A well insulated rigid tank contains 4 kg of argon gas at 450 kPa and 30 C. A valve is opened, allowing the argon to escape until the tank pressure drops to 200 kPa. Assuming that the argon remaining in the tank experiences a reversible adiabatic process, find the final mass of argon in the tank. Since you don't have argon gas tables, assume cp, cv, k as needed at some appropriate temperature(s).

Answer:

A. The biceps muscle contracts while the triceps muscle relaxes. Only then can the forearm move up to lift the mug.

Explanation:

Skeletal muscle is also known as voluntary muscle and it can be defined as a type of muscle connected with the skeleton by tendons, so as to form a mechanical system that enables the movement of the limbs and other body parts with respect to another.

Generally, skeletal muscles are only found in vertebrates.

The muscles in the arm work to lift a mug of coffee to your lips when the biceps muscle contracts while the triceps muscle relaxes.

Basically, the only way the forearm move up to lift a mug for example, is through the contraction of the biceps muscle and the relaxation of the triceps muscle.

This ultimately implies that, the elbow joint is straightened or bent due to the actions of the biceps muscle and triceps muscle against each other.

Answer:A

Explanation:

got it right on plato

Answer:

[tex]v_2=549.2 m/s\\[/tex]

Explanation:

Given:

[tex]P_1=2500kPa\\T_1=1500 k\\V_1=5 m/s\\P_2=100 kPa\\T_2=1400 k\\A_2=10 cm^2[/tex]

Solution:

For [tex]Co_2[/tex] y=1.4

Since Nozzle is adiababic

So,

[tex]h_1+\frac{V_1^2}{2}=h_2+\frac{V_2^2}{2}\\\frac{v_2^2}{2}=(h_2-h_2)+\frac{r^2}{2}\\v_2^2=2(h_1-h_2)+v_1^2\\v_2=\sqrt{2(h_1-h_2)+v_1^2}[/tex]

Now,

[tex]h_1-h_2=Cp_1T_1-CP_2T_2\\h_1-h_2=(1989-1838.2)*10^3\\ =150.8 * 10^3\\Cp for co_2\\C_{p1}=1.326 kj/kg\\C_{p2}=1.313 kj/kg\\v_2=\sqrt{301600+25}\\ =549.2 m/s[/tex]

Answer:

[tex]h_{1} = h_2} = 293.45 KJ/kg[/tex].

The quality of the refrigerant exiting the expansion valve is

[tex]x_{2}=0.193596[/tex].

Explanation:

Fluid given Ammonia.

Inlet 1:-

Temperature [tex]T_{1}[/tex] = [tex]24^{o} C[/tex].

Pressure [tex]P_{1}[/tex] = 10 bar.

Exit 2:-

Pressure [tex]P_{2}[/tex] = 1 bar.

Solution:-

Answer:

Explanation:

Given:

Q factor, =1000

natural frequency, [tex]f_n=2000~Hz[/tex]

Damping factor, [tex]\zeta=?[/tex]

Bandwidth, BW=?

We have the relation:

[tex]Q=\frac{1}{2\zeta}[/tex]

[tex]\zeta=\frac{1}{2Q}[/tex]

[tex]\zeta=\frac{1}{2\times 1000}[/tex]

[tex]\zeta=5\times 10^{-4}[/tex]

Bandwidth:

[tex]BW=\frac{f_n}{Q}[/tex]

[tex]BW=\frac{2000}{1000}[/tex]

[tex]BW=2~Hz[/tex]

Answer:

The Space Needle is a cut away with minimal residual deflection due to load transfer.

Answer:

a.  164 °F b. 91.11 °C c. 1439.54 kJ

Explanation:

a. [1 pts] How many degrees Fahrenheit (°F) must you raise the temperature?

Since the starting temperature is 48°F and the final temperature which water boils is 212°F, the number of degrees Fahrenheit we would need to raise the temperature is the difference between the final temperature and the initial temperature.

