By how many of powers of ten does density vary across all material classes (metals, ceramics, polymers, and composites)?

To solve this problem, we can use the ideal gas law and the equation for polytropic process.

The ideal gas law is a fundamental law of physics that describes the behavior of an ideal gas. It relates the pressure, volume, temperature, and number of particles of a gas using the following equation:

PV = nRT

a) First, we need to find the mass of the carbon dioxide in the tank. The ideal gas law is:

PV = mRT

where P is the pressure, V is the volume, m is the mass, R is the universal gas constant, and T is the temperature. Rearranging for the mass, we get:

m = PV / RT

Substituting the given values, we have:

m = (200 kPa)(0.3 m3) / [(0.287 kPam3/kgK)(27°C + 273.15)] = 3.87 kg

So the mass of the carbon dioxide in the tank is 3.87 kg.

b) To determine the work done by the gas during the process, we can use the equation for polytropic process:

P1V1^n = P2V2^n

where P1 and V1 are the initial pressure and volume, P2 and V2 are the final pressure and volume, and n is the polytropic index. Substituting the given values, we have:

(200 kPa)(0.3 m3)^n = (4)(0.6 m3)^n

Dividing both sides by (0.3 m3)^n and taking the logarithm of both sides, we get:

log(200) + nlog(0.3) = log(4) + nlog(0.6)

Solving for n, we get:

n = log(4/200) / log(0.6/0.3) ≈ 1.235

Using the polytropic work equation:

W = (P2V2 - P1V1) / (1 - n)

Substituting the given values, we have:

W = [(4 kPa)(0.6 m3) - (200 kPa)(0.3 m3)] / (1 - 1.235) = 233.7 kJ

So the work done by the gas during the process is 233.7 kJ.

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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 \(\overline T = 548 \ K\)

\(\rho = 7900 \ kg/m^3\)

\(K = 19.0 \ W/mk \\ \\ C_p = 545 \ J/kg.K\)

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

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

\(\theta_o = \dfrac{T_o-T_{\infty}}{T_i -T_\infty}} \\ \\ \theta_o = \dfrac{50-30}{500 -30}} \\ \\ \theta_o = 0.0426\\\)

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

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

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

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

Hence, for centerline temperature at 50 °C;

The surface temperature is:

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

Answer:

The answer is below

Explanation:

1) The synchronous speed of an induction motor is the speed of the magnetic field of the stator. It is given by:

\(n_s=\frac{120f_s}{p}\\ Where\ p\ is \ the \ number\ of\ machine\ pole, f_s\ is\ the\ supply \ frequency\\and\ n_s\ is \ the \ synchronous\ speed(speed \ of\ stator\ magnetic \ field)\\Given: f_s=60\ Hz, p=4. Therefore\\\\n_s=\frac{120*60}{4}=1800\ rpm\)

2) The speed of the rotor is the motor speed. The slip is given by:

\(Slip=\frac{n_s-n_m}{n_s}. \\ n_m\ is\ the \ motor\ speed(rotor\ speed)\\Slip = 0.05, n_s= 1800\ rpm\\ \\0.05=\frac{1800-n_m}{1800}\\\\ 1800-n_m=90\\\\n_m=1800-90=1710\ rpm\)

3) The frequency of the rotor is given as:

\(f_r=slip*f_s\\f_r=0.04*60=2.4\ Hz\)

4) At standstill, the speed of the motor is 0, therefore the slip is 1.

The frequency of the rotor is given as:

\(f_r=slip*f_s\\f_r=1*60=60\ Hz\)

Answer:

The mechanical advantage value is 1/2.

The wedge does not offer mechanical advantage because the value is less than 1.

Explanation:

The mechanical advantage of a wedge is given by the formula;

ME=length of slope/ width

slope is given as = 10 cm

width is given as = 20 cm

M.E =10/20 = 1/2

The wedge does not offer mechanical advantage because the value is less than 1.The mechanical advantage of a wedge should be greater than 1.

Answer:

None of the above cause thats what i put

Answer:

a. 27.65 N b. 5.93 cm/s c. 0.03n where n = 1,2,3....

Explanation:

a. Since Young's Modulus E = stress/strain = σ/ε and σ = εE, we find the stain in the steel wire by finding its strain at 20°C.

