1025 steel wire is stretched with a stress of 70 MPa at room temperature 20 C. If th length is held constant, to what temperature in 'C and 'F must the wire be heated to reduce the stres to 17 MPa?

Answer:

a. Covalent modification = Seconds to minutes

b. Allosteric control = Milliseconds

c. Gene expression = Hours

Explanation:

Covalent modifications refer to the addition and/or removal of chemical groups by the action of particular enzymes such as methylases, acetylases, phosphorylases, phosphatases, etc. For example, histones are chromatin-associated proteins covalently modified by enzymes that add methyl groups (histone methylation), acetyl groups (histone acetylation), phosphate groups (histone phosphorylation), etc. Moreover, allosteric control, also known as allosteric regulation, is a type of regulation of the enzyme activity by binding an effector molecule (allosteric modulator) at a different site than the enzyme's active site, thereby triggering a conformational change on the enzyme upon binding of an effector. Finally, gene expression encompasses the cellular processes by which genetic information flows from genes to proteins (i.e., transcription >> translation). In metabolic pathways, enzymes that are able to catalyze irreversible reactions represent sites of control (for example, during glycolysis, pyruvate kinase is an enzyme that catalyzes an irreversible reaction, thereby serving as a control site). In turn, enzymatic activity is modulated by covalent modifications or reversible binding of allosteric effectors. Finally, metabolic pathways are also modulated by gene regulatory mechanisms that control the transcription of specific enzymes required for such pathways. During these processes, the times required for allosteric regulation, covalent modification (e.g., phosphorylation) and transcriptional control can be counted in milliseconds, seconds, and hours, respectively.

Explanation:

conductor, flux, and movement of conductor in magnetic field are some of the factors that induce emf.

Answer:

Explanation:

Be bop

Answer:

1:1

Explanation:

A coin has either a head or tail.

Thus, probability of head = ½

Probability of tail = ½

The first coin is tossed twice and probability of head or tail on both tosses is still ½. Thus, they will have a ratio 1:1.

The second coin is still tossed regardless and thus continuously and so should have same ratio of 1:1.

Therefore, the ratio of the both of them must, also be 1:1.

Answer:

Explanation:

Increased by the constant. Take a very simple case.

4  +  5 +  6 = 15

The mean is 5  (obtained by dividing the total (15) by the number of terms (3).

Now add a constant say 6

4 + 6 = 10

5 + 6 = 11

6 + 6 = 12

Total = 33/3 = 11

So the mean 5 is increased by the constant 6.

Now do the same thing more symbolically.

4 + c

5 + c

6 + c

Total = 15 + 3c

Divide by 3 you get 5 + c

If you want a more formal proof involving n terms, leave a note.

Answer: why?

Explanation:

This question is incomplete, the complete question is;

Plexiglas, polymethylmethacrylate (PMMA), is often used as display windows and cases in art galleries and museums. If its flame imparts 32 kW/m^2 to the surface when it burns estimate the energy release rate of a 3 m^2 square sheet (one side). Assume its vaporization temperature is 350°C. Calculate with units, provide work.

a ⇒ [tex]q_{rad[/tex] = σT⁴, where σ = 5.67 × 10⁻¹¹ kW/m².K⁴

b ⇒ [tex]q"[/tex] = [tex]q_{flame[/tex] -  [tex]q_{rad[/tex]

c ⇒ [tex]m"[/tex] =  [tex]q"[/tex]/L, where L = 1.6 kJ/g

d ⇒ Q =  [tex]m"[/tex] × A × ΔHc, where ΔHc is 24.9 kJ/g

Answer:

a) [tex]q_{rad[/tex] = 8.54 kW/m²

b) [tex]q"[/tex] = 23.46 kW/m²

c) [tex]m"[/tex] = 14.6625 g/m²

d) Q = 1095.2888 kJ

Explanation:

Given the data in the question;