So, Δ°F = 212 °F - 48 °F = 164 °F

b. [2 pts] How many degrees Celsius (°C) must you raise the temperature?

To find the degree change in Celsius, we convert the initial and final temperature to Celsius.

°C = 5(°F - 32)/9

So, 48 °F in Celsius is

°C₁ = 5(48 - 32)/9

°C₁ = 5(16)/9

°C₁ = 80/9

°C₁ = 8.89 °C

Also, 212 °F in Celsius is

°C₂ = 5(212 - 32)/9

°C₂ = 5(180)/9

°C₂ = 5(20)

°C₂ = 100 °C

So, the number of degrees in Celsius you must raise the temperature is the temperature difference between the final and initial temperatures in Celsius.

So, Δ°C = °C₂ - °C₁ = 100 °C - 8.89 °C = 91.11 °C

c. [2 pts] How much energy is required to heat the four quarts of water from

48°F to 212°F (boiling)?

Since we require 15.8 kJ for every degree Celsius of temperature increase of the four quarts of water, that is 15.8 kJ/°C and it rises by 91.11 °C, then the amount of energy Q required is Q = amount of heat per temperature rise × temperature rise =  15.8 kJ/°C × 91.11 °C = 1439.54 kJ

Answer:

Walls Built Using Bricks and Wood

The brick house is more energy-efficient than the one built with wood.

Explanation:

Because of their high thermal mass, which gives bricks the ability to absorb heat and release it over time, bricks remain more energy-efficient than other building materials, including wood.  In summer, bricks leave your home cool.  In winter, they make it warm.  With these two advantages provided by bricks over other building materials, bricks are the most energy-efficient building material.

Answer:

The answers are in the explanation. The pH is 5.91

Explanation:

The CH3NH2 reacts with HCl as follows:

CH3NH2 + HCl → CH3NH3⁺ + Cl⁻

When 200mL of HCl are added, the moles of CH3NH2 and HCl are reacting completely producing CH3NH3+ and Cl-. That means the species present are:

no H+. All reacted

yes H2O. Because the water is present in the solutions of HCl and CH3NH2

yes Cl-. Is a product of the reaction

Yes CH3NH2. Is consumed in the reaction but comes from the equilibrium of CH3NH3+

yes CH3NH3+. Is the other product of the reaction. MAJOR SPECIES

When 300.00mL of HCl are added, 100mL are in excess:

yes H+. Is in excess: H+ + Cl- = HCl in water. MAJOR SPECIES. Determine the pH of the solution.

yes H2O. Is present because the reactants are diluted.  

yes Cl-. Is a product of reaction and comes from HCl.

Yes CH3NH2. The reactant is over but comes from the equilibrium of CH3NH3+

yes no CH3NH3+. Yes. Is a product and remains despite HCl is in excess.

To find the pH:

At equivalence point the ion that determines pH is CH3NH3+. Its concentration is:

0.100L * (0.200mol/L) = 0.0200 moles / 0.300L = 0.0667M CH3NH3+

The equilibrium of CH3NH3+ is:

Ka = Kw/kb = 1x10-14/4.4x10-4 = 2.273x10-11 = [H+] [CH3NH2] / [CH3NH3+]

As both [H+] [CH3NH2] comes from the same equilibrium:

[H+] =  [CH3NH2] = X

2.273x10-11 = [X] [X] / [0.0667M]

1.5159x10-12 = X²

X = 1.23x10-6M = [H+]

As pH = -log [H+]

pH = 5.91

The pH at the equivalent point for this titration is "5.91".

[tex]CH_3NH_2 = 0.200\ M\\\\ \text{volume} = 100.0\ mL = 0.100\ L\\\\HCl = 0.100\ M\\\\[/tex]

We must now quantify the pH well at the equivalence point.

We know that even at the point of equivalence, moles of acid and moles of the base are equivalent. As such, first, we must calculate the number of moles of the given base.

