Determine the critical frequency of the Sallen-Key low-pass
filter.
Example 1 Determine the critical frequency of the Sallen-Key low-pass filter 1.00 1.00 22μF ww 1.00

Answer:

  Research the company

Explanation:

It is a good idea to find out information about the company you are interviewing. Among other things, this will help you decide if you actually want to work for that company.

Telling your current boss can be risky to your job. A new outfit may be helpful for certain positions--especially those where manner of dress is important. Finishing school may or may not be an important requirement for the position you're seeking. Your company research will tell you.

Researching the company is one of the most important steps to take.

Answer:

agriculture,defense,industry,science

Explanation:

Edg 2021

RAM is called “random access” because any storage location on the computer can be accessed directly (as opposed to randomly).

The example of a primary function in MATLAB that uses a function handle and a nested function to compute the minimum of the function f(x) = 20*x^2 - 200*x + 12 over the range 0 ≤ x ≤ 10:

function min_value = compute_minimum()

   % Define the function handle

   f = (x) 20*x^2 - 200*x + 12;

   % Define the range of x values

   x_range = linspace(0, 10, 100); % 100 points between 0 and 10

   % Compute the minimum value using the nested function

   min_value = find_min(f, x_range);

   % Display the result

   fprintf('The minimum value of f(x) over the range 0 ≤ x ≤ 10 is: %f\n', min_value);

end

function min_val = find_min(f_handle, x_values)

   % Initialize the minimum value

   min_val = Inf;

   % Evaluate the function at each x value and update the minimum

   for i = 1:length(x_values)

       x = x_values(i);

       f_val = f_handle(x);

       if f_val < min_val

           min_val = f_val;

       end

   end

end

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To design the CFG rules for the given language (a+b)*aa(a+b)*, let's first consider the language itself. This language consists of all strings that begin and end with any combination of a's and b's, with exactly two a's in between them.

We can use this definition to come up with the following CFG rules:

S → AaaB | BAaaA → aA | bAB → Aa | bB | ε The above CFG rules generate all possible strings that are in the language (a+b)*aa(a+b)*.

Here's a breakdown of what each rule does:

S → AaaB | BThis rule states that any string in the language can either start with AaaB or simply be B. A → aA | bThis rule says that any string that starts with A can either be followed by an "a" and then another string that starts with A, or a "b". B → Aa | bB | ε

This rule says that any string that starts with B can either be followed by an "a" and then another string that starts with A, or a "b" and then another string that starts with B, or simply be an empty string.

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Answer:

Explanation:

Given that:

At state 1:

Pressure P₁ = 20 bar

Volume V₁ = 0.03 \(\mathbf{m^{3}}\)

From the tables at saturated vapour;

Temperature T₁ = 212.4⁰ C  ; \(v_1 = vg_1\) = 0.0996 \(\mathbf{m^{3}}\) / kg

The mass inside the cylinder is m = 0.3 kg, which is constant.

The specific internal energy u₁ = ug₁ = 2599.2 kJ/kg

At state 2:

Temperature T₂ = 200⁰ C

Since the 1 - 2 occurs in an isochoric process v₂ = v₁ = 0.099 \(\mathbf{m^{3}}\) / kg

From temperature T₂ = 200⁰ C

\(v_f_2 = 0.0016 \ m^3/kg\)  

\(vg_2 = 0.127 \ m^3/kg\)  

Since  \(vf_2 < v_2<vg_2\) , the saturated pressure at state 2 i.e. P₂ = 15.5 bar

Mixture quality \(x_2 = \dfrac{v_2-vf_2}{vg_2 -vf_2}\)

\(x_2 = \dfrac{(0.099-0.0016)m^3/kg}{(0.127 -0.0016) m^3/kg}\)

\(x_2 = \dfrac{(0.0974)m^3/kg}{(0.1254) m^3/kg}\)

\(\mathsf{x_2 =0.78}\)

At temperature T₂, the specific internal energy \(u_f_2 = 850.6 \ kJ/kg\) , also \(ug_2 = 2594.3 \ kJ/kg\)

Thus,

\(u_2 = uf_2 + x_2 (ug_2 -uf_2)\)

\(u_2 =850.6 +0.78 (2594.3 -850.6)\)

\(u_2 =850.6 +1360.086\)

\(u_2 =2210.686 \ kJ/kg\)

At state 3:

Temperature \(T_3=T_2 = 200 ^0 C ,\)

\(V_3 = 2V_1 = 0.06 \ m^3\)

Specific volume \(v_3 = 0.2 \ m^3/kg\)

Thus; \(vg_3 =vg_2 = 0.127 \ m^3/kg\) ,

SInce \(v_3 > vg_3\), therefore, the phase is in a superheated vapour state.