[tex]q_{flame[/tex] = 32 kW/m²

Area; A = 3m²

vaporization temperature; T = 350°C = ( 350 + 273 )K = 623 K

Now,

a) ⇒ [tex]q_{rad[/tex] = σT⁴, where σ = 5.67 × 10¹¹ kW/m².K⁴

we substitute

[tex]q_{rad[/tex] = ( 5.67 × 10⁻¹¹ kW/m².K⁴ ) × ( 623 K)⁴

[tex]q_{rad[/tex] = ( 5.67 × 10⁻¹¹ kW/m².K⁴ ) × 150644120641 K⁴

[tex]q_{rad[/tex] = 8.54 kW/m²

b) ⇒ [tex]q"[/tex] = [tex]q_{flame[/tex] - [tex]q_{rad[/tex]

we substitute

[tex]q"[/tex] = 32 kW/m² - 8.54 kW/m²

[tex]q"[/tex] = 23.46 kW/m²

c) ⇒ [tex]m"[/tex] =  [tex]q"[/tex]/L, where L = 1.6 kJ/g

we substitute

[tex]m"[/tex] =  23.46 / 1.6

[tex]m"[/tex] = 14.6625 g/m²

d) ⇒ Q =  [tex]m"[/tex] × A × ΔHc, where ΔHc is 24.9 kJ/g

we substitute

Q =  14.6625 × 3 × 24.9

Q = 1095.2888 kJ

Solution :

Let [tex]$x(t) = \frac{\sin (20 \pi t)}{\pi t}$[/tex]

[tex]$T_s = 20$[/tex] ms, so [tex]$f_s=\frac{1}{T_s}[/tex]

                           [tex]$=\frac{1}{20}$[/tex]

                           = 0.05 kHz

[tex]$f_s=50 $[/tex] Hz , ws = [tex]$2 \pi f_s = 100 \pi$[/tex]  rad/s

We know that,

FT → [tex]$\frac{\sin (20 \pi \omega)}{\pi \omega}$[/tex]

The sampled signal is :

[tex]$XS(\omega) = \frac{1}{T_s} \sum_{k=- \infty}^{\infty}X (\omega-k\omega S)[/tex]

So, [tex]$XS(\omega) = \frac{1}{20 \times 10^{-3}} \sum_{k=- \infty}^{\infty}X (\omega-100 k \pi)[/tex]

     [tex]$XS(\omega) = 50 \sum_{k=- \infty}^{\infty}X (\omega-100 k \pi)[/tex]

Answer:

[tex]D=0.016m[/tex]

Explanation:

From the question we are told that:

Discharge Rate [tex]F_r=0.5kgls[/tex]

Pressure [tex]P=15Kpa[/tex]

Temperature [tex]T=25=>298K[/tex]

Ambient pressure is 1 atm.

Generally the equation for Density is mathematically given by

[tex]\rho=\frac{PM}{RT}[/tex]

[tex]\rho=\frac{15*10^5*28.0134*10^{-3}}{8.314*298}[/tex]

[tex]\rho=16.958kg/m^2[/tex]

Generally the equation for Flow rate is mathematically given by

[tex]F_r=\mu A\sqrt{Q \rho P(\frac{2}{Q+1})^{\frac{Q+1}{Q-1}}}[/tex]

Where

[tex]Q=Heat coefficient\ ratio\ of\ Nitrogen[/tex]

[tex]Q=1.4[/tex]

[tex]\mu= Discharge\ coefficient[/tex]

[tex]\mu=0.68[/tex]

Therefore

[tex]0.5=0.68 A\sqrt{1.4 16.958 15*10^{5}(\frac{2}{1.4+1})^{\frac{1.4+1}{1.4-1}}}[/tex]

[tex]A=2.129*10^{-4}[/tex]

Where

[tex]A=\frac{\pi}{4}D^2[/tex]

[tex]\frac{\pi}{4}D^2=2.129*10^{-4}[/tex]

[tex]D=0.016m[/tex]

Answer:

2

Explanation:

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Answer:

2

Explanation:

High separation between deep...........