Calculating the Moles in [tex]CH_3NH_2 = 0.200\ M \times 0.100\ L = 0.0200\ moles[/tex]

Calculating the Moles in [tex]HCl = 0.0200 \ moles[/tex]

Calculating the volume of [tex]HCl[/tex]:

[tex]\to \text{Molarity} = \frac{ \text{moles}}{\text{volume \ (L)}} \\\\\to \text{Volume} = \frac{\text{moles}}{\text{molarity}}\\\\[/tex]

                [tex]= \frac{0.0200 \ moles}{ 0.100\ M}\\\\= 0.200 \ L\\\\= 200 \ mL\\\\[/tex]

Calculating the reaction among the acid and base:

[tex]CH_3NH_2 + HCl \longrightarrow CH_3NH_3^{+} + Cl^-[/tex]

[tex]0.0200 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.0200 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.0200[/tex]

Therefore the conjugate acid of the bases exists at the standard solution.

Then we must calculate the new molar mass of [tex]CH_3NH_3^+[/tex].

Total volume[tex]= 100 + 200 = 300\ mL = 0.300\ L[/tex]

[tex][CH_3NH_3^+] = \frac{0.0200\ mole}{ 0.300\ L}= 0.0667\ M[/tex]

Using the ICE table

[tex]CH_3NH_3^+ + H_2O \longrightarrow CH_3NH_2 + H_3O^+[/tex]

[tex]I \ \ \ \ \ \ \ \ \ \ 0.0667 \ \ \ \ \ \ \ \ \ 0\ \ \ \ \ \ \ \ \ 0\\\\C\ \ \ \ \ \ \ \ -x\ \ \ \ \ \ \ \ +x \ \ \ \ \ \ \ \ +x\\\\E \ \ \ \ \ \ \ \ \ \ \ \ 0.0667-x \ \ \ \ \ \ \ \ \ \ \ \ +x \ \ \ \ \ \ \ \ \ \ \ \+x\\\\\to Ka = \frac{[CH_3NH_2] [H_3O^+] }{[CH_3NH_3^+]}[/tex]

Calculating [tex]K_a[/tex] from [tex]K_b[/tex]

[tex]\to K_a \times K_b = 1\times 10^{-14}\\\\\to K_a = \frac{1\times 10^{-14}}{4.4\times 10^{-4}} = 2.27\times 10^{-11}\\\\[/tex]

                           [tex]= 2.27\times 10^{-11} \\\\= x\times \frac{x}{(0.0667-x)}[/tex]

The x in the 0.0667-x can be ignored since the Ka value is just too small and it also does not follow the five percent criteria.

[tex]\to 2.27 \times 10^{-11} \times 0.0667 = x_2\\\\\to x_2 = 1.515\times 10^{-12}\\\\\to x = 1.23\times 10^{-6}\ M\\\\\to [H_3O^+] = x = 1.23\times 10^{-6}\ M\\\\[/tex]

We have the formula to calculate pH.

[tex]\to pH = - \log [H_3O^+] = - \log 1.23\times 10^{-6}\ M= 5.91[/tex]

The pH at the equivalent point for this titration is "5.91".

Find out more information about the pH here:

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Answer: hello your question is incomplete below is the complete question

answer:

N010 GO2 X7.0 Y2.0 15.0 J2.0  ( option 1 )

Explanation:

Given that the NC machining has to be moved from point ( 5,4 ) to point ( 7,2 ) along a circular path

GO2 = circular interpolation in a clockwise path

G91 = incremental dimension

hence the correct option is :

N010 GO2 X7.0 Y2.0 15.0 J2.0  

Solution :

B. yes, the given solution assures confidentiality. The sender Alice encrypting his messages with its own private key PRA which provides authentication. Sender Alice further encrypts his messages with the receiver's public key PUB provides confidentiality.

C. So the given solution provides non repudiation. Alice and Bob who are exchanging messages. In one case, Alice denies sending a messages to Bob that he claims to have received being able to counter Alice's denial is caused non repudiation of origin.

