Let be a real-valued signal for which when . Amplitude modulation is preformed to produce the signal . A proposed demodulation technique is illustrated below where is the input, is the output and the ideal low-pass filter has a cutoff frequency of and a passband gain of 2. Determine . x(t) X(jω) = 0 g(t) = x(t)sin(2000πt) g(t) y(t) 2000π y(t) EENG 311: Signals and Systems Page 7 of 8 × lowpass filter y(t) cos(2000πt) g

Answer:

[tex]3.32\ \text{kW}[/tex]

Explanation:

[tex]T_c[/tex] = Outside temperature = [tex]-5^{\circ}\text{C}[/tex]

[tex]T_h[/tex] = Temperature of room = [tex]21^{\circ}\text{C}[/tex]

[tex]Q_h[/tex] = Heat loss = 135000 kJ/h = [tex]\dfrac{135000}{3600}=37.5\ \text{kW}[/tex]

Coefficient of performance of heat pump

[tex]\text{COP}=\dfrac{1}{1-\dfrac{T_c}{T_h}}\\\Rightarrow \text{COP}=\dfrac{1}{1-\dfrac{273.15-5}{273.15+21}}\\\Rightarrow \text{COP}=11.3[/tex]

Input power

[tex]W_i=\dfrac{Q_h}{\text{COP}}\\\Rightarrow W_i=\dfrac{37.5}{11.3}\\\Rightarrow W_i=3.32\ \text{kW}[/tex]

The minimum power required to drive this heat pump is [tex]3.32\ \text{kW}[/tex].

Answer:

On July 13, 1787, Congress enacts the Northwest Ordinance, structuring settlement of the Northwest Territory and creating a policy for the addition of new states to the nation. ... In 1781, Virginia began by ceding its extensive land claims to Congress, a move that made other states more comfortable in doing the same

Answer:

They divided up the Northwest Territory into acre squares and sold them.

Explanation:

Answer:

D, meter.

Explanation:

Rhythm is associated with meter, which identifies units of stressed and unstressed syllables.

If a poem has a regular rhythm throughout the poem, it has a meter. Option D is correct.

Meter, which distinguishes between stressed and unstressed syllables, is related to rhythm. The fundamental rhythmic framework of a stanza or a line of poetry is known as meter.

The number of feet in the poem serves as a measure of the poem's meter, which is the rhythm of the language.

Many traditional poem forms call for a certain verse meter or a group of meters that alternate in a specified pattern. Prosody refers to both the study of meters and other types of versification, as well as their practical application.

Therefore, option D is correct.

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70.40mol cuz

1g sodium sulfate = 0.00704mol

take 10kg × 1000 = 10,000g

10,000g × 0.00704

final answer 70.40mol

(as per my thinking)

Answer:

70.40mol cuz

1g sodium sulfate = 0.00704mol

take 10kg × 1000 = 10,000g

10,000g × 0.00704

final answer 70.40mol

You've asked an Incomplete question, lacking options. I answered based on the existing O*NET report.

Answer:

high school diploma

Explanation:

According to the Occupational Information Network (O*NET), most people who are Licensing Examiners and Inspectors typically have a high school diploma.

In other words, they do not seek to acquire a post-secondary school education.

Answer:

B

Explanation:

According to edge its answer B

associate's degree or on-the-job experience

got it right as a lucky guess as the O*net site is updated but edge doesn't bother to update their questions or links.

Click this link to view O*NET’s Education section for Licensing Examiners and Inspectors. According to O*NET, what is the most common level of education among Licensing Examiners and Inspectors?  

bachelor’s degree

associate's degree or on-the-job experience .......This is the correct answer.

some college, no degree

associate degree

Answer:

mechanical engineer is the best answer

Answer:

u do the same thing as part B but only add 100 k, I think, cuz I'm still in middle school but I mean if u see it asks u to do the same thing as B but C says that instead, u do it at half pressure and 100 k is higher temp so what its asking is to repeat b but the twist is u do it at half pressure and 100 k is the higher temp

hope this helps :)  

Answer:

Q = [ mCp ( ΔT) ] [tex]_{cooling water }[/tex]

(ΔT)[tex]_{cooling water}[/tex] and  Q  is given

[tex]m_{cooling water}[/tex]  = [tex]\frac{Q}{Cp[ T_{out} - T_{in} ] }[/tex]

next the rate of condensation of the steam

Q = [ m[tex]h_{fg}[/tex] ][tex]_{steam}[/tex]

