Design a second-order high-pass filter for each case below and state its transfer function H(s):
a) k=1, ω0= 1300 rad/s and Q=0.707
b) k=1, ω0= 950 rad/s and Q=0.8
Assume L=1H
Table: Second order RLC filters

Answer:

relating to the chemical processes and substances which occur within living organisms. .-.

Explanation:

Answer and Explanation:

Basically, the word "biochemical" means something connecting with the chemical process and/or substance that is happen internally within some sort of living thing.

Hope this helps.

This question is incomplete, the complete question is;

Illinois furniture , Inc produces all types of office furniture. The "Executive Secretary" is a chair that has been designed using ergonomics to provide comfort during long work hours the chair sells for $130. There are 480 minutes available during the day and the average daily demand has been 50 chairs. There are eight tasks.

TASK     PERFORMANCE TIME( MIN )     TASK MUST FOLLOW TASK LISTED

A                            4                                                     ---------

B                            7                                                     ----------

C                            6                                                        A,B

D                            5                                                          C

E                            6                                                          D

F                            7                                                           E

G                            8                                                          E

H                            8                                                          F,G

1) What is the cycle time for operation?

2) What is the theoretical minimum number of workstation?

Answer:

1) the cycle time for operation is 9.6 min

2) the theoretical minimum number of workstation is 5

Explanation:

Given the data in the question;

production time per day = 480 minutes

average daily demand = 50

Given the data in the question;

1) cycle time for operation

this is simply referred to as the total time for the process from start to finish.

cycle time = production time per day / units demand per day

we substitute

cycle time = 480 min / 50

cycle time = 9.6 min

Therefore, the cycle time for operation is 9.6 min

2) theoretical minimum number of workstation.

theoretical minimum number of workstation = total task time / cycle time

Total task time = ( 4 + 7 + 6 + 5 + 6 + 7 + 8 + 6 ) = 49 min

∴  theoretical minimum number of workstation = 49 min / 9.6 min

theoretical minimum number of workstation = 5.104 ≈ 5

Therefore, the theoretical minimum number of workstation is 5

Answer:

The wheel will go through 1146 revolutions in 5 minutes.

Explanation:

We can the formula:

\(\boxed{\omega = \frac{2 \pi}{T}}\)

where

ω ⇒ angular speed (24 rad/s),

T ⇒ time period (? s),

and solve for T to find the time it takes for the wheel to complete one revolution.

⇒  \(24 = \frac{2 \pi}{T}\)

⇒  \(T = \bf \frac{2 \pi}{24}\)  s

This means it takes \(\bf \frac{2 \pi}{24}\) seconds for the wheel to complete one revolution.

Now, using the unitary method,

In \(\frac{2 \pi}{24}\) seconds          ⇒    1 revolution completed

In 1 second              ⇒    1 ÷ \(\frac{2 \pi}{24}\)   =  \(\frac{24}{2 \pi}\)  revolutions completed

In (5 × 60 = ) 300s  ⇒     \(\frac{24}{2 \pi}\)   ×  300   = 1145.9

                                                            ≅ 1146 revolutions completed

Answer:A

Explanation:

Those who want to save money and will use the product for only a few years

Answer:

THE ANSWER is A - those who want to save money and will use the product for only a few years

Explanation: Got it right on edg 2021

Vacuum into a recovery/recycling machine, empty disposable R-134a refrigerant containers should be disposed of.

An area devoid of the matter is called a vacuum. The adjective vacuum, which means "empty" or "void," is the source of the word. A region with a gaseous pressure significantly below atmospheric pressure might be thought of as an approximate representation of such a vacuum. The term "partial vacuum" is used to describe an imperfect vacuum, such as one that might exist in a laboratory or in space, and is frequently used when physicists discuss the ideal test outcomes that would take place in a perfect vacuum, which they sometimes simply refer to as "vacuum" or free space. Contrarily, any area where the pressure is significantly lower than atmospheric pressure is referred to as a vacuum in engineering and applied physics. A space that envelops an object is referred to as being in vacuo in Latin.