D. The given solution provides integrity. Because it provides authentication and have not been changed.

E. It does not provide replay attacks because it does not captures the traffic. The client does not receive the messages twice.

Answer:

jsow

hfhcffnbxhdhdhdhdhdhdddhdhdgdhdhdhdhdhdhhhdhdjsksmalalaksjdhfgrgubfghhhhhhh

Explanation:

j

Answer:

[tex]\eta_{turbine} = 0.603 = 60.3\%[/tex]

Explanation:

First, we will find actual properties at given inlet and outlet states by the use of steam tables:

AT INLET:

At 4MPa and 350°C, from the superheated table:

h₁ = 3093.3 KJ/kg

s₁ = 6.5843 KJ/kg.K

AT OUTLET:

At P₂ = 125 KPa and steam is saturated in  vapor state:

h₂ = [tex]h_{g\ at\ 125KPa}[/tex] = 2684.9 KJ/kg

Now, for the isentropic enthalpy, we have:

P₂ = 125 KPa and s₂ = s₁ = 6.5843 KJ/kg.K

Since s₂ is less than [tex]s_g[/tex] and greater than [tex]s_f[/tex] at 125 KPa. Therefore, the steam is in a saturated mixture state. So:

[tex]x = \frac{s_2-s_f}{s_{fg}} \\\\x = \frac{6.5843\ KJ/kg.K - 1.3741\ KJ/kg.K}{5.91\ KJ/kg.K}\\\\x = 0.88[/tex]

Now, we will find [tex]h_{2s}[/tex](enthalpy at the outlet for the isentropic process):

[tex]h_{2s} = h_{f\ at\ 125KPa}+xh_{fg\ at\ 125KPa}\\\\h_{2s} = 444.36\ KJ/kg + (0.88)(2240.6\ KJ/kg)\\h_{2s} = 2416.088\ KJ/kg[/tex]

Now, the isentropic efficiency of the turbine can be given as follows:

[tex]\eta_{turbine} = \frac{h_1-h_2}{h_1-h_{2s}}\\\\\eta_{turbine} = \frac{3093.3\ KJ/kg-2684.9\ KJ/kg}{3093.3\ KJ/kg-2416.088\ KJ/kg}\\\\\eta_{turbine} = \frac{408.4\ KJ/kg}{677.212\ KJ/kg}\\\\\eta_{turbine} = 0.603 = 60.3\%[/tex]

Answer:

Veracity, Value and Visualization are not only the characteristics of Big Data but are also the characteristics of relational databases. Veracity of data is issue with smallest data stores this is the reason that it is important in relation...

Answer:

Explanation:

B. you would grab the plug closest to the outlet

If you don't have enough experience, it's always best to leave socket changing to the experts. If you make a mistake, you might inflict harm and potentially endanger yourself and other people. Read on if you're interested in learning how to change a socket safely. Thus, option D is correct.

Grip the plug, not the electrical cable, when taking an electrical plug out of its socket. Before handling an electrical switch, socket, or outlet, hands must be fully dry.

Reduce the extra so that it rests only on top of the existing plasterboard. If necessary, push it back a little by using your finger. Fill the dent with ready-mixed filler or powdered filler, whichever you want, and bring it flush with the surrounding wall. Allow to dry, then sand off any excess.

Therefore, It is acceptable to grasp the electrical cord when removing an electrical plug from its socket

Learn more about electrical plug here:

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#SPJ5

Answer:

Both are correct.

Explanation:

Graphing multi meter is used to verify signals that move from electrical components. Digital oscilloscope is an equipment which stores and analyzes input signals with digital technique.