From the tables of superheated vapour tables; at \(v_3 = 0.2 \ m^3/kg\) and T₃ = 200⁰ C

The pressure = 10 bar and v =0.206 \(\ m^3/kg\)

The specific internal energy \(u_3\) at the pressure of 10 bar = 2622.3 kJ/kg

The changes in the specific internal energy is:

\(u_2-u_1\)

= (2210.686 - 2599.2) kJ/kg

= -388.514 kJ/kg

≅ - 389 kJ/kg

\(u_3-u_2\)

= (2622.3 - 2210.686)  kJ/kg

= 411.614 kJ/kg

≅ 410 kJ/kg  

We can see the correct sketches of the T-v plot showing the diagrammatic expression in the image attached below.

The term that is not designed to absorb collision energy is dimples.

What is an energy-absorbing mechanism?

Energy-absorbing mechanism is a safety mechanism that absorbs the kinetic energy that is generated during an impact.

The shock-absorbing material, which is intended to be compressed in a controlled manner under the impact, is the most basic component of this system.

What is the purpose of the design of a crush zone?

Crush zones are engineered areas of a vehicle that are designed to absorb and dissipate energy during a collision.

They are often constructed from a variety of materials, including high-strength steel, aluminum, magnesium, and other composites.

O Slots, crush zones, and reinforcements are all safety features that are designed to absorb collision energy and lessen the impact of a collision on the vehicle and its occupants.

These energy-absorbing mechanisms allow the vehicle to deform or collapse in a controlled manner, slowing down the rate of deceleration and reducing the amount of force transferred to the vehicle's occupants.

Dimples, on the other hand, are not an energy-absorbing mechanism.

Dimples are tiny indentations or bumps on the surface of a material that is often used to reduce aerodynamic drag. They are not designed to absorb collision energy.

Therefore, the answer is dimples.

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Answer:

The answer is below

Explanation:

The complete question is attached.

a) Bernoulli equation is given as:

\(P+\frac{1}{2}\rho V^2+ \rho gz=constant\\\\\frac{P}{\rho g} +\frac{V^2}{2g} +z=constant\\\)

Where P = pressure, V = velocity, z = height, g = acceleration due to gravity and ρ = density.

\(\frac{P}{\rho g} +\frac{V^2}{2g} +z=constant\\\\\frac{P}{\gamma} +\frac{V^2}{2g} +z=constant\\\\\frac{P_1}{\gamma} +\frac{V_1^2}{2g} +z_1=\frac{P_2}{\gamma} +\frac{V_2^2}{2g} +z_2\\\\but \ z_1=z_2,P_1=0,V_1=0,V_2=60\ mph=88\ ft/s. Hence:\\\\\frac{P_2}{\gamma} =-\frac{V_2^2}{2g} \\\\P_2=\gamma*-\frac{V_2^2}{2g} =\rho g*-\frac{V_2^2}{2g} \\\\P_2=-\frac{V_2^2}{2}*\rho=-\frac{(88.8\ ft/s)^2}{2} * 0.00238\ slug/ft^3=-9.22\ lb/ft^2\\\\P_2+\gamma_{H_2O}h-\gamma_{oil}(1/12 \ ft)=0\\\\\)

\(\gamma_{oil}=0.9\gamma_{H_2O}=0.9*62.4\ lb/ft^3=56.2\ lb/ft^3\\\\Therefore:\\\\-9.22\ lb/ft^2+62.4\ lb/ft^3(h)-56.2\ lb/ft^3(1/12\ ft)=0\\\\h=0.223\ ft\)

b)

\(\frac{P}{\gamma} +\frac{V^2}{2g} +z=constant\\\\\frac{P_2}{\gamma} +\frac{V_2^2}{2g} +z_2=\frac{P_3}{\gamma} +\frac{V_3^2}{2g} +z_3\\\\but \ z_3=z_2,V_3=0,V_2=60\ mph=88\ ft/s. \\\\\frac{P_2}{\gamma}+\frac{V_2^2}{2g} = \frac{P_3}{\gamma}\\\\\frac{P_3-P_2}{\gamma}=\frac{V_2^2}{2g} \\\\P_3-P_2=\frac{V_2}{2g}*\gamma=\frac{V_2^2}{2g}*\rho g\\\\P_3-P_2=\frac{V_2}{2}*\rho=\frac{(88\ ft/s^2)^2}{2}*0.00238\ slg\ft^3\\\\P_3-P_2=9.22\ lb/ft^2\)

The more the gas is compressed, the less volume it takes up at a given pressure. There is a direct correlation between volume and temperature.