  [tex]m_{steam} = \frac{Q}{h_{fg} }[/tex]

Total resistance of the condenser is

R = [tex]\frac{Q}{change in T_{cooling water } }[/tex]

Explanation:

How will the rate of condensation of the steam and the mass flow rate of the cooling water can be determined

Q = [ mCp ( ΔT) ] [tex]_{cooling water }[/tex]

(ΔT)[tex]_{cooling water}[/tex] and  Q  is given

[tex]m_{cooling water}[/tex]  = [tex]\frac{Q}{Cp[ T_{out} - T_{in} ] }[/tex]

next the rate of condensation of the steam

Q = [ m[tex]h_{fg}[/tex] ][tex]_{steam}[/tex]

  [tex]m_{steam} = \frac{Q}{h_{fg} }[/tex]

Total resistance of the condenser is

R = [tex]\frac{Q}{change in T_{cooling water } }[/tex]

Answer:

the actual specific work required by the compressor to operate is 2425.88 kJ/kg

Explanation:

Given that;

h₁ = 1350 kJ/kg

h₂₅ = 3412 kJ/kg

compressor efficiency П_ise = 0.85

we know that;

compressor efficiency П_ise = isentropic work / actual work

П_ise = (h₂₅ - h₁) / (h₂ - h₁ )

so

0.85 =  (h₂₅ - h₁) / (actual work )

Actual work = (h₂₅ - h₁) / 0.85

Actual work = (3412 - 1350) / 0.85

Actual work = 2062 / 0.85

Actual work = 2425.88 kJ/kg

Therefore the actual specific work required by the compressor to operate is 2425.88 kJ/kg

Answer:

1.5510 mm

Explanation:

Plane strain fracture toughness = 45 MPa√m

failing stress ( б ) = 300 MPa

maximum length of surface crack ( a )= 0.95 mm

Determine maximum allowable surface crack length ( in mm )

we will make use of this relationship for  Design stress equation to determine the value of Y

б = [tex]\frac{k}{y\sqrt{\pi *a} }[/tex]    --------- ( 1 )

k = 45 MPa√m

б  = 300 MPa

a = 0.95 mm

y = ?

From equation 1 make Y subject of the equation ( also substitute values into equation 1 above )

hence ; y = 2.7457

Now determine maximum allowable surface crack when component is exposed to a stress of 300 MPa and made from another alloy with plane-strain fracture toughness of 57.5 MPa√m

we will apply the equation

б = [tex]\frac{k}{y\sqrt{\pi *a} }[/tex]    --------- ( 2 )

K = 57.5 MPa√m

б = 300 MPa

y = 2.7457

a ( maximum allowable surface crack ) = ?

from equation make a subject of the equation

a = [tex]\frac{1}{\pi } (\frac{k}{\alpha y} )^{2}[/tex]  

a = [tex]\frac{1}{\pi } (\frac{57.5}{300*2.7457} )^2[/tex]  =   1.5510 mm

Answer:

CLIMATE CHANGE HAS inexorably stacked the deck in favor of bigger and more intense fires across the American West over the past few decades, science has incontrovertibly shown. Increasing heat, changing rain and snow patterns, shifts in plant communities, and other climate-related changes have vastly increased the likelihood that fires will start more often and burn more intensely and widely than they have in the past.

Explanation:

Answer:

Maybe when there is a fire there can be fire drones that can take it out. and it can also resuce people who are stuck there.

Explanation:

Answer:

Para x=0:

[tex]\phi=1.2 W/m^{2}[/tex]  

Para x=30 cm:

[tex]\phi=-2.4 W/m^{2}[/tex]  

Explanation

Podemos utilizar la ley de Fourier par determinar el flujo de calor:

[tex]\phi=-k\frac{dT}{dx}[/tex](1)

Por lo tanto debemos encontrar la derivada de T(x) con respecto a x primero.

Usando la ley de potencia para la derivda, tenemos:

[tex]\frac{dT(x)}{dx}=300x-30[/tex]

Remplezando esta derivada en (1):

[tex]\phi=-0.04(300x-30)[/tex]

Para x=0:

[tex]\phi=0.04(30)[/tex]

[tex]\phi=1.2 W/m^{2}[/tex]  

Para x=30 cm:

[tex]\phi=-0.04(300*0.3-30)[/tex]

[tex]\phi=-2.4 W/m^{2}[/tex]    

Espero que te haya ayudado!