The following should be done by the team member:

Learn more: brainly.com/question/17429689

Answer:

a) 1.9%

b) T2s = 505.5 k = 232.5°C

Explanation:

P1 = 95 kPa

T1 = 27°C  = 300 k

P2 = 600 kPa

T1 = 277°c  = 550 k

Table used : Table ( A - 17 ) Ideal gas properties of air

a) determining the isentropic efficiency of the compressor

Л = ( h2s - h1 ) / ( h2a -  h1 ) ---- ( 1 )

where ; h1 = 300.19 kJ/kg , T1 = 300 K , h2a = 554.74 kJ/kg , T2 = 550 k

To get h2s we have to calculate the the value of Pr2 using Pr1(relative pressure)

 Pr2 = P2/P1 * Pr = ( 600 / 95 ) * 1.306  hence; h2s = 500.72 kJ/kg

back to equation1

Л = 0.019 = 1.9%

b) Calculate the exit temperature of the air if compressor is reversible

if compressor is reversible the corresponding exit temperature

T2s = 505.5 k = 232.5°C

given that h2s = 500.72 kJ/kg

Answer:

can you please give us the option for this question

Answer:

The code is as follows:

<form name = "myForm">

       <div>

           <input type="radio" name="vehicle" value="D0" id="D0"/>

           <label for="D0">Car</label>

       </div>

       <div>

           <input type="radio" name="vehicle" value="D1" id="D1"/>

           <label for="D1">Bike</label>

       </div>

   </form>

Explanation:

This defines the first button

           <input type="radio" name="vehicle" value="D0" id="D0"/>

           <label for="D0">Car</label>

This defines the second button

           <input type="radio" name="vehicle" value="D1" id="D1"/>

           <label for="D1">Bike</label>

The code is self-explanatory, as it follows all the required details in the question

Answer:

H(s) = 20 / [ 1 + s / 10^5 ]^2

Explanation:

Given data:

cutoff frequency = 100 kHz

stopband attenuation rate = 40 dB/decade

nominal passband gain = 20 dB

new nominal passband gain at cutoff = 14 dB

Represent the transfer function H(s)

The attenuation rate show that there are two(2) poles

H(s) = k / [ 1 + s/Wc ]^2  ----- ( 1 )

where : Wc = 100 kHz = 10^5 Hz , K = 20 log k = 20 dB ∴ k = 20

Input values into equation 1

H(s) = 20 / [ 1 + s / 10^5 ]^2

Answer:

Each of the metal specimens HAS an indentation near the center to ensure that the fracture point would occur in this region. Tension tests WERE CONDUCTED as follows. Two pieces of reflective tape WERE PLACED approximately 1 inch apart in the center of the specimen where the indentation 4 WAS LOCATED. The width and the thickness of the specimen at this location WAS MEASURED using a Vernier caliper. Then the specimen WAS SECURED in the MTS Load Frame. A laser extensometer WAS PLACED into position to measure the deformation of the specimen. The laser extensometer WAS USED to measure the original distance between the pieces of reflective tape. The MTS WAS SET to elongate the specimen one tenth of an inch every minute.

Answer:

[tex]a=45.31m/s^2[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=5.74[/tex]

Force [tex]F=317N[/tex]

Gravitational Acceleration [tex]g=9.81m/s^2[/tex]

Generally the equation for Force is mathematically given by

 [tex]F-mg=ma[/tex]

 [tex]317-5.74*9.81=5.74 a[/tex]

 [tex]a=\frac{260.7}{5.74}[/tex]

 [tex]a=45.31m/s^2[/tex]

Answer:

a)  Δh = 2 cm,  b) Δh = 0.4 cm

Explanation:

Let's start by using Bernoulli's equation for the Pitot tube, we define two points 1 for the small entry point and point 2 for the larger diameter entry point.

