(TCO 3) Below is the frequency response of the IIR realization of an analog band-pass filter. What type of filter is this?

Answer:

  12.44Ω

Explanation:

For a question such as this, the assumption must be that the resistance is inversely proportional to the square of the diameter. The resistance will be ...

  20Ω × (1.262/1.6)² ≈ 12.44Ω

Answer:

[tex]ohms[/tex]

Explanation:

hope it helps

Answer:

Oxygen is an important part of the atmosphere and is necessary to sustain terrestrial life. Because it comprises most of the mass in water, it also comprises most of the mass of living organisms. All major classes of structural molecules in living organisms, such as proteins, carbohydrates, and fats, contain oxygen, as do the major inorganic compounds that comprise animal shells, teeth, and bone. Elemental oxygen (O2) is produced by cyanobacteria, algae, and plants through the process of photosynthesis, and is used in cellular respiration by most living organisms on earth. Oxygen is toxic to obligate anaerobic organisms (organisms which need a lack of oxygen for survival), which were the dominant form of early life on Earth, until O2 began to accumulate in the atmosphere.

Explanation:

Answer:

They do not cause phase-distortion of an input signal. ( option 2 )

Explanation:

Linear-phase filters are desirable in many applications simply because  they do not cause phase-distortion of an input signal.  

The signals inputted into a Linear-phase filter ( FIR type ) are delayed equally as frequency is delayed in the pass band but they are are distorted.

The main special function of the Linear-phase filter is to reduce signal distortion.

Answer: walk dog on leash for one and two you csnt park in a car you park the car and walk the dog on a leash

Explanation: did you phrase the question wrong?

Answer:fall arrest harness

Explanation:cuz it’s just right

Answer: True

Explanation:

Answer:

- the force of the rope on the tugboat is 14.87 kN

- Force acting on the tugboat is 24.79 kN

Explanation:

 Given the data in the question;

tugboat increases its speed uniformly to 25 km/h

v₁ = 25 km/h = (25 × 1000) / ( 1 × 60min × 60sec )

= 25000m / 3600s

=  6.94 m/s

Now, lets determine the force on the rope using the following relation;

T = [tex]m_b[/tex]v₁ / t₁

[tex]m_b[/tex] is mass of barge( 75 Megagram = 75 × 10³ Kilogram  ), time t₁ is 35 s and v₁ is 6.94 m/s

so we substitute

T = [(75 × 10³) × 6.94 ]  / 35

T = 520500 / 35

T = 14871.43 N

T = 14871.43 / 1000

T = 14.87 kN

Therefore, the force of the rope on the tugboat is 14.87 kN

Now, to determine F acting on the tugboat;

[tex]Ft_1[/tex] = ( [tex]m_t[/tex] + [tex]m_b[/tex] )[tex]v_1[/tex]

we solve for F

F = ( [tex]m_t[/tex] + [tex]m_b[/tex] )[tex]v_1[/tex] / [tex]t_1[/tex]

where [tex]m_t[/tex] is mass of tugboat (50 Megagram = 50 × 10³ Kilogram  )

so we substitute

F = [( (50 × 10³) + (75 × 10³) )6.94] / 35

F = [ 125000 × 6.94 ) / 35

F = 867500 / 35

F = 24785.7 N

F = 24785.7 / 1000

F = 24.79 kN

Therefore, Force acting on the tugboat is 24.79 kN

Answer:

I’m sorry I don’t understand this is there more steps?‍♀️

Explanation:

Answer:

skate-themed video games

library research

interviews with skaters

documentaries on skater culture

(I got a 90% on the test so I dont  know if this was the one that gave me the 90% but it said it was right)

Answer:

Ans: To see if the results are similar

Answer:

i) 25.04 W/m^2 .k

ii) 23.82 minutes = 1429.2 secs

Explanation:

Given data:

Diameter of steel ball = 15 cm

uniform temperature = 350°C

p = 8055 kg/m^3

Cp = 480 J/kg.k

surface temp of ball drops to  250°C

average surface temperature = ( 350 + 250 ) / 2 = 300°C

i) Determine the average convection heat transfer coefficient during the cooling process

Note : Obtain the properties of air at 1 atm at average film temp of 30°C from the table  " properties of air "  contained in your textbook

average convection heat transfer coefficient = 25.04 W/m^2 .k

ii) Determine how long this process has taken

Time taken by the process = 23.82 minutes = 1429.2 seconds

Δt = Qtotal / Qavg = 683232 / 477.92 = 1429.59 secs

attached below is the detailed solution of the given question

Answer:

the required expression is [tex]E_2 - E_1[/tex] = 4[ 1 - [tex]e^{(-0.05t)}[/tex] ]