Many everyday facts allow you to observe this behaviour of gases, such as the fact that the volume of trapped air in a syringe decreases with pressure when you press down on the plunger while covering the exit hole. Boyle's law for ideal gases, which states that PV = constant and that P is inversely linked to volume, can also be used to demonstrate this. The average density will drop, making the submarine float since the volume of the tanks and the submarine isn't changing (stay constant) and the object's mass (the submarine with tanks) is decreasing (due to the discharge of a significant amount of air).

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The given triangular waveform with 50 Hz frequency and 1 V peak is to be converted into a TTL-compatible pulse waveform with fundamental frequency 50 Hz. TTL-compatible pulse waveform has high and low voltage levels of 5 V and 0 V respectively.

The basic idea of conversion is to compare the input triangular waveform with a reference voltage level of 2.5 V (halfway between 5 V and 0 V) and create a pulse waveform such that output is high (5 V) when the input waveform is above 2.5 V and low (0 V) when the input waveform is below 2.5 V.

Here, we can use a simple NAND gate.The logic gate will produce a high output (5 V) only when both its inputs are low (0 V). Therefore, we can connect the comparator output to one input of the NAND gate and a 5 V source to the other input of the NAND gate. This will give a high output when the input waveform is below 2.5 V and low output when the input waveform is above 2.5 V. Thus, we will get a TTL-compatible pulse waveform.The circuit diagram is as shown below:And the input-output waveforms are shown below:

Therefore, we have successfully designed a circuit that can convert a 50 Hz triangular wave with 1V peak into a TTL-compatible pulse wave with a fundamental frequency of 50Hz.

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The pipe that presents water at a particular pressure to all branching pipes in a pipeline system is called the main or feeder pipe. This pipe is responsible for delivering the water from the source to all the branching pipes, which then distribute the water to the desired locations. The main or feeder pipe is typically larger in diameter than the branching pipes to accommodate the larger volume of water flow.

The pipe that presents water at a particular pressure to all branching pipes is called the "header" pipe. In a pipeline system, the header pipe is typically a large diameter pipe that connects to the main water source and supplies water to the entire system. The header pipe is designed to maintain a certain pressure throughout the system, which ensures that all of the branching pipes receive a consistent flow of water.The header pipe is usually located at the highest point in the system to take advantage of gravity to maintain the pressure. Additionally, the header pipe is designed to minimize friction losses and ensure that the flow of water through the system is as efficient as possible.The header pipe may also be equipped with various valves, meters, and other components to monitor and control the flow of water through the system. These components help to maintain the pressure and flow of water at optimal levels and ensure that the pipeline system operates efficiently and reliably. The pipe that presents water at a particular pressure to all branching pipes in a pipeline system is called the "main distribution pipe" or "primary feed pipe." This large diameter pipe ensures a consistent flow of water from the main source to the branching pipes, maintaining adequate pressure throughout the system.

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The flow of water from a main source enters a pipeline system through a large diameter pipe.

The pipe that presents water at a particular pressure to all branching pipes is called the "main distribution pipe" or "main supply line".

This main distribution pipe maintains consistent water pressure and ensures adequate water flow to all connected branching pipes within the system.

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The component that moves in the brake caliper to push the brake pads up against the brake rotor is called the brake piston.

The ability of the brake pads to make contact with the braking rotor and stop a vehicle is provided by the pistons. The master cylinder releases braking fluid into the brake caliper when the brake pedal is depressed, and the pistons can be found in both floating and fixed brake calipers.

On the inboard side of the braking rotor, a floating caliper typically has one or two large diameter pistons. The piston(s) pull both pads toward the brake disc when they are engaged. Behind the brake pads, pistons are forced outward.

The brake pedal and master cylinder of the car generate hydraulic pressure in the braking fluid that engages the caliper. In this assembly, friction is produced by pressing the brake pads up against the disc rotor surface.

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Answer: Photo lines

Explanation: made more sense

Answer:

Utilizing flush valves for toilets and urinals and/or low-flow versions of each, as well as installing low-flow faucets, or aerators can significantly reduce water use.

Explanation:

Answer:

sorry im answering questions for the points cuz im built dfferent

Explanation:

Answer: That happens to me too sometimes. I hate when people give the wrong answers...like if u don't know the answer, then don't respond

Answer:

Lack of fluid oil – lack of fluid oil in your vehicle, or a fluid leakage, can lead to heavy steering. If there is a lack of fluid oil, or a leak, this can reduce the pressure in the system, meaning the steering wheel does not receive enough supply of fluid to perform freely.