            P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

Point 1 is called the stagnation point where the fluid velocity is reduced to zero (v₁ = 0), in general pitot tubes are used  in such a way that the height of point 2 of is the same of point 1

           y₁ = y₂

subtitute

           P₁ = P₂ + ½ ρ v₂²

           P₁ -P₂ = ½ ρ v²

where ρ is the density of fluid  

now we measure the pressure on the included beforehand as a pair of communicating tubes filled with mercury, we set our reference system at the point of the mercury bottom surface

           ΔP =ρ_{Hg} g h - ρ g h

           ΔP =  (ρ_{Hg} - ρ) g h

as the static pressure we can equalize the equations

          ΔP = P₁ - P₂

         (ρ_{Hg} - ρ) g h = ½ ρ v²

         v = [tex]\sqrt{\frac{2 (\rho_{Hg} - \rho) g}{\rho } } \ \sqrt{h}[/tex]

in this expression the densities are constant

        v = A  √h

       A =[tex]\sqrt{\frac{2(\rho_{Hg} - \rho ) g}{\rho } }[/tex]

They indicate the density of mercury rhohg = 13600 kg / m³, the density of dry air at 20ºC is rho air = 1.29 kg/m³

we look for the constant

        A = [tex]\sqrt{\frac{2( 13600 - 1.29) \ 9.8}{1.29} }[/tex]

        A = 454.55

we substitute

       v = 454.55 √h

to calculate the uncertainty or error of the velocity

         h = [tex]\frac{1}{454.55^2} \ v^2[/tex]

       Δh = [tex]\frac{dh}{dv}[/tex]   Δv

       [tex]\frac{\Delta h}{h } = 2 \ \frac{\Delta v}{v}[/tex]

Suppose we have a height reading of h = 20 cm = 0.20 m

a) uncertainty 2.5 m / s ( 0.05)

        [tex]\frac{\delta v}{v} = 0.05[/tex]

       [tex]\frac{\Delta h}{h}[/tex] = 2 0.05  

       Δh = 0.1 h

       Δh = 0.1  20 cm

       Δh = 2 cm

b) uncertainty 0.5 m / s ( Δv/v= 0.01)

        [tex]\frac{\Delta h}{h}[/tex] =  2 0.01

        Δh = 0.02 h

        Δh = 0.02 20

        Δh = 0.1 20 cm

        Δh = 0.4 cm = 4 mm

Answer:

53 sec / cycle

Explanation:

Displayed red time = 35 seconds

Start up lost time = 2 seconds

Yellow time = 4 seconds

Red time = 1 second

Total lost time L = 2n + r

L = lost time

n = number of phase

R = red time

35+2+4+4*1

= 45

L = 2x4+45

= 53 sec/cycle

The total lost time is typically calculated as 53 seconds per cycle

Answer: Can these nuts fit in your mouth?

Explanation:

im just here for the points >:)

Answer:

[tex]T_o=82.1sec[/tex]

Explanation:

From the question we are told that:

Lost Time [tex]t=12secs[/tex]

Sum of critical flow ratios [tex]X=0.72[/tex]

Generally the Webster Method's equation for Optimum cycle time is is mathematically given by

 [tex]T_o=\frac{1.5t+5}{1-x}[/tex]

 [tex]T_o=\frac{1.5*12+5}{1-0.72}[/tex]

 [tex]T_o=82.1sec[/tex]

Answer:

Tensile stress = 0.1855Kpsi

Total deformation = 0.0012243 in

Unit strain =  1.855 *10^-5   or  18.55μ

Change in the rod diameter = 5.02 * 10^ -6 in

Explanation:

Data given: D= 0.82 in

                   L = 5.5 ft * 12 = 66 in

load (p) = 3000 (Ibf) /32.174 = 93.243 Ibm

Area = (π/4) D² = (π/4) 0.82²  = 0.502655 in²

∴ Tensile stress Rt = P/A = 93.243/0.502655 = 185.50099 pound/in²

                           Rt = 0.1855 Kpsi

∴ Total deformation = PL / AE = Rt * L/ Eal

                                 = 0.1855 * 10³  * 66 / 10000 * 10³

                                 = 0.0012243 in

∴the unit strains = total deformation / L = 0.0012243/ 66

                          =0.00001855 = 1.855 *10^-5

                         = 18.55μ

∴ Change in rod   Δd/ d = μ ΔL/L

                           = (0.33) 1.855 *10^-5 * 0.82

                           = 5.02 * 10^ -6 in