Answer:

generalization

Explanation:

Please mark me brainliest I need to level up

Answer:

FOLLOWED BY...  2 6 5 3 5 8 9

UwU

Answer:

The ability to read electrical schematics is a really useful skill to have. To start developing your schematic reading abilities, it's important to memorize the most common schematic symbols. ... You should also be able to get a rough idea of how the circuit works, just by looking at the schematic.

Explanation:

Answer:a

Ieieksdjd snsnsnsnsksks

Technicians A and B both are right with respect to their thinking and determination. Thus, the correct option is C.

Collision energy may be defined as a circumstance in which two or more bodies or particles come together with a resulting exchange of energy and alteration of direction.

Both pillar and floor pan reinforcement may be designed to transfer collision energy through the most prominent way to explain this mechanism of collisions.

A pillar exerts complete pressure on the floor in order to support the roof of the building, while floor pan reinforcement performs the same mechanism of collision energy with a different mechanism of action.

Therefore, technicians A and B both are right with respect to their thinking and determination. Thus, the correct option is C.

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A power washer is being used to clean the siding of a house. Water at the rate of 0.1 kg/s enters at 20°c and 1 atm, with the velocity 0.2m/s. The jet of water exits at 23°c, 1 atm with a velocity 20m/s at an elevation of 5m. At steady state, the magnitude of the heat transfer rate from power unit to the surroundings is 10% of the power input. Determine the power input to the motor in kW.

Answer:

Net power of 1.2 KW is being extracted

Explanation:

We are given;

Mass flow rate; m' = 0.1kg/s

Inlet temperature; T1 = 20°C = 293K

Inlet pressure; P1 = 1 atm = 10^(5) pa

Inlet velocity; v1 = 0.2 m/s

Exit Pressure; P2 = 1 atm = 10^(5) pa

Exit Temperature; T2 = 1 atm = 296K

Exit velocity; V2 = 20m/s

Change in elevation; h = Z2 - Z1 = 5m

We are told that the magnitude of the heat transfer rate from the power unit to the surroundings is 10% of the power input.

Thus;

Q = -0.1W

From Bernoulli equation;

Q - W = ∆Potential energy + ∆Kinetic energy + ∆Pressure energy

Where;

∆Potential energy = mg(z2 - z1)

∆Kinetic energy = ½m(v2² - v1²)

∆Pressure energy = mc_p(T2 - T1)

Thus;

-0.1W - W = [m'g(z2 - z1)] + [½m'(v2² - v1²)] + [m'c_p(T2 - T1)]

Where C_p is specific heat capacity of water = 4200 J/Kg.k

Plugging in the relevant values, we have;

-1.1W = (0.1 × 9.81 × 5) + (½ × 0.1(20² - 0.2²)) + (0.1 × 4200 × (296 - 293))

-1.1W = 4.905 + 19.998 + 1260

-1.1W = 1284.903

W = -1284.903/1.1

W ≈ -1168 J/s ≈ -1.2 KW

The negative sign means that work is extracted from the system.

Answer:

Explanation:

The cylinder head sits on the engine and closes off the combustion chamber. The gap that remains between the cylinder head and the engine is completed by the head gasket. Another task of the cylinder head is to ensure the constant lubrication of the cylinder        

Answer:

Less Friction Error

Explanation:

The friction error is very less in the moving

iron instrument because their torque weight

ratio is high.

hope it helps!

Answer:

Universal Use -The MI instrument is independent of the of current and hence for both AC and DC

Explanation:

Mark as Brainlist Answer

Answer:

a) for back pressures of (a) 500 kPa

- mass flow rate is 2.127 kg/s

- exit Mach number is 1.046

b) for back pressures of (a) 784 kPa

- mass flow rate is 1.793 kg/s

- exit Mach number is 0.6

Explanation:

Given that;

