T/F: webwork can't solve for adaptive parameters

Answer:

C. throttle

Explanation:

see the picture hope this helps someone

The feed axis system, shown in the figure below, is driven by a motor with the specifications given in the table below.

Where I is the moment of inertia of the axis, α is the angular acceleration, Fr is the force of resistance to motion, and r is the radius of the axis .Where μ is the coefficient of friction, and Fn is the normal force.

The normal force is the force exerted on the axis by the load, and is equal to the weight of the load. In this case, the weight of the load is 100 N, and therefore the normal force is also 100 N. The coefficient of friction is given as 0.3.

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Answer:

\(T_o = 141.81 ^0C\)

Explanation:

Given that;

Thermal diffusivity \(\alpha = 5 \times 10 ^{-6} m^2/s\)

Thermal conductivity \(k = 20 \ W/m.K\)

Heat transfer coefficient h = ( we are to assume the imposed surface temperature ) = 20 W/m².K

Initial temperature = 150 ° C = (150+273) K = 423 K

Then coolant temperature with which the casting is exposed to = 20° C = (20+273)K = 293 K

Time = 40 seconds

Length = 20mm = 0.02 m

The objective is to determine the  temperature at the surface  at a depth of 20 mm after 40 seconds.

\(Bi = \dfrac{hL}{k}\)

\(Bi = \dfrac{20*0.02}{20}\)

Bi == 0.02

\(\tau = \dfrac{\alpha t}{L^2}\)

\(\tau= \dfrac{5*10^{-6 }* 40}{0.020^2}\)

\(\tau = 0.5\)

For a wall at 0.2 Bi

\(A_1 = 1.0311\)

\(\lambda _1 = 0.4328\)

Therefore;

\(\dfrac{T_o - T_{\infty}}{T_i - T_{\infty}}= A_1 e ^{-( \lambda_1^2 \ \tau)\)

\(\dfrac{T_o - 293 }{423 - 293}= 1.0311 \times e ^{-( 0.438^2 \times 0.5 )\)

\(\dfrac{T_o - 293 }{423 - 293}= 1.0311 \times e ^{-( 0.0959 )\)

\(\dfrac{T_o - 293 }{130}= 1.0311 \times 0.9085\)

\(\dfrac{T_o - 293 }{130}= 0.937\)

\(T_o - 293= 0.937 \times 130\)

\(T_o - 293= 121.81\)

\(T_o = 121.81+ 293\)

\(T_o = 414.81 \ K\)

\(T_o = (414.81 - 273)^0C\)

\(T_o = 141.81 ^0C\)

|H(omega)| = 1/(sqrt(1 + (omegaRC) ² )) is the magnitude of the frequency response, which is a standard form for a first-order low-pass filter, and the cutoff frequency is given by omega_c = 1/(R  ˣ C).

What is the frequency response of the given circuit and how is it calculated?

To determine the frequency response of the given circuit, we need to find the transfer function H(omega) = V/V_s, where V is the output voltage and V_s is the input voltage.

We can start by analyzing the circuit using the voltage divider rule. The impedance of the resistor R is simply R, while the impedance of the capacitor C is given by 1/(jˣ omegaˣ C), where j is the imaginary unit. Thus, the voltage across the capacitor can be expressed as:

V_c = V_s   ˣ  1/(1 + j  ˣ omega  ˣ R  ˣ C)

The output voltage V is simply the voltage across the capacitor, so:

V = V_c = V_s  ˣ 1/(1 + j  ˣ omega  ˣ R  ˣ C)

Now we can calculate the frequency response by taking the magnitude of H(omega):

|H(omega)| = |V/V_s| = 1/(sqrt(1 + (omega  ˣ R  ˣ C) ² ))

This is a standard form for the frequency response of a first-order low-pass filter, where the cutoff frequency is given by:

f_c = 1/(2  ˣ pi  ˣ R  ˣ C)

In terms of omega, the cutoff frequency is:

omega_c = 1/(R  ˣ C)

To sketch |H(omega)| versus omega, we can plot the function 1/(sqrt(1 + (omega  ˣ R  ˣ C) ² )) for omega > 0. The plot will look like a decreasing curve starting from 1 at omega = 0 and approaching 0 as omega goes to infinity. The cutoff frequency omega_c marks the point where the magnitude of the frequency response drops to 1/sqrt(2) or -3 dB.