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Answer:

see attached

Explanation:

if you are looking for the correct measurement... see attached image

Answer:

Shut off cylinder valve before turning off forklift.

Explanation:

To reduce the pressure in the fuel line, the forklift operator should keep the engine running with the garage door closed and sealed.

In warehouses and on construction sites, a forklift driver is in charge of delivering, moving, loading, and unloading a range of commodities.

Always release the pressure in the LPG lines before disconnecting while filling LPG tanks. To accomplish this, shut off the tank valve and let the engine run until the lines are completely empty.

Before opening the valve on the new tank, shut off the engine and make sure the LPG tanks are fastened to the forklift.

Therefore, the forklift driver should keep the engine running with the garage door closed and sealed in order to lower the pressure in the fuel line.

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Answer:

Email etiquette is important

Explanation:

It is important to understand how to use correct email etiquette because it helps you communicate more clearly. It also makes you seem a bit more professional too. For example depending in who you're emailing like say you're emailing your teacher for help then here's how it'd go:

Dear(teacher name, capitalize, never use first name unless they allow it)

Hello (teacher name), my name is (first and last name) from your (number class) and I was wondering if you could please help me out with (situation, be clear on what you need help with otherwise it won't get through to them)? If you could that would be greatly appreciated!

Sincerely,

(your name first and last)

The asymptotes of the open loop transfer are:

Horizontal: y = 0

Vertical: x = -10 and x = -100

The open loop transfer function is given as:

f(s) = 100(s + 1)/((s + 10)(s + 100))

Set the numerator of the function to 0.

So, we have:

f(s) = 0/((s + 10)(s + 100))

Evaluate

f(s) = 0

This means that, the vertical asymptote is y = 0

Set the denominator of the function to 0.

(s + 10)(s + 100)  0

Split

s + 10 = 0 and s + 100 = 0

Solve for s

s = -10 and s = -100

This means that, the horizontal asymptotes are s = -10 and s = -100

See attachment for the graph of the asymptotes.

Therefore, The asymptotes of the open loop transfer are:

Horizontal: y = 0

Vertical: x = -10 and x = -100

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Objective function: Minimize the total cost of crude oilCost = Cost per barrel * Number of barrelsMinimize:  Cost =  (Cost per barrel of Brent * x1) + (Cost per barrel of Dubai * x2) + (Cost per barrel of WTI * x3)

After setting up the spreadsheet, you would use Solver, an add-in in Excel, to find the optimal solution. Solver will adjust the values in the x1, x2, and x3 cells to minimize the objective function while satisfying all the constraints. The Solver's answer report will provide information on the optimal solution, including the values for x1, x2, and x3, as well as the minimum cost achieved.

b. To set up the corresponding LP in Excel and run Solver, you would simply exclude the availability constraint for Brent oil. The objective function, cost per barrel, and composition constraints would remain the same as in Q1. By running Solver, you can find the new optimal blending plan with unlimited Brent oil availability, which would result in a potentially lower total cost.

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They were made of stone blocks that were held together with lime mortar. Some were plastered over and painted.

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Yes, in SolidWorks, you can fully define the shape and size of a sketch using only dimensions and geometric relations.

These dimensions and relations allow you to control the exact placement and orientation of each element within the sketch, ensuring that it meets your design requirements. By defining the sketch in this way, you can easily make changes to the design by simply adjusting the dimensions and relations, rather than having to recreate the entire sketch from scratch. This makes SolidWorks a powerful tool for creating precise and accurate 3D models.

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To prove the relationship between the mean coefficient of friction (μ) and the shear angle (x), given that the frictional force (F) on the tool rake face is equal to KtA, where K is a constant, A is the area of the cross-section of the chip, and a is the rake angle, we can follow these steps:

Step 1: Express the frictional force in terms of the mean coefficient of friction:

The frictional force, F, is equal to the product of the mean coefficient of friction (μ) and the normal force (N), which is equal to KtA:

F = μN = μKtA.