A₂ = 0.001 m²

P₁ = 1 MPa

T₁ = 360 K

k = 1.4

P₂ = 500 Kpa

(1000/500)^(1.4-1 / 1.4) = 360 /T₂

2^(0.4/1.4) = 360/T₂

1.219 = 360 / T₂

T₂ = 360 / 1.219

T₂ = 295.32 K

CpT₁ + V₁²/2000 = CpT₂ + V₂²/2000

we substitute

CpT₁ + V₁²/2000 = CpT₂ + V₂²/2000

1.005 × 360 =  1.005 × 295.32 + v₂²/2000

v₂ = 360.56 m/s²

p₂v₂ = mRT₂

500 × (0.001 × 360.56) = m × 0.287 × 295.32

m = 2.127 kg/s

so Mach Number = V₂ / Vc

Vc = √( kRT) = √( 1.4 × 287 × 295.32) = 344.47 m/s

So Mach Number =  V₂ / Vc  =  360.56 / 344.47 = 1.046

Therefore for back pressures of (a) 500 kPa

- mass flow rate is 2.127 kg/s

- exit Mach number is 1.046

b)

AT P₂ = 784 kPa

(1000/784)^(1.4-1 / 1.4) = 360/T₂

T₂ = 335.82 K

now

V₂²/2000 = 1.005( 360 - 335.82)

V₂ = 220.45 m/s

P₂V₂ = mRT₂

784 × (0.001 × 220.45) = m( 0.287) ( 335.82)

172.83 = 96.38 m

m = 172.83 / 96.38

m = 1.793 kg/s

just like in a)

Vc = √( kRT) = √( 1.4 × 287 × 335.82) = 367.32 m/s

Mach Number = V₂ / Vc = 220.45 / 367.32 = 0.6

Therefore for back pressures of (a) 784 kPa

- mass flow rate is 1.793 kg/s

- exit Mach number is 0.6

Following are the

Given:

[tex]A_2 = 0.001\ m^2\\\\P_1 = 1\ MPa\\\\T_1 = 360\ K\\\\k = 1.4\\\\P_2 = 500\ Kpa\\\\[/tex]

To find:

Flow rate of mass, and Mach number

Solution:

For point a)

Using formula:

[tex]\to P^{\frac{r-1}{r}} \alpha T\\\\\to (\frac{1000}{500})^(\frac{1.4-1}{1.4}) = \frac{360}{T_2}\\\\\to 2^(\frac{0.4}{1.4}) = \frac{360}{T_2}\\\\\to 1.219 = \frac{360}{T_2}\\\\\to T_2 = \frac{360}{1.219}\\\\\to T_2 = 295.32\ K\\\\[/tex]

[tex]\to C_pT_1 + \frac{V_1^2}{2000} = C_pT_2 + \frac{V_2^2}{2000}\\\\\to 1.005 \times 360 = 1.005 \times 295.32 + \frac{v_2^2}{2000}\\\\\to v_2 = 360.56 \ \frac{m}{s^2} \\\\\to p_2v_2 = mRT_2\\\\\to 500 \times (0.001 \times 360.56) = m \times 0.287 \times 295.32\\\\\to m = 2.127\ \frac{kg}{s}\\\\[/tex]

Mach Number [tex]= \frac{V_2}{V_c}\\\\[/tex]

[tex]\to V_c = \sqrt{( kRT)} = \sqrt{( 1.4 \times 287 \times 295.32)} = 344.47 \ \frac{m}{s}\\\\[/tex]

[tex]\to \frac{V_2}{V_c} = \frac{360.56}{344.47} = 1.046[/tex]

For point b)

[tex]\to P_2 = 784\ kPa\\\\\to (\frac{1000}{784})^{(\frac{0.4}{1.4})} = \frac{360}{T_2}\\\\\to T_2 = 335.82\ K\\\\[/tex]

now

[tex]\to \frac{V_2^2}{2000} = 1.005( 360 - 335.82)\\\\\to V_2 = 220.45 \frac{m}{s}\\\\\to V_c = \sqrt{( kRT)} = \sqrt{( 1.4 \times 287 \times 335.82)} = 367.32\ \frac{ m}{s}\\\\\to c= \frac{220.45}{ 367.32} = 0.6\\[/tex]

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Answer:

Air at 1 atm and 25◦C blows across a large concrete surface 20 m wide maintained

at 60◦C. The flow velocity is 6 m/s. Calculate the convection heat loss from the

surface.

This is heat transfer convection, mechanical engineering

please solve this question guys I'm gonna really really be appreciate it for you guys

Answer:

false

Explanation:

Answer:

7.47 lb. s/ft

22.42  lb. s/ft

Explanation:

Without mincing words let us dive straight into the solution to the above problem.