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The net heat transfer rate by radiation from Body 1 to Body 2 per square meter is approximately 10.35 W/m^2.

The problem concerns the radiation heat transfer between two parallel disks. The disks have radii of 4 cm and emissivities of 0.6 and 0.8. They are maintained at temperatures of 200°C and 100°C respectively, with a distance of 5 cm between them. The objective is to determine the net heat transfer rate per square meter from Body 1 to Body 2, assuming no convective heat transfer occurs between the disks.

To calculate the heat transfer rate by radiation, the Stefan-Boltzmann law is applied:

q'' = σA1A2 (T1^4 - T2^4) / πD^2

Where:

q'' is the heat transfer rate by radiation

σ is the Stefan-Boltzmann constant (5.67 x 10^-8 W/m^2-K^4)

A1 and A2 are the areas of the two disks

T1 and T2 are the temperatures of the two disks

D is the distance between the two disks

From the equation, it is evident that the heat transfer rate by radiation is directly proportional to the area of the disks and the temperature difference between them, while inversely proportional to the square of the distance between them.

Calculating the area of each disk, A, we find A = πr^2 = 3.14 x 4^2 / 10000 = 0.000502 m^2.

Substituting the given values into the Stefan-Boltzmann law, we obtain:

q'' = 5.67 x 10^-8 x 0.000502 x 0.000502 x (473.15^4 - 373.15^4) / (π x 0.05^2)

≈ 10.35 W/m^2

Therefore, the net heat transfer rate by radiation from Body 1 to Body 2 per square meter is approximately 10.35 W/m^2.

Answer: 10.35 W/m^2.

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The repair order must detail why the vehicle is in for repairs or what the customer is complaining about. Note that this on the repair order is known as Problem Description.

A repair order is a document used in an automotive repair shop or service center to document the details of a repair job.

It contains information about the vehicle, the customer, the problem or reason for the repair, the work done, and the repair cost. It is used as a record of the work done and for billing and warranty purposes.

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Explanation:

Geometric Shapes can be defined as figure or area closed by a boundary which is created by combining the specific amount of curves, points, and lines.

Geomorphic

Explanation:

Because it's pretty easy if you think of the right ways of geomorphic ways to process the equation

Answer:

flashlight???

Explanation:

Answer:

a rock is made up of two or more minerals but a mineral is a natural substance with chemical and physical properties

Fourth-order Low Pass  Filter If the gain of both filters is set at 1.586, the voltage gain will be down 6 dB at the cutoff frequency. We can get a more flat response by choosing different values of voltage gain for both stages.

It is possible to create a 4th order low-pass filter with just one Op Amp. Due to the interconnected nature of all filter parameters, component tolerance sensitivity needs to be confirmed. The 4th order Sallen-Key topology is too sensitive to component tolerances, making it unsuitable for application.

 Formula fc= 1/(2πRC).

The attenuation rate for a low-pass or high-pass filter will be -20 times the filter order, expressed in decibels per decade. A first-order filter, for instance, will attenuate at a rate of -20 dB/decade, but a fourth-order filter will attenuate at a rate of about -80 dB/decade.

Attenuation (dB)= 10 X log(PI/PO)

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We can calculate the real and reactive power supplied to the motor from the input current and power factor, the output power from the efficiency and input power,

(a) The total real power supplied to the motor is

240 V x 15 A x 0.8 pf = 2.88 kW

and the reactive power supplied to the motor is

240 V x 15 A x sin(arccos(0.8)) = 2.135 kVAR.

(b) The output power of the motor is the input power multiplied by the efficiency, so it is

2.88 kW x 0.95 = 2.736 kW.

(c) The balanced, parallel, delta-connected impedance that allows the motor to draw the same current at unity pf is equal to the impedance of the motor at its operating point divided by the power factor at the operating point. This can be calculated using the formula

\(Z = V^2 / (S * pf)\) , where Z is the impedance, V is the voltage, S is the apparent power, and pf is the power factor.