Step 2: Calculate the normal force, N:

The normal force can be determined using trigonometry. Since a is the rake angle, the component of the normal force in the direction of the shear force is N cos(x - a).

Step 3: Equate the frictional force with the calculated normal force:

Setting F = N cos(x - a), we get:

μKtA = N cos(x - a).

Step 4: Substitute the expression for the normal force:

μKtA = (N cos(x - a)).

Step 5: Rearrange the equation to solve for μ:

Divide both sides by N cos(x - a):

μ = KtA / (N cos(x - a)).

Step 6: Express N in terms of K and A:

Using trigonometry, we find that N = K sin(x - a).

Step 7: Substitute N into the equation:

μ = KtA / ((K sin(x - a)) cos(x - a)).

Step 8: Simplify the expression:

μ = K cos^2(x - a) / (K sin(x - a) cos(x - a) + 1).

Therefore, we have derived the relationship between the mean coefficient of friction (μ) and the shear angle (x) as μ = K cos^2(x - a) / (K sin(x - a) cos(x - a) + 1), where K is a constant, A is the area of the cross-section of the chip, and a is the rake angle.

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Answer:

ser gay

Explanation:

no es malo y se fjfjdj gracias

The acceleration of gravity is g = 32.2 ft/s2.

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The air pressure surrounding tank b is lower than the pressure inside tank b, the pressure gauge a reads 5 psig (vacuum). The water column in the U-tube manometer is 10 inches high, which translates to a pressure difference of 10 inches of mercury, or 5 psig, when the water is in contact with the air in tank b. As a result, the absolute pressure inside tank b is 14.7 psia minus 5 psig = 9.7 psia, whereas the absolute pressure inside tank an is 9.7 psia plus 5 psig = 14.7 psia. This is the same as the pressure in the air around tank b.

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The concentration of the salt in the w crude oil (with salt) to water used is 4:1 is 0.0061.

What is crude oil?

Crude oil is a hydrocarbon mixture that exists in liquid form in natural subterranean reservoirs and stays liquid at atmospheric pressure after passing through surface separation facilities.

The salt content of "spent" wash water is 15%.

5% of crude oil is salt.

The crude oil (with salt) to water ratio utilised is 4: 1.

Crude oil fraction:

95% of crude oil = 0.95

0.95F1 = x1E1

0.95(100) = x1E1

95= x1E1

E1 = 95/X1

wash water fraction:

15% of wash water = 100% -15% = 85%

= F2 = 0.85E2

salts fraction:

0.05F1 = (1-X1)E1 + 0.15E2

0.05F1 = (1-X)95/X1 + 0.15(F2/0.85)

0.05F1 = 95/X1 - 95 + 0.15(F2/0.85) (SINCE = 95.588 = 95/X1 = 99.38)

0.05(100) = 95/X1 - 95 + 17.61(1/4F1)

5 = 95/X1 - 95 + 4.41

Fraction of salts = 1- X1 = 0.0061

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Perforated wavy fins in heat pump type air conditioners have been studied both through modeling and experimental investigations to analyze their comprehensive performance under frosting and non-frosting conditions.

Perforated wavy fins play a crucial role in enhancing the performance of heat pump type air conditioners, particularly in frost-prone environments. These fins are designed with a wavy pattern and small perforations, which provide several benefits. Firstly, the wavy shape increases the surface area of the fins, facilitating better heat transfer between the refrigerant and the surrounding air. This results in improved overall efficiency of the heat pump system.

Secondly, the perforations on the fins help to mitigate frost formation. When operating in frost-prone conditions, the fins tend to accumulate frost, which reduces their heat transfer effectiveness. However, the perforations allow some of the frost to pass through, preventing excessive buildup and maintaining the performance of the fins.

Furthermore, the comprehensive performance analysis of these perforated wavy fins involves studying parameters such as heat transfer coefficient, pressure drop, energy consumption, and defrosting effectiveness. This analysis aids in optimizing the design and operation of heat pump type air conditioners for different climatic conditions.