Explanation:

Given the data in the question;

Q = -0.2[ 1 - [tex]e^(-0.05t)[/tex] ]

ω = 100 rad/s

Torque T = 18 N-m

Electric power input = 2.0 kW

now, form the first law of thermodynamics;

dE/dt = dQ/dt + dw/dt = Q' + w'

dE/dt = Q' + w'  ------ let this be equation 1

w' is the net power on the system

w' = [tex]w_{elect[/tex] - [tex]w_{shaft[/tex]

[tex]w_{shaft[/tex] = T × ω

we substitute

[tex]w_{shaft[/tex]= 18 × 100

[tex]w_{shaft[/tex] = 1800 W

[tex]w_{shaft[/tex] = 1.8 kW  

so

w' = [tex]w_{elect[/tex] - [tex]w_{shaft[/tex]

w' = 2.0 kW - 1.8 kW

w' = 0.2 kW

hence, from equation 1, dE/dt = Q' + w'

we substitute

dE/dt = -0.2[ 1 - [tex]e^{(-0.05t)[/tex] ] + 0.2

dE/dt = -0.2 + 0.2[tex]e^{(-0.05t)[/tex] ] + 0.2

dE/dt = 0.2[tex]e^{(-0.05t)[/tex]

Now, the change in total energy, increment E, as a function of time;

ΔE = [tex]\int\limits^t_0}\frac{dE}{dt} . dt[/tex]

ΔE = [tex]\int\limits^t_0} 0.2e^{(-0.05t)} dt[/tex]

ΔE = [tex]\int\limits^t_0} \frac{0.2}{-0.05} [e^{(-0.05t)}]^t_0[/tex]

[tex]E_2 - E_1[/tex] = 4[ 1 - [tex]e^{(-0.05t)}[/tex] ]

Therefore, the required expression is [tex]E_2 - E_1[/tex] = 4[ 1 - [tex]e^{(-0.05t)}[/tex] ]

Answer:

3

Explanation:

it is compulsory to have a bachelor's degree

Answer:

24 i believe

Explanation:

Answer:

please give me brainlist and follow

Explanation:

Mild steel can be converted into high carbons steel by which of the following heat treatment process? Explanation: Case hardening, also referred as carburizing increases carbon content of steel, thus, imparting hardness to steel.

Answer:

missing question; How long must the heated section be

answer : 5.898 meters

Explanation:

Given parameters:

Diameter of steel pipe = 1.5 inches = 0.0381 m

velocity of air = 2.10 ft/s = 0.64 m/s

Inlet temperature = 25°C

Outlet temperature = 95.5°C

pipe wall temperature = 121.1°C

Average temperature = ( 25 + 95.5 ) / 2 = 60.25°

determine Reynolds number of air at 60.25°

= VD / v = ( 0.64 * 0.0381 ) / ( 1.83 * 10^-5 )

              = 1332.46  this means the flow is laminar

hence mass flow rate ( m ) = eAv = 1.08 * πd^2 * velocity

= 1.08 * 0.0046 * 0.64 = 0.0032 g/s

attached below is the remaining part of the solution

Answer:

A) Not possible, B) Posible, C) Possible, D) Not possible.

Explanation:

The maximum theoretical efficiency for any thermal engine is defined by Carnot's cycle, whose energy efficiency ([tex]\eta[/tex]), no unit, is expressed below:

[tex]\eta = 1-\frac{T_{L}}{T_{H}}[/tex] (1)

Where:

[tex]T_{L}[/tex] - Cold reservoir temperature, in Kelvin.

[tex]T_{H}[/tex] - Hot reservoir temperature, in Kelvin.

If we know that [tex]T_{L} = 250\,K[/tex] and [tex]T_{H} = 750\,K[/tex], then the maximum theoretical efficiency for the thermal engine is:

[tex]\eta = 1-\frac{T_{L}}{T_{H}}[/tex]

[tex]\eta = 0.667[/tex]

For real thermal engines, the following inequation is observed:

[tex]0 \le \eta_{r} \le \eta[/tex] (2)

Where [tex]\eta_{r}[/tex] is the efficiency of the real heat engine, no unit.

There are two possible criteria to determine if a given heat engine is real:

Efficiency

[tex]\eta_{r} = 1 - \frac{Q_{L}}{Q_{H}}[/tex] (3)

Where:

[tex]Q_{L}[/tex] - Heat rejected to the cold reservoir, in kilojoules.