Solving for Z, we get

\(Z = 240^2 / (15 *0.8) = 2304ohm\)

In summary, we can calculate the real and reactive power supplied to the motor from the input current and power factor, the output power from the efficiency and input power, and the impedance for unity power factor from the voltage, current, and power factor at the operating point.

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When a wire with length l and area of cross-section a is pulled, its length is doubled and its area of cross-section is halved. This results in the wire becoming longer and thinner, but the volume of the wire remains the same.

This is due to the fact that the volume of a wire is equal to its length multiplied by its area of cross-section, and since the volume remains constant, the product of length and area of cross-section must remain constant too.The resistance of a wire is directly proportional to its length and inversely proportional to its area of cross-section. Thus, when the wire is pulled and its length is doubled, its resistance also doubles. Similarly, when its area of cross-section is halved, its resistance becomes four times its initial value.The formula for resistance of a wire is R = ρl/a, where ρ is the resistivity of the material. Since the resistivity of the material remains constant, we can use this formula to calculate the new resistance of the wire after it has been pulled. If the initial resistance was R1, then the new resistance R2 can be calculated as follows:
R2 = ρ(2l)/(a/2)
  = ρ(4l)/a
Therefore, the new resistance is four times the initial resistance.

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Answer:

I think its True

it should be True

Answer:

Tech B

Explanation:

Question: "Pontoon type floating roofs shall have sufficient buoyancy to remain afloat on a liquid with a specific gravity of:"

Pontoon type floating roofs shall have sufficient buoyancy to remain afloat on a liquid with a specific gravity of A) 1.0.

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Answer:

C(h,w)=5w²+8(h*w)

Explanation:

Let w be the name of the width of this storage.

Let h be the height of the storage.

Constraint: w²h=8 cubic meters.

Taking into acount that the base of this storage is a square, width=length.

Area of the base: w²

•The cost of the base is given by:

C(b)=$5*w²=

5w²

•The cost of the sides is given by:

C(s)={$2*(h*w)}*4

We multiply by 4 because there are 4 sides.

C(s)=8(h*w)

•The total cost should be given by:

C(h,w)=5w²+8(h*w)

-------------------------------------------------------------------------------------------------------

To verify this answer let's illustrate an example and solve it.

Let's say we have a storage with a base of 9m² (3 by 3, each side=3) and then the height of the storage is 4m.

Let's plug the values into the formula and try to find the total cost:

C(4,3)=5(3)²+8(4*3)= $141.

Area of the base was 9, 9*5= $45.

Area of the 4 sides should be 4*3*4= 48, 48*2= $96

$45+$96= $141, that's correct!

If a mechanic replaces a car's springs with stiffer springs, the car's bounce will be less frequent than originally.

This is because stiffer springs are designed to absorb and dampen the shocks that a car experiences on the road, which means that the car's body will bounce less when going over bumps or rough terrain. However, it is important to note that stiffer springs can also make the ride feel more rigid and less comfortable, as the car will not absorb as much of the shock. Additionally, the handling of the car may also be affected, as stiffer springs can reduce body roll during cornering, making the car feel more stable and responsive. Ultimately, the decision to replace a car's springs with stiffer ones depends on the driver's preference and the type of driving they plan on doing.

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An airplane flying across the sky experience drag force determined by the factors including the speed of flight, coefficient of skin friction and the reference surface area

The boundary layer thickness is approximately 0.233 cm

The total skin friction drag, is approximately 265 N

Reason:

First part:

Given parameters are;

Chord length, L = 3 m

Velocity of the plane, V = 200 m/s

Density of the air, ρ = 1.225 kg/m³

Viscosity of the air, μ = 1.81 × 10⁻⁵ kg/(m·s)

The Reynolds number is given as follows;

\(R_{eL} = \dfrac{\rho \times V \times L}{\mu}\)

Therefore;

\(R_{eL} = \dfrac{1.255 \times 200 \times 3}{1.81 \times 10^{-5}} = 4.16022099448 \times 10^7 \approx 4.16 \times 10^7\)

Boundary layer thickness, \(\delta_L\), for laminar flow, is given as follows;