STEP ONE: calculate or determine the frequency.

The frequency can be calculated by making use of the formula given below;

frequency, w = √k/m. Thus, frequency = √ 12 × 15/ [40/32.2] = 12 rad/s.

STEP TWO: Determine or calculate for the viscous damping coefficient, c  for damping ratio of 0.5 and 1.5 respectively.

The viscous damping coefficient, c for 0.5 = [ 12 × 0.5 × 2 ×[ 40/32.2] / 2= 7.47 lb. s/ft.

The viscous damping coefficient, c for 1.5= [ 12 × 1.5 × 2 ×[ 40/32.2] / 2 = 22.42  lb. s/ft.

Answer:

When are resistors in series? Resistors are in series whenever the flow of charge, called the current, must flow through devices sequentially. For example, if current flows through a person holding a screwdriver and into the Earth, then  

R

1

 in Figure 1(a) could be the resistance of the screwdriver’s shaft,  

R

2

 the resistance of its handle,  

R

3

 the person’s body resistance, and  

R

4

 the resistance of her shoes.

Figure 2 shows resistors in series connected to a voltage source. It seems reasonable that the total resistance is the sum of the individual resistances, considering that the current has to pass through each resistor in sequence. (This fact would be an advantage to a person wishing to avoid an electrical shock, who could reduce the current by wearing high-resistance rubber-soled shoes. It could be a disadvantage if one of the resistances were a faulty high-resistance cord to an appliance that would reduce the operating current.)

Two electrical circuits are compared. The first one has three resistors, R sub one, R sub two, and R sub three, connected in series with a voltage source V to form a closed circuit. The first circuit is equivalent to the second circuit, which has a single resistor R sub s connected to a voltage source V. Both circuits carry a current I, which starts from the positive end of the voltage source and moves in a clockwise direction around the circuit.

Figure 2. Three resistors connected in series to a battery (left) and the equivalent single or series resistance (right).

To verify that resistances in series do indeed add, let us consider the loss of electrical power, called a voltage drop, in each resistor in Figure 2.

According to Ohm’s law, the voltage drop,  

V

, across a resistor when a current flows through it is calculated using the equation  

V

=

I

R

, where  

I

 equals the current in amps (A) and  

R

 is the resistance in ohms  

(

Ω

)

. Another way to think of this is that  

V

 is the voltage necessary to make a current  

I

 flow through a resistance  

R

.

So the voltage drop across  

R

1

 is  

V

1

=

I

R

1

, that across  

R

2

 is  

V

2

=

I

R

2

, and that across  

R

3

 is  

V

3

=

I

R

3

. The sum of these voltages equals the voltage output of the source; that is,

V

=

V

1

+

V

2

+

V

3

.

This equation is based on the conservation of energy and conservation of charge. Electrical potential energy can be described by the equation  

P

E

=

q

V

, where  

q

 is the electric charge and  

V

 is the voltage. Thus the energy supplied by the source is  

q

V

, while that dissipated by the resistors is

q

V

1

+

q

V

2

+

q

V

3

.

Explanation:

Answer:

Hello your question is incomplete  below is the complete question

Electronic components are often mounted with good heat conduction paths to a finned aluminum base plate, which is exposed to a stream of cooling air from a fan. The sum of the mass times specific heat products for a base plate and components is 5000 J/K, and the effective heat transfer coefficient times surface area product is 10 W/K. The initial temperature of the plate and the cooling air temperature are 295 K when 300 W of power are switched on. 1) Find the plate temperature after 10 minutes.

answer ; 311.36 k

Explanation:

Given data :

sum of mass * specific heat products for a base plate and components ( Mcp )

= 5000 J/K

effective heat transfer coefficient * surface area ( hA )  = 10 W/K

Initial temperature of plate and cooling air temperature( Tc ) = 295 k

power ( Q = W ) = 300 W

a) Determine plate temperature after 10 minutes

10 mins = 600 secs ( t )

heat supplied = change in temp + heat loss

          Q * t    = mCp ( ΔT ) + hA ( ΔT ) t

   300*600   =  5000 * ( T -295 )  + 10 ( T -295 ) * 600

therefore ; T - 295 = 16.363

                           T = 311.36 K