find a regular expression for the language of all binary strings containing an odd number of 1’s (and any number of 0’s)

Complete Question:

Problem 8 Determine the horsepower required to compress 1 lbm/min of ethylene oxide from 70 °F and 1 atm to 250 psia. The compressor has an efficiency of 75%. The molar heat capacity of ethylene oxide is given by

C_p=10.03+0.0184T  C_p[=]Btu/lbmole- "F ; T[=] °F C,

Answer:

\(P'=0.377hp\)

Explanation:

From the question we are told that:

Initial Temperature T_1=70 F

Final Temperature \(T_2=250pisa =114.94F\)

Efficiency \(E=75\%=0.75\)

Generally the equation for Work-done  is mathematically given by

 \(W=\int C_pT\)

 \(W=10.03(114.94-70 )+0.0184((114.94)^2-70^2 )\)

 \(W=527.21btu/ibmole\)

 \(W=11.982btu/ibm\)

Generally the equation for Efficiency  is mathematically given by

 \(E=\frac{isotropic Power}{Actual P'}\)

 \(E=\frac{P}{P'}\)

Since

Isotropic Power

  \(P=0.0167*11.982btu/ibm\)

 \(P=0.2btu/s\)

Therefore

 \(P'=\frac{0.2}{0.75}\)

 \(P'=0266btu/s\)

Since

 \(1btu/s=1.4148hp\)

Therefore

 \(P'=0.377hp\)

The difference between the baseband and the broadband connection is made with the transmission of signal over the frequency.

The data transmission is given as the transfer of the data from one digital device to another. The transmission can take place with the baseband transmission and the broadband transmission.

The major difference between the two transmissions is that the baseband utilizes a complete frequency for transmission of signal, while the transmission with the broadband is made with multiple signals over multiple frequencies.

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Answer:

To calculate the steady state flux of atomic hydrogen through a steel vessel, we need to use Fick's first law, which states that the flux (J) is equal to the diffusivity (D) multiplied by the concentration gradient (dC/dx).

First, we need to calculate the concentration gradient by dividing the difference in hydrogen concentration between the inside and outside surfaces by the wall thickness of the vessel. The inside surface is kept saturated with hydrogen at a concentration of 4.5 moles/m3, and the outside surface is exposed to the atmosphere, which has a hydrogen concentration of 0 moles/m3. Therefore, the concentration gradient is (4.5 - 0) moles/m3 / (4 mm) = 1.125 moles/m3 mm.

Next, we need to substitute this value into Fick's first law along with the diffusivity of hydrogen in steel, which is given as 0.1 mm2/s. This gives us the steady state flux as J = (-0.1 mm2/s) * (1.125 moles/m3 mm) = -0.01125 moles/s mm2.

Finally, we need to convert the units of the flux from moles/s mm2 to moles/s m2. To do this, we can multiply the flux by 1,000 to convert the units of millimeters to meters, giving us a final steady state flux of -0.01125 moles/s mm2 * 1,000 = -1.125 moles/s m2.

IF THE VESSEL CONTAINS 20 MOLES OF HYDROGEN, CALCULATE THE TIME TAKEN TO DISSIPATE ALL OF THE HYDROGEN OF THAT THE VESSEL HAS A SURFACE AREA OF 3 M2.

To solve this problem, we need to first calculate the flux of atomic hydrogen through the vessel using Fick's first law:

J = -D(dC/dx)

where J is the flux, D is the diffusivity of hydrogen in steel, and dC/dx is the concentration gradient.

Given that the diffusivity of hydrogen in steel is 0.1 mm2/s, the inside concentration is 4.5 moles/m3, and the outside concentration is 0, the concentration gradient is 4.5 moles/m3.

Plugging these values into the equation above, we get:

J = -0.1 mm2/s * 4.5 moles/m3 = -0.45 moles/s-m2

Next, we need to calculate the time it takes to dissipate all 20 moles of hydrogen from the vessel. We can do this by dividing the total number of moles of hydrogen by the flux:

t = 20 moles / (-0.45 moles/s-m2) = 44.44 s

So it would take approximately 44.44 seconds to dissipate all of the hydrogen from the vessel.

Explanation:

SELF EXPLANATORY

The time taken is 44.44 seconds to dissipate all of the hydrogens from the vessel.

To solve this problem, we need to first calculate the flux of atomic hydrogen through the vessel using Fick's first law:

J = -D(dC/dx)

where J is the flux, D is the diffusivity of hydrogen in steel, and dC/dx is the concentration gradient.

Given that the diffusivity of hydrogen in steel is 0.1 mm²/s, the inside concentration is 4.5 moles/m³ and the outside concentration is 0, the concentration gradient is 4.5 moles/m³.

Plugging these values into the equation above, we get:

J = -0.1 mm²/s * 4.5 moles/m³ = -0.45 moles/s-m²

Next, we need to calculate the time it takes to dissipate all 20 moles of hydrogen from the vessel. We can do this by dividing the total number of moles of hydrogen by the flux:

t = 20 moles / (-0.45 moles/s-m2) = 44.44 s

So it would take approximately 44.44 seconds to dissipate all of the hydrogen from the vessel.

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Using the normalized distribution it is possible to calculate the value of Z score and for cases where there is a 99% chance, so Zscore is -3.6 and X= 16.45.

This curve determines the probability of the event associated with it occurring. The Gaussian distribution is the most common, hence it is known as the Normal Distribution. The area under the distribution curve is always equal to 1.0.

The formula for this distribution can be described as:

\(Z score = \frac{x-\mu}{\sigma}\)

Where:

In this case,

\(Z score =\frac{1-10}{2.5} \\Z score = -3.6\)

P-value from Z-Table will be P(x<1) = 0.00015911

For 99% we have that z will be 2.58, so

\(2.58=\frac{x-10}{2.5}\\x = 16.45\)

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Answer:

(i) \(\frac{2}{3}\)\(\pi\)rh

(ii) \(\frac{16}{3}\)\(\pi\)

Explanation:

Given:

V = \(\frac{1}{3}\)\(\pi\)r²h

Where;

V = volume of a right circular cone.

r = radius of the cone

h = height of the cone.

(i) The rate of change of V with respect to r if r changes and h remains constant is \(\frac{dV}{dr}\), and is given by finding the differentiation of V with respect to r as follow:

\(\frac{dV}{dr}\) = \(\frac{d}{dr}\)[\(\frac{1}{3}\)\(\pi\)r²h]

\(\frac{dV}{dr}\) = \(\frac{2}{3}\)\(\pi\)rh     --------------------(i)

(ii)

Given;

h = 2

r = 4

Substitute these values into equation (i) as follows;

\(\frac{dV}{dr}\) = \(\frac{2}{3}\)\(\pi\)(4 x 2)

\(\frac{dV}{dr}\) = \(\frac{2}{3}\)\(\pi\)(8)

\(\frac{dV}{dr}\) = \(\frac{16}{3}\)\(\pi\)

\(\frac{dV}{dr}\) = \(\frac{16}{3}\)\(\pi\)

A right circular cone is one where the axis of cones is the line connecting the vertex to circular base's midway, the volume of right circular cone as follows:

Formula:

\(V = \frac{1}{3} \pi r^2h\)

Where;

V = right circular cone volume  

r = Cone radius.

h = Cone height.

The calculation for part 1:

\(\frac{dV}{dr}\) is indeed the rate of change of V with reference to r when r changes but h remains constant, and it is calculated via calculating the differentiation of V with respect to r as follows:

\(\to \frac{dV}{dr} =\frac{d}{d}r [ \frac{1}{3} \pi r^2h] =\frac{2}{3} \pi r h\)

The calculation for part 2:

When  h = 2  and r = 4 then substituting the value into the part 1 equation then:

\(\to \frac{dV}{dr} = \frac{2}{3} \pi (4 \times 2) = \frac{2}{3} \pi (8) = \frac{16}{3} \pi\)

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Employ a number of physical security measures to prevent tailgating, including strong physical security measures.

When an unauthorized person enters a guarded location by trailing an authorized person, this is known as tailgating, also referred to as piggybacking. Data breaches and other cyber-attacks may occur as a result of the physical risk posed by tailgating. Tailgaters are intruders with the capacity to cause a lot of harm to a company. According to a 2020 Pokemon Institute report, physical security compromises account for 10% of malicious breaches. Tailgating is when someone illegally enters a building or another restricted area while posing as a real employee, contractor, visitor, etc. Tailgating, or unauthorized physical access, can lead to both cyber-attacks and physical property damage.

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The suffix comes from the Greek skopos, meaning "to look at" or "to examine." It is often used in medical and scientific terms, but it can also be found in everyday language.

The compound suffix form meaning an instrument for examining is -scope. A suffix is a group of letters that can be attached to the end of a word to change its meaning or create a new word. In the case of -scope, it means an instrument for examining.

The most common examples include: Microscope - an instrument for examining small objects such as cells, bacteria, and other microorganisms.Telescope - an instrument for examining objects that are far away, such as stars and planets.Stethoscope - an instrument used to listen to sounds inside the body, such as the heart or lungs.

Otoscope - an instrument used to examine the ear.Endoscope - an instrument used to examine internal organs or cavities of the body like the colon or stomach. This suffix comes from the Greek skopos, meaning "to look at" or "to examine." It is often used in medical and scientific terms, but it can also be found in everyday language.

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Answer:

There are five (5) unknown support reactions in this problem.

Explanation:

A roller joint rotates and translates along the surface on which the roller rests. The resulting reaction force is always a single force that is perpendicular to, and away from, the surface. This allows the roller to move in a single plane along the surface where it rests.

A cable support provides support in one direction, parallel, and in opposite direction to the load on it. There exists a single reaction from the cable pointed upwards.

A ball-and-socket joint have  reaction forces in all 3 cardinal  directions. This allows it to move in the x-y-z plane.

The total unknown reactions on the member are five in number.

The four key subsystems of von Neumann architecture are the central processing unit (CPU), memory, input/output (I/O) devices, and the system bus.

The CPU is responsible for executing instructions and performing calculations.
Memory stores data and instructions that the CPU needs to execute.
I/O devices allow the computer to interact with the outside world, such as keyboards and monitors.
The system bus connects all of the subsystems and allows them to communicate with each other.
The purpose of each subsystem is to work together to process and execute instructions, store and retrieve data, and interact with the user and other devices.The Von Neumann architecture consists of a single, shared memory for programs and data, a single bus for memory access, an arithmetic unit, and a program control unit. The Von Neumann processor operates fetching and execution cycles seriously

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Answer:

C. WordArt

Explanation:

WordArt feature of MS word is used for styling the text n order to make it look neater or more noticeable. These features are found in the top home menu. They feature text size, text style, text color, and options to make text bold, italic, underlined, etc. There are also many features to make text flashy, colorful, and add special effects. It can be shaded, neon, have shadows, etc.

Given the differential equation Y''' = 12We have to compute the y function by using differential equations.Using the third order differential equation

we can write three first order differential equations as follows;Let Y1

= Y, Y2

= Y' and Y3

= Y''Y1' = Y'

= Y2Y2'

= Y''

= Y3Y3'

= Y''' = 12Y1'

= Y2Y2'

= Y3Y3'

= 12 Integrating the third equation with respect to x, we get;Y3

= ∫12 dx

= 12x + C1Where C1 is a constant of integration.Differentiating with respect to x, we get;Y2

= Y'

= dY/dx

= d(Y1)/dx

= Y1'

= Y2

= dy/dxY1'

= Y2

= Y'

= dy/dxAgain differentiating with respect to x, we get;Y3

= Y''

= d²Y/dx²

= d(Y2)/dx

= d²(Y1)/dx²

= Y1''

= Y3

= d²y/dx²Y1'

= Y2

= dy/dxY2'

= Y3

= d²y/dx²Using the first two equations, we can obtain a second-order differential equation as follows;Y2'

= Y3Y1'

= Y2Y2'

= Y3Substituting the values of Y2' and Y3 in the above equation, we get;Y1'

= Y2dY2/dx

= 12d²y/dx²Simplifying the above equation, we get;d²y/dx²

= (1/12)d(Y')/dxIntegrating the above equation with respect to x, we get;dY/dx

= (1/12) d(Y')/dx + C2Where C2 is a constant of integration.Again integrating the above equation with respect to x, we get;Y = (1/12) d(Y')/dx x² + C2x + C3Where C3 is a constant of integration. Hence, the solution to the differential equation Y'''

= 12 is given by;Y

= (1/24)x² + C1x² + C2x + C3Where C1, C2, C3 are constants of integration.

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Answer:

10

Explanation:

From the information given:

The input for a fixed voltage range for SQNR when N = 7 is 32 dB

Required: To find the minimum value of N such that SQNR ≥ 46 dB

For an ADC:

SQNR = 6.02(N) + 1.76 dB

Comparing both SQNR such that SQNR₁ = 32 dB and SQNR₂ = 46 dB

Taking the ratio:

\(\dfrac{32 - 1.76}{46-1.76}= \dfrac{7}{N_{minimum}}\)

\(\dfrac{30.24}{44.24}= \dfrac{7}{N_{minimum}}\)

\(N_{minimum} \times 30.24 = 44.24 \times 7\)

\(N_{minimum} = \dfrac{44.24 \times 7}{30.24}\)

\(N_{minimum} = 10.24\)

\(N_{minimum} \simeq 10\)

Explanation:

The main sources of error in the collection of data are as follows : Due to direct personal interview. Due to indirect oral interviews. Information from correspondents may be misleading. Mailed questionnaire may not be properly answered. Schedules sent through enumerators, may give wrong information.

The inequality which can be used to determine s, the number of sets of spoons Christopher could buy for the restaurant to have enough spoons is

35 + 10s ≥ 75

An inequality is simply used to show the relationship between the expressions that aren't equal.

The restaurant needs at least 75 spoons and there are currently 35 spoons and each set on sale contains 10 spoon.

Let each set be represented as s. This will be illustrated as:

35 + (10 × s) ≥ 75

35 + 10s ≥ 75

Collect like terms

10s ≥ 75 - 30

10s ≥ 45

Divide

s ≥ 45/10

s ≥ 4.5

The restaurant must have at least 5 sets.

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The inequality that can be used to determine s, the number of sets of spoons Christopher could buy for the restaurant, is 35 + 10s ≥ 75.

Inequality is a term used to show two relations between two values that are not equal to each other.

Given, the restaurant needs at least 775 spoons. They have 355 spoons and each set contains 10 spoons.

Then the expression is ;

35 + (10 × s) ≥ 75

35 + 10s ≥ 75

Like terms are extracted

10s ≥ 75 - 30

10s ≥ 45

s ≥ 45/10

s ≥ 4.5

Thus, the inequality in the number of sets of spoons is 35 + 10s ≥ 75.

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Answer:

871.8008 KJ/kg

2573.42085 KJ/kg

0.3743

Explanation:

Solution:-

- We are to analyze an ideal Rankine cycle, where the condenser and boiler operating temperatures are defined.

- We start of by evaluating the properties of water at each state before and after the process.

State 1: Condenser Exit / Pump Inlet

           T1 = 40°C   ---> P1,sat = 7.3851 KPa , v1 = vf = 0.001008 m^3/kg

            sat liquid         h1 = hf = 167.53 KJ/kg , s1 = sf = 0.5724 KJ/kg.K

            P1 = condenser pressure

State 2: Pump Exit / Boiler Inlet

           P2 = P3 = Psat,300°C = 8587.9 KPa

Process 1: " Isentropic Compression - constant volume "

The work done by pump in the compression process is:

          wp = v1* ( P2 - P1 )

          wp = ( 0.001008 ) * ( 8587.9 - 7.3851 )

          wp = 8.64915 KJ /kg

Determine the enthalpy at " State 2 " by energy balance on pump ( control  Volume) :

          h2 = h1 + wp

          h2 = 167.53 + 8.64915

          h2 = 176.17915 KJ/kg

State 3: Boiler Exit / Turbine Inlet

          T1 = 300°C   ---> P3,sat = 8587.9 KPa

           sat vapor         h3 = hg =  2749.6 KJ/kg , s3 = sg =  5.7059 KJ/kg.K

           P3 = Boiler pressure

Process 2: " Heat Addition - constant pressure "

The heat supplied in the boiler is:

          qb = h3 - h2

          qp = ( 2749.6 - 176.17915 )

          qb = 2573.42085 KJ /kg    .... Answer ( b )

State 4: Turbine Exit / Condenser Inlet ( Isentropic )

          P4 = P1 = 7.3851 KPa         ..... sfg = 7.685 KJ/kg.K

          s4 = s3 = 5.7059 KJ/kg.K       sf = 0.5716994 KJ/kg.K

                                                           hf = 170.32524 KJ/kg

                                                          hfg = 2406.11 KJ/kg

Process 3: Isentropic Expansion - Determine the quality of liquid-vapor mixture phase ( x ) at state (4):

          x = (s4 - sf) / sfg

          x = (5.7059 - 0.5716994) / 7.685

          x = 0.66808

          h4 = hf + x*hfg

          h4 = 170.32524 + 0.66808*2406.11

          h4 = 1777.79920 KJ/kg

- The work-done by the turbine in the isentropic expansion process ( wt ) is:

          wt = h3 - h4

          wt = 2749.6 - 1777.7992

          wt = 971.8008 KJ/kg   ... Answer ( a )

- To determine the thermal efficiency ( nth ) of the rankine cycle. We need to determined the net work produced by the cycle ( wn ). The net work is the energy balance between the isentropic compression ( work done - pump ) and isentropic expansion ( work produced - turbine ):

          wn = wt - wp

          wn = 971.8008 - 8.64915

          wn = 963.15165 KJ/kg

- The thermal efficiency of a power cycle is the ratio of net work-produced ( wn ) and the heat supplied to the working fluid in the boiler ( qb ) as follows:

         nth = wn / qb

         nth = 963.15165 / 2573.42085

         nth = 0.3743    ..... Answer ( c )

The selection of shaft keys is critical in preventing premature failure of keyed joints. Shaft keyways and keys are used to transmit torque from shafts to mechanical transmission elements such as gears, pulleys, and so on.

A key to preventing early failure of keyed joints is choosing the right shaft key. The use of a keyed joint allows for the transmission of torque from shafts to mechanical transmission components like gears, pulleys, etc. They can be produced using a standard stock material, like key stock, or they can be specially machined to fit the application.

According to various standards like BS4235, the nominal shaft diameter is typically used to specify the key size, and the commonly accessible rectangular key is used for the majority of applications. As a result, the standards do not specify the key material or joint limitations, and a keyed joint is oversized to support all loads.

There are four main groups of shaft keys:

Stainless steel or medium carbon steel are typically used to make shaft keys. To suit various application environments, they can be made from a wide variety of materials, including bronze, copper, brass, and aluminum alloy. For example, stainless steel grade for use in food service equipment and brass or bronze keys for marine propeller shafts.

Key steel is typically supplied in accordance with BS46 and BS4235 and is a medium carbon steel that is unalloyed and has a respectable tensile strength. Due to their ideal blend of strength, toughness, and favorable machining properties, unalloyed medium carbon steels with carbon contents ranging from 0.25% to 0.60% are used.

The table below lists a few popular shaft key materials along with their Ultimate Tensile Strength (UTS). ↓↓↓

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A project is meant to achieve clearly defined goals within present constraints

The correct option for the source of work in a project which is NOT a source of authorized work to be performed is option (a)

a. Scope creep

The reason for the selecting the above option is as follows;

In project management, alongside work planned in a work package, authorized work to be performed can come from;

A scope creep, however, is generally is due to the inclusion of the end

users of the project at a time that is behind schedule, insufficient analysis of

the requirements, and project complexity underestimation. and consists of

changes made after a project has commenced

A scope creep therefore, changes the amount of work to be done which

therefore alters the project timeline, and due to project, budgetary, and

end user constraints, CANNOT be a source of authorized work to be

performed

The correct option for the source that is NOT a authorized work to be performed is option a. Scope creep

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Answer:

it involves the detection of light emitted by a sample after it has absorbed light of a specific wavelength.

Step-by-step:

Fluorescence spectroscopy is more sensitive than UV-Vis absorption spectroscopy because it involves the detection of light emitted by a sample after it has absorbed light of a specific wavelength. This emission of light is called fluorescence.

In UV-Vis absorption spectroscopy, the amount of light absorbed by a sample is measured as a function of wavelength. The amount of light absorbed is proportional to the concentration of the absorbing species in the sample, but the sensitivity of this technique is limited by the intensity of the incident light and the path length of the sample.

On the other hand, fluorescence spectroscopy measures the emission of light that occurs when the excited molecules return to their ground state. This emitted light is typically at a longer wavelength than the absorbed light, and it is much weaker than the incident light. However, the sensitivity of fluorescence spectroscopy is enhanced by the fact that the emitted light is measured at right angles to the excitation light, which reduces background noise from scattered light and improves the signal-to-noise ratio.

Additionally, fluorescence spectroscopy can be more selective than UV-Vis absorption spectroscopy because it can detect specific molecular species based on their unique fluorescence spectra. This selectivity is due to the fact that the fluorescence emission spectra of different molecules can be quite distinct, even for molecules with similar UV-Vis absorption spectra.

Overall, the increased sensitivity and selectivity of fluorescence spectroscopy make it a powerful technique for the detection and quantification of trace amounts of fluorescent molecules in complex samples.

Hope this helps!

Answer:

Fluorescence and UV-Vis absorption spectroscopy are two commonly used analytical techniques in the field of chemistry. Both techniques rely on the interaction of light with molecules to provide information about their electronic structure and chemical properties. However, fluorescence is generally considered to be more sensitive than UV-Vis absorption spectroscopy for several reasons.

Firstly, fluorescence is a more selective technique than UV-Vis absorption spectroscopy. When a molecule absorbs light, it undergoes a transition from its ground state to an excited state. This transition can occur via a number of different pathways, depending on the energy of the absorbed light and the electronic structure of the molecule. In contrast, fluorescence occurs when a molecule emits light after being excited by light of a specific wavelength. This means that fluorescence only occurs when certain conditions are met, such as the presence of specific functional groups or the correct excitation wavelength. As a result, fluorescence is generally more selective than UV-Vis absorption spectroscopy, which can detect any absorbing species in a sample.

Secondly, fluorescence is generally more sensitive than UV-Vis absorption spectroscopy because it produces a larger signal-to-noise ratio. In UV-Vis absorption spectroscopy, the signal is proportional to the concentration of absorbing species in the sample. However, the signal is also affected by other factors such as path length, instrument sensitivity, and background noise. In contrast, fluorescence produces a much stronger signal because it involves emission of light at a different wavelength than that used for excitation. This means that background noise and other interfering factors are less likely to affect the signal-to-noise ratio in fluorescence measurements.

Finally, fluorescence is more sensitive than UV-Vis absorption spectroscopy because it can detect lower concentrations of analyte in a sample. This is because fluorescence is an amplification process – each absorbed photon can result in multiple emitted photons if the molecule undergoes multiple cycles of excitation and emission. This amplification effect means that even low concentrations of fluorescent analyte can produce a measurable signal, whereas in UV-Vis absorption spectroscopy, the signal becomes weaker as the concentration of analyte decreases.

In summary, fluorescence is more sensitive than UV-Vis absorption spectroscopy because it is a more selective technique, produces a larger signal-to-noise ratio, and can detect lower concentrations of analyte in a sample.

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Answer:

rain

Explanation:

evaoration causes clouds

clouds condense and rain

Answer:

rain

Explanation:

The velocity after 5 seconds can be calculated using the equation:

v = v0 + at, where v0 is the initial velocity (0 m/s) and a is the acceleration defined by the equation a = 9.81[1 - v^2 (10^-4)].

To find the terminal velocity, we need to set the acceleration to zero and solve for v.

0 = 9.81[1 - v^2 (10^-4)]

v^2 = 9.81 / 10^-4

v = sqrt (9.81 / 10^-4) = 99.97 m/s

(a) The velocity when t = 5 s is approximately 99.97 m/s.

(b) The body's terminal or maximum attainable velocity is 99.97 m/s

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Answer:

10.5

Explanation:

Convert to an equation for example P%* X=Y

P is 7.5% X is 140, so the equation Is 7.5 percent * 14= Y

convert 7.5% Into a decimal by removing the percent sign and deviding by 7.5/100= 0.075

Substitute 0.075 for 7.5% in the equation: 7.5%*140=Y becomes 0.075*140= 10.5

Answer:

hey. its a big question. solved from *c hegg

Explanation:

A force of 50 pounds applied over a distance of 20 feet in 2 minutes generates a power of 500 foot-pounds per minute. This indicates the rate at which work is done or energy is transferred in the system.

Force = 50 pounds

Distance = 20 feet

Time = 2 minutes

To calculate power, we use the formula:

Power = (Force × Distance) / Time

Substituting the given values in the formula:

Power = (50 pounds × 20 feet) / 2 minutes

Power = (1000 foot-pounds) / 2 minutes

Power = 500 foot-pounds per minute

Therefore, the correct answer is option a: 500.

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Answer:

yes it is very hard you should find a reccomended doctor to aid in your situation. But in the meantime how about you give me that lil brainliest thingy :p

Answer:

Energy Saved = 6.93 x 10⁹ Btu

Cost Saved = $ 30145.5

Explanation:

The energy generated by each boiler can be given by the following formula:

\(Annual\ Energy = (Heat\ In)(Combustion\ Efficiency)(Operating\ Hours)\)

Now, the energy saved by the increase of efficiency through tuning will be the difference between the energy produced before and after tuning:

\(Energy\ Saved = (Heat\ In)(Efficiency\ After\ Tune - Efficiency\ Before\ Tune)(Hours)\)\(Energy\ Saved = (5.5\ x\ 3\ x\ 10^{6}\ Btu/h)(0.8-0.7)(4200\ h)\)

Energy Saved = 6.93 x 10⁹ Btu

Now, for the saved cost:

\(Cost\ Saved = (Energy\ Saved)(Unit\ Cost)\\Cost\ Saved = (6.93\ x\ 10^{9}\ Btu)(\$4.35/10^{6}Btu)\\\)

Cost Saved = $ 30145.5

The Cisco rollover standard is typically used in a rollover cable.

A rollover cable is a cable with an RJ-45 connector at one end and a console connector at the other end used to connect to a console port on networking equipment, such as routers and switches.

The console port is a special serial port used for device configuration and management. Rollover cables are wired differently than other types of Ethernet cables and typically follow the Cisco rollover standard (also known as Yost). This standard uses the following pinout configuration: Pin 1 to Pin 8, Pin 2 to Pin 7, Pin 3 to Pin 6, Pin 4 to Pin 5, Pin 5 to Pin 4, Pin 6 to Pin 3, Pin 7 to Pin 2, Pin 8 to Pin 1.

This allows the transmit and receive pins to be crossed over, enabling the device and console to communicate.

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Answer:

Question 1 A 1020 Cold-Drawn steel shaft is to transmit 20 hp while rotating at 1750 rpm. Calculate the transmitted torque in lbs. in. Ignore the effect of friction. Answer with three decimal points. 60.024 Question 2 Based on the maximum-shear-stress theory, determine the minimum diameter in inches for the shaft in Q1 to provide a safety factor of 3. Assume Sy = 57 Kpsi. Answer with three decimal points. 0.728 Question 3 If the shaft in Q2 was made of ASTM 30 cast iron, what would be the factor of safety? Assume Sut = 31 Kpsi, Suc = 109 Kpsi 0 2.1 O 2.0 O 2.5 0 2.4 2.3 O 2.2

Explanation:

hope it helps

Answer:

the theoretical fracture strength of the brittle material is 5.02 × 10⁶  psi

Explanation:

Given the data in the question;

Length of surface crack α = 0.25 mm

tip radius ρ\(_t\) = 1.2 × 10⁻³ mm

applied stress σ₀ = 1200 MPa

the theoretical fracture strength of a brittle material = ?

To determine the the theoretical fracture strength or maximum stress at crack tip, we use the following formula;

σ\(_m\) = 2σ₀\((\) α / ρ\(_t\) \()^{\frac{1}{2}\)

where α is the Length of surface crack,

ρ\(_t\) is the tip radius,

and σ₀ is the applied stress.

so we substitute

σ\(_m\) = (2 × 1200 MPa)\((\) 0.25 mm / ( 1.2 × 10⁻³ mm ) \()^{\frac{1}{2}\)

σ\(_m\) = 2400 MPa × \((\) 208.3333 \()^{\frac{1}{2}\)

σ\(_m\) = 2400 MPa × 14.43375

σ\(_m\) = 34641 MPa

σ\(_m\) = ( 34641 × 145 )psi

σ\(_m\) = 5.02 × 10⁶  psi

Therefore, the theoretical fracture strength of the brittle material is 5.02 × 10⁶  psi