybeans contain 18.5 wt% oil and the remainder insoluble solids, and the hexane is fed at a rate corresponding to 2.0 kg hexane per kg beans. The residual solids leaving the extraction unit contain 35.0 wt% hexane, all of the non-oil solids that entered with beans, and 1.0% of the oil that entered the beans. For a feed rate of 1000 kg/h of dried flaked soybeans, calculate mass flow rates of extract and residual solids and the composition of extract.

Answers

Answer: its c

Explanation:

In the context of mechanical systems, what does the term efficiency mean? OA the factor by which a machine multiplies a force B. the ratio of a machine's power to the force of its input Ос. the rate at which a machine performs work D. the rate at which a machine consumes energy E. the ratio of the work output of a machine to the work input​

Answers

Answer:

E

Explanation:

I have a big brain and I just took the test and got it correct.

A cylindrical specimen of this alloy 12 mm in diameter and 188 mm long is to be pulled in tension. Assume a value of 0.34 for Poisson's ratio.Calculate the stress (in MPa) necessary to cause a 0.0105 mm reduction in diameter.

Answers

This question is incomplete, the missing image in uploaded along this answer below.

Answer:

The required stress is 200 Mpa

Explanation:

Given the data in the question;

diameter D = 12 mm = 12 × 10⁻³ m

Length L = 188 mm = 188 × 10⁻³ m

Poisson's ratio v = 0.34

Reduction in diameter Δd = 0.0105 mm = 0.0105 × 10⁻³ m

The transverse strain will;

εˣ = Δd / D

εˣ = -0.0105 × 10⁻³ /  12 × 10⁻³ m

εˣ = -0.00088

The longitudinal strain will be;

[tex]E^z[/tex] = - ( εˣ  / v )

[tex]E^z[/tex] = - ( -0.00088  / 0.34 )

[tex]E^z[/tex] = - ( - 0.002588 )

[tex]E^z[/tex] = 0.0026

Now, Using the values for strain, we get the value of stress from the graph provided in the question, ( first image uploaded below.

From the graph, in the Second image;

The stress is 200 Mpa

Therefore, The required stress is 200 Mpa

The unit of solar radiation?

Answers

Answer: The solar irradiance is measured in watt per square metre (W/m2) in SI units. Solar irradiance is often integrated over a given time period in order to report the radiant energy emitted into the surrounding environment (joule per square metre, J/m2) during that time period.

Explanation: hope that helped!

Watts is the unit of solar radiation
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