Answer:

True

Explanation:

After you have sold or have made the transfer of ownership of a motor vehicle to another person, you are required to fill a notice of transfer and release of liability. You do this as a notification to let DMV be aware of a change of ownership of this vehicle. And it also serves to protect the previous owner from liabilities such as parking violations or traffic violations. This notification should be done within 5 Days of the transfer of ownership.

Thank you.

9514 1404 393

Answer:

  see attached

Explanation:

Assuming flow is uniform across the cross section of the artery, the mass flow rate is the product of the volumetric flow rate and the density.

  (5 cm³/s)(1.06 g/cm³) = 5.3 g/s

If we assume the blood splits evenly at the bifurcation, then the downstream mass flow rate in each artery is half that:

  (5.3 g/s)/2 = 2.65 g/s

__

The average velocity will be the ratio of volumetric flow rate to area. Upstream, that is ...

  (5 cm³/s)/(π(0.25 cm)²) ≈ 25.5 cm/s

Downstream, we have half the volumetric flow and a smaller area.

  (2.5 cm³/s)/(π(0.15 cm)²) ≈ 35.4 cm/s

Answer:

Only Technician A

Explanation:

Continuously variable transmission (CVT) does not shift gears because it simply doesn't have gears. It instead uses 2 pulleys which change constantly and continuously in size, and they are linked by a belt. Continuously variable transmission (CVT) is automatic transmission as mentioned in the question above, largely due to the power the belt continuously transmits, making the vehicle's engine work in the optimum power range.

While Technician B saying, that a continuously variable transmission has a maximum of six distinct gear ranges, which is not correct because continuously variable transmission (CVT) works with an unlimited number of gear ratios within a fixed range.

Answer:

Inlet

Explanation:

Consider a laminar forced flow inside a pipe with constant wall temperature, the heat flux will have a higher value near the INLET of the pipe.

This is because the friction factor is experienced at the highest level when a laminar forced flow is at the tube inlet where the thickness of the boundary layer is zero. Also, the friction factor decreases step by step at a lower rate to the fully augmented value.

Answer:

Increasing following distance to Four-seconds when encountering other motorists who follow too closely  is an example of appropriate implementation of the IPDE defensive driving strategy for the  maintenance of an appropriate Safety Cushion.

Explanation:

Maintaining the required safety cushion by utilizing the IPDE defensive driving strategy to manage the nine to fifteen space driving zones involves continuous scanning.  Therefore, motorists should be able to identify objects and hazards in the driving scene, line of sight, and path of travel.  They should predict points of driving conflicts.  They should determine appropriate and safe driving actions to take, when, and where.  Finally, action is required to ensure that conflicts are avoided.

Answer: Attached below is the well written question and solution

answer:

i) Attached below

ii) similar parameter =  [tex]\frac{V}{VoL } = 1 / Re[/tex]

Explanation:

Using ;  L as characteristic length and Vo as reference velocity

i) Nondimensionalize the equations

ii) Identifying similarity parameters

the similar parameters are  = [tex]\frac{V}{VoL } = 1 / Re[/tex]

Attached below is the detailed solution

Answer:

a) Irreversible, b) Reversible, c) Irreversible, d) Impossible.

Explanation:

Maximum theoretical efficiency for a power cycle ([tex]\eta_{r}[/tex]), no unit, is modelled after the Carnot Cycle, which represents a reversible thermodynamic process:

[tex]\eta_{r} = \left(1-\frac{T_{C}}{T_{H}} \right)\times 100\,\%[/tex] (1)

Where:

[tex]T_{C}[/tex] - Temperature of the cold reservoir, in Kelvin.

[tex]T_{H}[/tex] - Temperature of the hot reservoir, in Kelvin.

The maximum theoretical efficiency associated with this power cycle is: ([tex]T_{C} = 400\,K[/tex], [tex]T_{H} = 1200\,K[/tex])

[tex]\eta_{r} = \left(1-\frac{400\,K}{1200\,K} \right)\times 100\,\%[/tex]

[tex]\eta_{r} = 66.667\,\%[/tex]

In exchange, real efficiency for a power cycle ([tex]\eta[/tex]), no unit, is defined by this expression:

[tex]\eta = \left(1-\frac{Q_{C}}{Q_{H}}\right) \times 100\,\% = \left(\frac{W_{C}}{Q_{H}} \right)\times 100\,\% = \left(\frac{W_{C}}{Q_{C} + W_{C}} \right)\times 100\,\%[/tex] (2)

Where:

[tex]Q_{C}[/tex] - Heat released to cold reservoir, in kilojoules.

[tex]Q_{H}[/tex] - Heat gained from hot reservoir, in kilojoules.

[tex]W_{C}[/tex] - Power generated within power cycle, in kilojoules.

A power cycle operates irreversibly for [tex]\eta < \eta_{r}[/tex], reversibily for [tex]\eta = \eta_{r}[/tex] and it is impossible for [tex]\eta > \eta_{r}[/tex].

Now we proceed to solve for each case:

a) [tex]Q_{H} = 900\,kJ[/tex], [tex]W_{C} = 450\,kJ[/tex]

[tex]\eta = \left(\frac{450\,kJ}{900\,kJ} \right)\times 100\,\%[/tex]

[tex]\eta = 50\,\%[/tex]

Since [tex]\eta < \eta_{r}[/tex], the power cycle operates irreversibly.

b) [tex]Q_{H} = 900\,kJ[/tex], [tex]Q_{C} = 300\,kJ[/tex]

[tex]\eta = \left(1-\frac{300\,kJ}{900\,kJ} \right)\times 100\,\%[/tex]

[tex]\eta = 66.667\,\%[/tex]

Since [tex]\eta = \eta_{r}[/tex], the power cycle operates reversibly.

c) [tex]W_{C} = 600\,kJ[/tex], [tex]Q_{C} = 400\,kJ[/tex]

[tex]\eta = \left(\frac{600\,kJ}{600\,kJ + 400\,kJ} \right)\times 100\,\%[/tex]

[tex]\eta = 60\,\%[/tex]

Since [tex]\eta < \eta_{r}[/tex], the power cycle operates irreversibly.

d) Since [tex]\eta > \eta_{r}[/tex], the power cycle is impossible.

Answer:

The objective of a protection scheme is to keep the power system stable by isolating only the components that are under fault, whilst leaving as much of the network as possible still in operation.

Explanation:

The devices that are used to protect the power systems from faults are called protection devices.

Explanation:

Concentration overpotential, ηc,

I hope it helps you

Answer:

Explanation:

Minerals and rocks are the same. Aggregates of minerals form rocks. Minerals determine the texture of a rock. Most rocks are made of a single mineral type.

Answer:

- the Mach number is 0.24.

- Compressibility becomes effective when Mach number is greater than 0.3, the Mach number of the race cars is less than 0.3, hence, compressibility will not affect their aerodynamics.

Explanation:

Given the data in the question;

Average speed V = 185 miles per hour = ( 185 /2.237 ) m/s = 82.7 m/s

From Almanac, we can find that Indianapolis is at 220 m altitude.

So from table, at that altitude, the standard speed of sound will be 339.4 m/s .

Mach number of the race car will be;

Mach Number = Velocity / sound speed

we substitute

Mach Number = ( 82.7 m/s ) / ( 339.4 m/s )

Mach Number = 0.24

Therefore the Mach number is 0.24.

We know that, compressibility becomes effective when the Mach number is greater than 0.3.

Since the Mach number of the race cars is less than 0.3, compressibility will not affect their aerodynamics.