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

Power output

[tex]W = Q_{H}-Q_{L}[/tex] (4)

Where [tex]W[/tex] is the power output, in kilojoules.

Now we proceed to verify each case:

A) [tex]Q_{H} = 900\,kJ[/tex], [tex]Q_{L} = 600\,kJ[/tex], [tex]W_{m} = 400\,kJ[/tex]

[tex]\eta_{r} = 0.333[/tex]

[tex]0 \le \eta_{r} \le \eta[/tex]

[tex]W = 300\,kJ[/tex]

[tex]W \ne W_{m}[/tex]

This engine is not possible.

B) [tex]Q_{H} = 900\,kJ[/tex], [tex]Q_{L} = 500\,kJ[/tex], [tex]W_{m} = 400\,kJ[/tex]

[tex]\eta_{r} = 0.444[/tex]

[tex]0 \le \eta_{r} \le \eta[/tex]

[tex]W = 400\,kJ[/tex]

[tex]W = W_{m}[/tex]

The engine is possible.

C) [tex]Q_{H} = 900\,kJ[/tex], [tex]Q_{L} = 300\,kJ[/tex], [tex]W_{m} = 600\,kJ[/tex]

[tex]\eta_{r} = 0.667[/tex]

[tex]0 \le \eta_{r} \le \eta[/tex]

[tex]W = 600\,kJ[/tex]

[tex]W = W_{m}[/tex]

The engine is possible.

D) [tex]Q_{H} = 900\,kJ[/tex], [tex]Q_{L} = 100\,kJ[/tex], [tex]W_{m} = 800\,kJ[/tex]

[tex]\eta_{r} = 0.889[/tex]

[tex]\eta_{r} > \eta[/tex]

[tex]W = 800\,kJ[/tex]

[tex]W = W_{m}[/tex]

The engine is possible.

Answer:

B is correct

Explanation:

Answer:

  entering drivers

Explanation:

Drivers entering a traffic circle must yield the right-of-way to those already in the traffic circle. (Among other things, this rule helps avoid gridlock.)

Answer:

Answer:

Solving

Explanation:

Answer:

Product Teardown 28 pieces (1) Plastic packaging: protect and display product for purchase. (4) Exterior screws: hold case halves together. (1) Right case half: acts as part of a handle and contains the rest of the parts. (1) Left case half: acts as part of a handle and contains the rest of the parts.

Explanation:

A product teardown process is an orderly way to know about a particular product and identify its parts, system functionality to recognize modeling improvement and identify cost reduction opportunities. Unlike the traditional costing method, tear down analysis collects information to determine product quality and price desired by the consumers.

Answer:

?

Explanation:

Answer:

Explanation:

Given that:

The Inside pressure (p) = 1402 kPa

= 1.402 × 10³ Pa

Force (F) = 13 kN

= 13 × 10³ N

Thickness (t) = 18 mm

= 18 × 10⁻³ m

Radius (r) = 306 mm

= 306 × 10⁻³ m

Suppose we choose the tensile stress to be (+ve) and the compressive stress to be (-ve)

Then;

the state of the plane stress can be expressed as follows:

[tex](\sigma_ x) = \dfrac{Pd}{4t}+ \dfrac{F}{2 \pi rt}[/tex]

Since d = 2r

Then:

[tex](\sigma_ x) = \dfrac{Pr}{2t}+ \dfrac{F}{2 \pi rt}[/tex]

[tex](\sigma_ x) = \dfrac{1402 \times 306 \times 10^3}{2(18)}+ \dfrac{13 \times 10^3}{2 \pi \times 306\times 18 \times 10^{-3} \times 10^{-3}}[/tex]

[tex](\sigma_ x) = \dfrac{429012000}{36}+ \dfrac{13000}{34607.78467}[/tex]

[tex](\sigma_ x) = 11917000.38[/tex]

[tex](\sigma_ x) = 11.917 \times 10^6 \ Pa[/tex]

[tex](\sigma_ x) = 11.917 \ MPa[/tex]

[tex]\sigma_y = \dfrac{pd}{2t} \\ \\ \sigma_y = \dfrac{pr}{t} \\ \\ \sigma _y = \dfrac{1402\times 10^3 \times 306}{18} \ N/m^2 \\ \\ \sigma _y = 23.834 \times 10^6 \ Pa \\ \\ \sigma_y = 23.834 \ MPa[/tex]

When we take a look at the surface of the circular cylinder parabolic variation, the shear stress is zero.

Thus;

[tex]\tau _{xy} =0[/tex]