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Answer: A. Makes a commodity available for sale or exchange

Explanation: hope it helps ^w^

Answer:

- the process should operate each day for 7.9286 hours

- the amount of brine that must be fed to the evaporator is 94.594 kg/hr

Explanation:

Given that;

concentration of brine = 26%

so water concentration will be 100% - 26% = 74%

for evaporation of 70kg water per hour, residual is pure salt( 0% moisture)

so mass flow rate of brine = 70/(74%) = 70/0.74 = 94.5945 kg/hr

Amount of pure dry salt produced(0%) = 94.5945 - 70 = 24.5945 kg/hr

Now for production of 195kg of pure dry salt, number of hours required will be

T = 195 / 24.5945 = 7.9286 hrs

Therefore the process should operate each day for 7.9286 hours.

Total brine solution required ( 26% salt conc.) = 195/0.26 = 750 kg

Feed rate of brine solution ( 26% salt conc.) = 750 / 7.9286 = 94.594 kg/hr

Therefore the amount of brine that must be fed to the evaporator is 94.594 kg/hr

Answer:

150

Explanation:

Mark me Brainliest

Answer:

you have told the answer already as 150 cm

Explanation:

here is your answer hope you will enjoy mark me as brain list

thank you

The general solution to the differential equation is y =  c1 e^(x/2) cos(√3x/2) + c2 e^(x/2) sin(√3x/2) + (1/10) e^(x/2) cos(x/2) - (3/10) e^(x/2) sin(x/2)

What is the general solution of the equation

To solve the differential equation y''-2y+4y = e^(x/2) cos(x/2), we first need to find the characteristic equation. The characteristic equation is obtained by assuming that the solution to the homogeneous part of the equation (y''-2y+4y=0) is of the form y = e^(rx).

Substituting this into the homogeneous part of the equation, we get:

r^2 e^(rx) - 2re^(rx) + 4e^(rx) = 0

Dividing both sides by e^(rx) (since e^(rx) is non-zero), we get:

r^2 - 2r + 4 = 0

Solving for r using the quadratic formula, we get:

r = (2 ± √(4-4(1)(4))) / 2

r = 1 ± i√3

Therefore, the homogeneous solution is y_h = c1 e^(x/2) cos(√3x/2) + c2 e^(x/2) sin(√3x/2).

Now, we need to find a particular solution to the non-homogeneous part of the equation, which is of the form e^(x/2) cos(x/2).

We can use the method of undetermined coefficients to find a particular solution. Since the non-homogeneous part of the equation is of the form e^(x/2) cos(x/2), we assume that the particular solution is of the form:

y_p = Ae^(x/2) cos(x/2) + Be^(x/2) sin(x/2)

Taking the first and second derivatives of y_p, we get:

y_p' = (A + B/2) e^(x/2) cos(x/2) + (-A/2 + B) e^(x/2) sin(x/2)

y_p'' = (-A/2 - B/4) e^(x/2) cos(x/2) + (A/4 - B/2) e^(x/2) sin(x/2)

Substituting these expressions into the original equation and equating coefficients of like terms, we get:

(-A/2 - B/4 + A) e^(x/2) cos(x/2) + (A/4 - B/2 + B) e^(x/2) sin(x/2) - 2(Ae^(x/2) cos(x/2) + Be^(x/2) sin(x/2)) + 4(Ae^(x/2) cos(x/2) + Be^(x/2) sin(x/2)) = e^(x/2) cos(x/2)

Simplifying and solving for A and B, we get:

A = 1/10

B = -3/10

Therefore, the particular solution is:

y_p = (1/10) e^(x/2) cos(x/2) - (3/10) e^(x/2) sin(x/2)

The general solution to the differential equation is then given by the sum of the homogeneous and particular solutions:

y = y_h + y_p

= c1 e^(x/2) cos(√3x/2) + c2 e^(x/2) sin(√3x/2) + (1/10) e^(x/2) cos(x/2) - (3/10) e^(x/2) sin(x/2)

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Spheres come to rest in contact:

- both gray, charges -q and +q

- both orange, charges -q and +q

- both orange, charges +Q and -q

- gray, charge -Q, orange, chargef +q

Spheres come to rest separated

- both gray, +Q and +q

- both gray, -Q and +q

- both orange, -q and -Q

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The weight of the spheres is small enough that electrostatic forces are significant. Spheres come to rest in contact:

- both gray, charges -q and +q

- both orange, charges -q and +q

- both orange, charges +Q and -q

- gray, charge -Q, orange, charge f +q

Spheres come to rest separated

- both gray, +Q and +q

- both gray, -Q and +q

- both orange, -q and -Q

The distance between two charges creates an electrostatic force. The distance between two charges and their respective magnitudes determine the magnitude of the electrostatic force. When two charges, either positive or negative, are brought together, they repel one another.

The phenomena and properties of electric charges that are either stationary or moving slowly are the focus of the physics branch known as electrostatics. Coulomb's law describes electrostatic phenomena, which are caused by the forces that electric charges exert on one another.

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Answer:

a

Explanation:

becasue it is

The corresponding temperature at the center of the sphere after approximately 0.309 seconds is approximately 198.83°C.

To determine the time required for a sphere near the inlet of the system to accumulate 90% of the maximum possible thermal energy, we can use the concept of thermal diffusion through a sphere. The time required can be calculated using the equation for the thermal diffusion time constant:

τ = (ρ * c * r^2) / (4 * k)

where:

τ is the thermal diffusion time constant,

ρ is the density of the sphere material (in this case, aluminum) = 2700 kg/m^3,

c is the specific heat capacity of the sphere material (in this case, aluminum) = 950 J/kg⋅K,

r is the radius of the sphere (diameter/2) = 75 mm / 2 = 37.5 mm = 0.0375 m,

k is the thermal conductivity of the sphere material (in this case, aluminum) = 240 W/m⋅K.

Substituting these values into the equation, we can calculate the thermal diffusion time constant:

τ = (2700 kg/m^3 * 950 J/kg⋅K * (0.0375 m)^2) / (4 * 240 W/m⋅K)

Now we can solve for τ:

τ ≈ 0.309 seconds

The thermal diffusion time constant represents the time required for a sphere to reach approximately 63.2% of the maximum possible thermal energy. To calculate the time required to accumulate 90% of the maximum energy, we can use the following relation:

t = τ * ln((90% - 63.2%) / (100% - 63.2%))

Substituting the values into the equation:

t = 0.309 seconds * ln((90% - 63.2%) / (100% - 63.2%))

t ≈ 0.309 seconds * ln(0.271 / 0.368)

t ≈ 0.309 seconds * ln(0.736)

t ≈ 0.309 seconds * (-0.305)

t ≈ -0.094 seconds

The negative value obtained implies that 90% of the maximum possible thermal energy cannot be accumulated within a sphere near the inlet of the system. This suggests that the system may need additional time or adjustments to reach the desired energy level.

To calculate the corresponding temperature at the center of the sphere, we can use the concept of one-dimensional transient heat conduction through a sphere. The equation for this scenario is:

T = Ti + (Tg,i - Ti) * (1 - exp(-t / τ))

where:

T is the temperature at the center of the sphere at time t,

Ti is the initial temperature of the spheres = 25°C,

Tg,i is the gas temperature at the inlet of the system = 300°C,

t is the time,

τ is the thermal diffusion time constant calculated earlier.

Let's calculate the corresponding temperature at the center of the sphere when t = 0.309 seconds

T = 25°C + (300°C - 25°C) * (1 - exp(-0.309 / 0.309))

T ≈ 25°C + 275°C * (1 - exp(-1))

T ≈ 25°C + 275°C * (1 - 0.3679)

T ≈ 25°C + 275°C * 0.6321

T ≈ 25°C + 173.8275°C

T ≈ 198.8275°C

Therefore, the corresponding temperature at the center of the sphere after approximately 0.309 seconds is approximately 198.83°C.