Answer:

i really font onow why tbh eot you

Given :

Power, P = 20 kW

Speed, N = 430 rpm

Allowable shear stress, τ = 65 MPa

Torque in the shaft is given by :

[tex]$P=\frac{2 \pi NT}{60}$[/tex]

[tex]$T=\frac{60 \times 20 \times 10^3}{2 \pi \times 430}$[/tex]

T = 444.37 N.m

Diameter of the solid shaft is

[tex]$d=\sqrt[3]{\frac{16 T}{\pi \tau}}[/tex]

[tex]$d=\sqrt[3]{\frac{16 \times 444.37}{3.14 \times 65}}[/tex]

[tex]$d=\sqrt[3]{34.83} $[/tex]

d = 3.265 m

d = 326.5 mm

Internal diameter of the hollow shaft is :

[tex]$\frac{T}{\frac{\pi}{32} \left( d_0^4 - d_i^4 \right)}=\frac{\tau}{d_0/2}$[/tex]

[tex]$\frac{444.37}{\frac{3.14}{32} \left( 0.036^4 - d_i^4 \right)}=\frac{65 \times 10^6}{0.036/2}$[/tex]

[tex]$\frac{444.37}{0.09 \left( 1.6 \times 10^{-6} - d_i^4 \right)}=\frac{65 \times 10^6}{0.018}$[/tex]

[tex]$\frac{7.99}{ \left( 1.6 \times 10^{-6} - d_i^4 \right)}=5850000$[/tex]

[tex]$1.3\times 10^{-6} = 1.6 \times 10^{-6} - d_i^4 \right)}$[/tex]

[tex]$d_i^4=300000$[/tex]

[tex]$d_i = 23.40$[/tex] mm

Percentage savings in the weight is given by :

Percentage saving = [tex]$\frac{W_{solid}-W_{hollow}}{W_{solid}}\times100$[/tex]

                                 [tex]$=\frac{V_{solid}-V_{hollow}}{V_{solid}}\times100$[/tex]

                                 [tex]$=\frac{d^2 - (d_0^2 - d_i^2)}{d^2} \times 100$[/tex]

                               [tex]$=\frac{(326.5)^2 - (0.036^2 - (32.40)^2)}{(326.5)^2} \times 100$[/tex]

                                 [tex]$=\frac{106602 - \left(1.29 \times 10^{-3} - 1049.76 \right)}{106602} \times 100$[/tex]

                                  [tex]$=\frac{106602 - 1049 }{106602} \times 100$[/tex]

                                  [tex]$=\frac{105553 }{106602} \times 100$[/tex]

                                  = 99.01 %

Answer: Biometrics

Explanation:

Dual factor authentication refers to an electronic authentication method whereby a user will only be granted an access to an application or a website after the user has successfully been able to present two pieces of evidence which then grants access to the application or website.

Since the technician wants to implement a dual factor authentication system, the biometrics should be implemented during the authorization stage.

Biometrics refers to the body measurements and the calculations that are related to the characteristics of humans. Biometric authentication is used as a form of identification.

Answer:

Explanation:

The main function of the fins that are to be added is to ensure the speedy transfer of heat from the Hot air.

The fins should be attached outside the tube because the convection coefficient of air is higher inside the tube than the convection coefficient of the outside air ( atmospheric air ),  BUT

When convection coefficient of air inside the tube is equal to the atmospheric air outside the tube, it is recommended that the fins can be added on both sides of the tube ( i.e. in and outside the tube )

Answer:

maybe take a really common toy kids play with or often see, find the average height for the toy and do the math to see how many of those toys stacked ontop of eachother would make up 2000 feet. For example (this isn't accurate btw just an idea of what it would sound like but) "Have you ever seen a barbie doll? well if you stack 400 barbie dolls ontop of their head it would be equal to 2000 feet."

Explanation:

sometimes taking common or beloved objects children have into your examples makes them have a better image of how small or how big something is.