\(\dfrac{ \delta_L }{L}=\dfrac{5.0}{\sqrt{R_{eL} } }\)

\({ \delta_L }=\dfrac{5.0 \times L}{\sqrt{R_{eL} } }\)

Which gives;

\({ \delta_L }=\dfrac{5.0 \times 3}{\sqrt{4.16 \times 10^{7}} } \approx 2.33 \times 10^{-3 }\)

The boundary layer thickness, \(\delta_L\) ≈ 2.33 × 10⁻³ m = 0.233 cm

Second Part

The total skin friction is given as follows;

\(Dynamic \ pressure, q = \dfrac{1}{2} \cdot \rho \cdot V^2\)

Therefore;

\(q = \dfrac{1}{2} \times 1.225 \times 200^2 = 24,500\)

The dynamic pressure, q = 24,500 N/m²

Skin friction drag coefficient, \(C_D\), is given as follows;

\(C_D = \dfrac{1.328}{\sqrt{R_{eL} } }\)

Therefore;

\(C_D = \dfrac{1.328}{\sqrt{4.16 \times 10^7 } } \approx 2.06 \times 10^{-4}\)

Skin friction drag, \(D_f\), is given as follows;

\(D_f\) = q × \(C_D\) × A

Where;

A = The reference area

∴ \(D_f\) = 24,500 N/m² × 2.06 × 10⁻⁴ × 3 m × 17.5 m = 264.9675 N ≈ 265 N

The total skin friction drag, \(D_f\) ≈ 265 N

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When it becomes impracticable or unwise to continue to or land at the specified landing aerodrome, an aircraft may go to an alternate airfield that has the required services and facilities, can accommodate aircraft performance requirements, and is operational at the anticipated time of us.

If the departure aerodrome's meteorological conditions are below the required landing minima or if returning to the departure aerodrome is not feasible for any other reason, a take-off alternate airfield should be mentioned in the operational flight plan.

In the case of two-engine aircraft, such a backup shall be within one hour's flight time at the one-engine-inoperative rate of speed specified in the Aerodrome Operating Manual (AOM) or comparable document.

Therefore,  assuming the aircraft mass is the actual mass at take-off and ISA and still-air conditions are in effect.

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Answer:

In a single-family residence, a line-voltage thermostat is primarily used to control heating and cooling systems.

Answer:

Get ready and comfortable.

List all of the tasks you need to accomplish over the next week. .

Next schedule everything.

Get a planner/calender.

Cut those tasks that do not fit into your

A set of control signals is a group of inputs to a digital circuit that determine its response. These control signals govern the operation of the circuit and determine its logical functionality.

The ideal set of control signals that are used in a digital circuit depends on the requirement of the circuit. As given, we have only 2-input NAND, NOR, XOR, XNOR gates, and inverters, which means we cannot make all other gates, and we have to select the appropriate set of control signals.Among the given gates, NAND and NOR gates can be used to implement all other logic gates. XOR gate can be obtained by combining NAND gates and inverters, while XNOR gate can be obtained by using NOR gates and inverters. However, when we use only NAND and NOR gates, the size of the circuit increases. Therefore, the preferred set of control signals among the given gates should be the combination of XOR and inverters.The XOR gate requires less logic than the implementation of XOR using NAND and NOR gates. The XOR gate can be used as a universal gate and is also used in various arithmetic and encryption circuits. Therefore, the preferred set of control signals should be XOR and inverters.

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With this in mind, the answer to the provided student question is as follows:When parking a car on a level street, it is important to turn your front wheels towards the curb.

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When in regular operation, ash and other debris should be removed daily from the combustion chamber of a wood-burning heater.

A wood stove radiates heat as it warms up via the stove's top and walls. The nearby region is warmed by the radiant heat, which can also warm other areas of the house thanks to the natural airflow in the building.

Fans using electric or convection power can help move this heat around to warm a bigger space. Some wood stoves use a convection chamber that wraps around the firebox to combine radiant and convection heat into a single appliance. Cool air is drawn into this convection chamber where it is heated and then circulated once again around the space.

Ash and other debris should be taken out of a wood-burning heater's combustion chamber every day when it is in regular use.

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