Now, let's consider the advantage of using copper instead

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Strength tests are conducted by engineers to ascertain how much weight a material can support.

Thus, Some materials are more ductile than others, which means they will bend under strain before breaking more readily than brittle materials will. When subjected to the greatest stress, brittle materials simply snap.

The yield strength of ductile materials is used to calculate the safety factor. The ultimate strength is used by brittle materials.

A safety factor lowers the possibility of a product failing and raises people's sense of security. The safety component is crucial when it comes to fall protection and safety equipment. There is a chance of harm and death from a building failure, as well as financial loss for a business.

Thus, Strength tests are conducted by engineers to ascertain how much weight a material can support.

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Answer: the minimum mass flow rate of air required to generate a power output of 105 MW is 238.2 kg/s

Explanation:

from the T-S diagram, we get the overall pressure ratio of the cycle is 9

Calculate the pressure ratio in each stage of compression and expansion. P1/P2 = P4/P3  = √9 = 3

P5/P6 = P7/P8  = √9 =3  

get the properties of air from, "TABLE A-17 Ideal-gas properties of air", in the text book.

At temperature T1 =300K

Specific enthalpy of air h1 = 300.19 kJ/kg

Relative pressure pr1 = 1.3860  

At temperature T5 = 1200 K

Specific enthalpy h5 = 1277.79 kJ/kg

Relative pressure pr5 = 238  

Calculate the relative pressure at state 2

Pr2 = (P2/P1) Pr5

Pr2 =3 x 1.3860 = 4.158  

get the two values of relative pressure between which the relative pressure at state 2 lies and take the corresponding values of specific enthalpy from, "TABLE A-17 Ideal-gas properties of air", in the text book.  

Relative pressure pr = 4.153

The corresponding specific enthalpy h = 411.12 kJ/kg  

Relative pressure pr = 4.522

The corresponding specific enthalpy h = 421.26 kJ/kg  

Find the specific enthalpy of state 2 by the method of interpolation

(h2 - 411.12) / ( 421.26 - 411.12) =  

(4.158 - 4.153) / (4.522 - 4.153 )

h2 - 411.12 = (421.26 - 411.12) ((4.158 - 4.153) / (4.522 - 4.153))  

h2 - 411.12 = 0.137

h2 = 411.257kJ/kg  

Calculate the relative pressure at state 6.

Pr6 = (P6/P5) Pr5

Pr6 = 1/3 x 238 = 79.33  

Obtain the two values of relative pressure between which the relative pressure at state 6 lies and take the corresponding values of specific enthalpy from, "TABLE A-17 Ideal-gas properties of air", in the text book.  

Relative pressure Pr = 75.29

The corresponding specific enthalpy h = 932.93 kJ/kg  

Relative pressure pr = 82.05

The corresponding specific enthalpy h = 955.38 kJ/kg  

Find the specific enthalpy of state 6 by the method of interpolation.

(h6 - 932.93) / ( 955.38 - 932.93) =  

(79.33 - 75.29) / ( 82.05 - 75.29 )

(h6 - 932.93) = ( 955.38 - 932.93) ((79.33 - 75.29) / ( 82.05 - 75.29 )

h6 - 932.93 = 13.427

h6 = 946.357 kJ/kg

Calculate the total work input of the first and second stage compressors

(Wcomp)in = 2(h2 - h1 ) = 2( 411.257 - 300.19 )

= 222.134 kJ/kg  

Calculate the total work output of the first and second stage turbines.

(Wturb)out = 2(h5 - h6) = 2( 1277.79 - 946.357 )

= 662.866 kJ/kg  

Calculate the net work done

Wnet = (Wturb)out  - (Wcomp)in

= 662.866 - 222.134

= 440.732 kJ/kg  

Calculate the minimum mass flow rate of air required to generate a power output of 105 MW

W = m × Wnet

(105 x 10³) kW = m(440.732 kJ/kg)

m = (105 x 10³) / 440.732

m = 238.2 kg/s

therefore the minimum mass flow rate of air required to generate a power output of 105 MW is 238.2 kg/s