Now Δl = lαΔθ where l = length of steel wire = 1.000 m, α = coefficient of linear expansion of steel = 11.0 × 10⁻⁶ C⁻¹ and Δθ = change in temperature = 20 °C - 60 °C = -40 °C

The strain  ε = Δl/l = αΔθ = 11.0 × 10⁻⁶ C⁻¹ × -40 °C = -4.4 × 10⁻⁴

σ = T/A= εE, where T = tension in steel wire and A = cross-sectional area of steel wire

T = εEA = εEπd²/4 where d = diameter of wire = 2.000 mm = 2 × 10⁻³ m

So,

T = εEπd²/4

= 4.4 × 10⁻⁴ × 2× 10¹⁰ Pa × π ×  (2 × 10⁻³ m )²/4

= 27.65 N

b. The velocity of the travelling wave at this lower temperature is

v = √(T/μ) where μ = mass per unit length = ρl where ρ = density of steel wire = 7.86 × 10³ kg/m³ and l = length of wire = 1.000 m

v = √(T/μ)

= √(T/ρl)

= √(27.65 N/(7.86 × 10³ kg/m³ × 1.000 m))

= √(27.65 N/7.86 × 10³ kg/m²)

= √(0.3517 × 10⁻² N/kg/m²)

= 0.05931 m/s

≅ 0.0593 m/s

= 5.93 cm/s

c. To find the allowable frequencies, we first find the fundamental frequency f₀ = v/2l

= 0.0593 m/s ÷ (2 × 1 m)

= 0.0593 m/s ÷ 2

= 0.0297 Hz

≅ 0.03 Hz

So, the allowable modes of frequency are multiples of f₀, that is fₙ = nf₀ where n = 1,2,3....

So, fₙ = 0.03n Hz where n = 1,2,3....

a) A Turing machine that decides the language {w|w contains an equal number of Os and 1s} can work as follows:

b) A Turing machine that decides the language {w|w contains twice as many Os as 1s} can work as follows:

c) A Turing machine that decides the language {w|w does not contain twice as many Os as 1s} can work as follows:

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Answer:

12 mins

Explanation:

The summation of the forces in vertical direction

= Fb - Fd - w = 0 ∴ Fd = Fb - w  ----- ( 1 )

Fb ( buoyant force ) = Pair * g * Vballoon ------- ( 2 )

Pair = air density , Vballoon = volume of balloon

Vballoon = \(\frac{\pi D^3}{6}\) ,  where D = 4  ∴  Vballoon = 33.51 ft^3

g = 32.2 ft/s^2

From property tables

Pair = 2.33 * 10^-3 slug/ft^3

μ ( dynamic viscosity ) = 3.8 * 10^-7 slug/ft.s

Insert values into equation 2

Fb = ( 2.33 * 10^-3 ) * ( 32.2 ) *( 33.51 )  = 2.514 Ib

∴ Fd = 2.514 - 2.5 = 0.014 Ib ( equation 1 )

Assuming that flow is Laminar  and  RE < 1

Re = (Pair * vd) / μair   -------- ( 3 )

where:  Pair = 2.33 * 10^-3 slug/ft^3 ,  vd = ( 987 * 4 ) ft^2/s ,   μair = 3.8 * 10^-7 slug/ft.s

Insert values into equation 3

Re = 2.4 * 10^7 (  this means that the assumption above is wrong )

Hence we will use drag force law

Assume Cd = 0.5

Express  Fd using the relation below

Fd = 1/2* Cd * Pair * AV^2

therefore V = 1.39 ft/s

Recalculate  Reynold's number using  v = 1.39 ft/s

Re = 34091

from the figure Cd ≈ 0.5 at Re = 34091  

Finally calculate the rise time ( time taken to reach an altitude of 1000 ft )

t = h/v

  = 1000 / 1.39 = 719 seconds ≈ 12 mins

The bending moments for the rectangular solid are:

a) M = 256,000 in-lb

b) M = 227,730 in-lb

a) To calculate the allowable bending moment for a solid rectangular timber beam, we can use the formula below:

M = (allowable stress)*(section modulus)

where M is the allowable bending moment and the section modulus is a measure of the beam's resistance to bending.

For a solid rectangular beam with nominal dimensions of 6 inches by 16 inches, the section modulus can be calculated as:

S = (b*h²) / 6

where b is the width of the beam and h is the height.

Substituting the values, we get:

S = (6*6²) / 6

S = 256 in³

Now, we can calculate the allowable bending moment as:

M = (allowable stress)*(section modulus)

M = 1000 psi*256 in³

M = 256,000 in-lb

Therefore, the allowable bending moment for the timber beam with nominal dimensions is 256,000 in-lb.

(b) For dressed dimensions, we need to adjust the section modulus calculation by subtracting the amount of wood that is removed during the dressing process. Assuming that the dressing process removes 1/8 inch from each side, the width and height of the dressed beam would be 5 3/4 inches and 15 1/2 inches, respectively.

Using these dimensions, the section modulus can be calculated exactly like above, but just with the different numbers:

S = (b*h²) / 6

S = (5.75*15.5²) / 6

S = 227.73 in³

Now replace that in the formula

M = 1000 psi x 227.73 in³

M = 227,730 in-lb

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Answer:

See explanation.

Explanation:

Since no figure was given I solved a problem that was similar to the one you described that I worked in my mechanics of materials class. The method should be very similar for your figure. See attached image for my work.

If it is subjected to the couple determine the maximum shear stress in regions AC and CB of the shaft. G st = 75 GPa. Than the answer will be 52Mpa.

We need to perform a two step process to obtain the maximum shear stress on the shaft. For the solid shaft,

P=2×pi×N×T/60 or T=60×p/2×pi×N

Where P=power transmitted by the shaft=50×10³W

N=rotation speed of the shaft in rpm=730rpm

Pi=3.142

T is the twisting moment

By substituting the values for pi, N and P, we get

T=654Nm or 654×10³Nmm

Also, T=pi×rho×d³/16 or rho=16×T/pi×d³

Where rho=maximum shear stress

T = twisting moment=654×10³Nmm

d= diameter of shaft= 40mm

By substituting T, pi and d

Rho=52Mpa

b. For a hollow shaft, the value for rho is unknown

T=pi×rho(do⁴-di⁴/do)/16

Rho=T×16×do/pi×(do⁴-di⁴)

Where

T= twisting moment=654×10³Nmm gotten above

do=outside shaft diawter=40mm

di= inside shaft diameter =30mm

Pi=3.142

Substituting values for pi, do, di and T.

Rho=76Mpa

Therefore, If it is subjected to the couple determine the maximum shear stress in regions AC and CB of the shaft. G st = 75 GPa. Than the answer will be 52Mpa.

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Answer: 50

Explanation:

The maximum value of rolling resistance over which a loaded scraper can maintain in a speed of 11 mph is 275 lb/ton

The maximum rolling resistance is the maximum rolling friction that can cause the motion of a body or a substance to resist the force when rolled on a flat surface.

Using Rimpull Performance Chart for wheel-tractor scraper;

The total resistance = (gross+rolling) resistance

25% = 0% - rolling resistance

rolling resistance = 25%

The maximum rolling resistance = rolling resistance percentage × speed

The maximum rolling resistance \(\mathbf{=25\% ( \dfrac{11 \ lb/ton}{1 \% \ resistance})}\)

The maximum rolling resistance = 275 lb/ton

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Answer:

Fix it

Explanation:

Use take or something to secure it and make sure no water comes out of the leak

Answer:

FLEX TAPE BOIZ

Explanation:

Answer:

Adding the length of the tape measure case.

Answer:

the perimeter of a square room is 48m. what is the area of the room? What cost will incur if the rate of carpeting the room is Rs. 10 per m square

The lack of reciprocity from Jeannie has ended Martin's willingness to be warm and pleasant toward her.

Reciprocity refers to the mutual exchange of positive behavior, such as kindness, generosity, and respect, between two people. In any relationship, whether it's a friendship, romantic partnership, or professional collaboration, reciprocity is crucial for building trust, intimacy, and satisfaction.

The lack of reciprocity has caused Martin to feel unappreciated and disheartened, and he no longer sees the value in investing his time and energy into a relationship that doesn't bring him joy or fulfillment. It's possible that Jeannie is unaware of how her behavior is affecting Martin, or that she has her own reasons for being distant.

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The timing for the first leg outbound in a nonstandard holding pattern should begin at the time of the initial clearance.

Which is when the ATC controller grants permission for the aircraft to depart and enter the airspace for the holding pattern. This clearance will also include the altitude, heading, speed, and other instructions needed to enter the pattern.  The aircraft should enter the pattern at the designated time, or as soon as possible depending on the operating environment. Once the aircraft is in the pattern, the time for the next leg will be determined by the controller. The aircraft should then continue to follow the instructions from ATC to maintain the desired spacing and separation in the pattern.

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Answer:

I would say that the hardest part about learning Thermodynamics is the generality and abstractness of it all.

Explanation:

Answer:

a) 4 hectómetros cuadrados equivalen a 400 decámetros cuadrados.

b) 21345 centímetros cuadrados equivalen a 2,135 metros cuadrados.

c) 0,592 kilómetros cuadrados equivalen a 592000 metros cuadrados.

d) 0,102 metros cuadrados equivalen a 1020 centímetros cuadrados.  

e) 23911 kilómetros cuadrados equivalen 2391100 hectómetros cuadrados.

Explanation:

a) 4 hectómetros cuadrados a decámetros cuadrados:

Según las unidades de área y sus escalas utilizadas por el Sistema Internacional de Pesos y Medidas, un hectómetro cuadrado equivale a 100 decámetros cuadradps. Entonces, obtenemos el dato equivalente por la siguiente regla de tres simple:

\(x = 4\,Hm^{2}\times\frac{100\,Dm^{2}}{1\,Hm^{2}}\)

\(x = 400\,Dm^{2}\)

4 hectómetros cuadrados equivalen a 400 decámetros cuadrados.

b) 21345 centímetros cuadrados a metros cuadrados:

Según las unidades de área y sus escalas utilizadas por el Sistema Internacional de Pesos y Medidas, un metro cuadrado equivale a 10000 centímetros cuadrados. Entonces, obtenemos el dato equivalente por la siguiente regla de tres simple:

\(x = 21345\,cm^{2}\times \frac{1\,m^{2}}{10000\,cm^{2}}\)

\(x = 2,135\,m^{2}\)

21345 centímetros cuadrados equivalen a 2,135 metros cuadrados.

c) 0,592 kilómetros cuadrados a metros cuadrados:

Según las unidades de área y sus escalas utilizadas por el Sistema Internacional de Pesos y Medidas, un kilómetro cuadrado equivale a 1000000 metros cuadrados. Entonces, obtenemos el dato equivalente por la siguiente regla de tres simple:

\(x = 0,592\,km^{2}\times \frac{1000000\,m^{2}}{1\,km^{2}}\)

\(x = 592000\,m^{2}\)

0,592 kilómetros cuadrados equivalen a 592000 metros cuadrados.

d) 0,102 metros cuadrados a centímetros cuadrados:

Según las unidades de área y sus escalas utilizadas por el Sistema Internacional de Pesos y Medidas, un metro cuadrado equivale a 10000 centímetros cuadrados. Entonces, obtenemos el dato equivalente por la siguiente regla de tres simple:

\(x = 0,102\,m^{2}\times \frac{10000\,cm^{2}}{1\,m^{2}}\)

\(x = 1020\,cm^{2}\)

0,102 metros cuadrados equivalen a 1020 centímetros cuadrados.

e) 23911 kilómetros cuadrados a hectómetros cuadrados:

Según las unidades de área y sus escalas utilizadas por el Sistema Internacional de Pesos y Medidas, un kilómetro cuadrado equivale a 100 hectómetros cuadrados. Entonces, obtenemos el dato equivalente por la siguiente regla de tres simple:

\(x = 23911\,km^{2}\times \frac{100\,Hm^{2}}{1\,km^{2}}\)

\(x = 2391100\,Hm^{2}\)

23911 kilómetros cuadrados equivalen 2391100 hectómetros cuadrados.

Answer:

about half a pound i think

Explanation:

To calculate the excavation and backfill for the attached trench, we first need to determine the excavation quantity of the earth and backfill stone. The last step is to calculate the backfill quantity of the earth.

Determination of the excavation quantity of the earth is done by measuring the length, width, and depth of the trench and multiplying these values together. For example, if the trench is 10 feet long, 2 feet wide, and 3 feet deep, the total excavation quantity would be 60 cubic feet of earth.

Next, we need to calculate the backfill stone. Assuming the pipe does not require a deduction, the backfill stone quantity is equal to the excavation quantity of earth. Therefore, for the example above, the total backfill stone quantity would be 60 cubic feet.

Finally, we need to calculate the backfill quantity of earth. This is done by subtracting the backfill stone quantity from the excavation quantity. In the example above, the backfill quantity of earth would be 0 cubic feet, since the backfill stone quantity is equal to the excavation quantity.

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Answer:

answer is C. his seat belt

Answer:

Explanation:

Work, U, is equal to the force times the distance:

U = F · r

Force needed to lift the weight, is equal to the weight: F = W = m · g

so:

U = m · g · r

   = 20.4kg · 9.81 \(\frac{N}{kg}\) · 1.50m

   = 35.316 \(\frac{N}{m}\)

   = 35.316 W

PRIVACY 1920I/BAA3613 Issue No. 1 Show that 200 mg/L of CaCO3 and 234 mg/L of NaCl are the same. A sample of water was subjected to an ion analysis, and the outcomes are displayed in Table

Concentration in chemistry refers to the quantity of a material in a certain area. The ratio of the solute in a solution to the solvent or whole solution is another way to define concentration. In order to express concentration, mass per unit volume is typically used. The solute concentration can, however, alternatively be stated in moles or volumetric units. Concentration may be expressed as per unit mass rather than volume. Although concentration is typically used to describe chemical solutions, it may be computed for any mixture. The term "concentration" in chemistry refers to the elements that make up a combination or solution. Here's how concentration is defined and how it can be calculated using various techniques.

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