A digital multimeter can not be used to test charging system voltage,So Technician A and Technician B is not correct.

Neither technician is entirely correct.

Technician A is partially correct in that it is generally recommended to have a battery that is at least 75% charged before testing the charging system. However, this is not an absolute requirement, and some testing can still be performed with a battery that has a lower charge.

Technician B is incorrect in stating that a digital multimeter cannot be used to test charging system voltage. In fact, a digital multimeter is one of the most commonly used tools for this purpose, and it can provide accurate readings of the voltage output from the alternator.

Therefore, the correct answer is neither A nor B.

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Answer: There are no major convergent boundaries in the Atlantic, which is where these events are more likely to occur

Explanation:

Answer:

33.56 ft^3/sec.in

Explanation:

Duration = 6 hours

drainage area = 185 mi^2

constant baseflow = 550 cfs

Derive the unit hydrograph using the inverse procedure

first step : calculate for the volume of direct runoff hydrograph using the details in table 2 attached below

Vdrh = sum of drh *  duration

        = 29700 * 6 hours ( 216000 secs )

        = 641,520,000 ft^3.

next step : Calculate the volume of runoff in equivalent depth

Vdrh / Area = 641,520,000  / 185 mi^2

                    = 1.49 in

Finally derive the unit hydrograph

Unit of hydrograph = drh /  volume of runoff in equivalent depth

                                = 50 ft^3 / 1.49 in  =  33.56 ft^3/sec.in

Answer:

Make sure both cars are in park or neutral with the parking brake engaged. This makes sure that all electrical functions (headlights, radios, etc.) are turned off.

When jump starting a car, make sure it is in the right distance of the cables to reach battery.

Thus, when jump-starting a car, always remember the cables are in the right position.

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Gross errors produce data that differ unreasonably from data from similar samples and apply to the given category. The correct option is B.

Gross errors, often known as "outliers," are errors that are not systematic or random. They are inherently unpredictable and frequently huge.

Gross errors result from experimenter negligence or device malfunction. These “outliers” are typically ignored when analyzing data because they are far above or below the genuine value.

Therefore, the correct option is B: gross errors produce data that differ unreasonably from data from similar samples.

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Answer:

Explanation:

he metallic rod is fixed into the dies by using a die holder and then a drawing head is used in which the metallic rod is fixed through a jaw mechanism. And then the bar is stretched and slides between the surfaces of the die

When air undergoes adiabatic expansion, the process is governed by the equation PV^γ = constant, where γ is the ratio of specific heats. In this case, the value of γ for air is 1.4. We are given that one pound of air undergoes an adiabatic expansion from 200 psia to 50 psia, with an initial volume of 4 ft3/lbm and PV^1.4 = constant.

Let's calculate the final volume of the air using the initial and final pressures and the initial volume. Using the formula P1V1^γ = P2V2^γ and substituting the given values, we have: 200(4)^1.4 = 50(V2)^1.4V2 = (200(4)^1.4 / 50)^(1/1.4)V2 = 11.14 ft^3/lbmThe work done by the air is given by the equation W = ∆E + Q, where ∆E is the change in internal energy and Q is the heat added to or removed from the system. Since the process is adiabatic (Q = 0), the work done is equal to the change in internal energy. Let's calculate the work done:W = ∆E = C_v (T2 - T1)where C_v is the specific heat at constant volume, and T1 and T2 are the initial and final temperatures, respectively. The specific heat at constant volume for air is 0.1715 Btu/lbm·R. Let's calculate the final temperature of the air using the initial and final pressures and volumes and the equation P1V1^γ/T1 = P2V2^γ/T2.200(4)^1.4/T1 = 50(11.14)^1.4/T2T2 = T1 * (P2V2^γ / P1V1^γ)T2 = 1183.3 RLet's substitute the values into the equation for work done to get:W = C_v (T2 - T1)W = 0.1715 Btu/lbm·R (1183.3 R - 527.7 R)W = 0.1715 Btu/lbm·R (655.6 R)W = 112.3 Btu/lbmThe change in internal energy is also 112.3 Btu/lbm, since Q = 0. The change in temperature is T2 - T1 = 1183.3 R - 527.7 R = 655.6 R.Answer: The work done by the air is 112.3 Btu/lbm, and the change in internal energy and temperature of the gas are also 112.3 Btu/lbm and 655.6 R, respectively.

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Answer: At least I’m not the only one looking for answers lol

Explanation: