Given a transfer function H(w)= jw/(jw+1000), find the gain (V/V) at a frequency of 0.19 kHz. Enter your answer to 3 signficant figures. 2 points Save Answer
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Answer:

SpaceX

Explanation:

SpaceX successfully launches 60 more Starlink satellites as it continues towards 2020 service debut. SpaceX has launched another big batch of Starlink satellites, the low Earth orbit spacecraft that will provide connectivity for its globe-spanning high-bandwidth broadband internet network.

Answer:

Problem definition

Explanation:

The engineering design is a series of step that the engineer gradually take in other to create a functioning product or process. There are many different models of the engineering design process , but they can be shrunk into four basic steps which are

The students are within the first step, and the problem here is 'an appropriate means for the students to keep their drinks cold, for 3 hours for the ball game.'

Answer:

(The diagram of the question is given in Attachment 1)

The largest load which can be applied is:

P=67.62 kN

Explanation:

All the forces are shown in the diagram in Attachment 2.

∑ F(y)= 0 (Upwards is positive)

\(F_{BC}sin\theta-P=0\\\frac{4}{5}F_{BC} - P=0\\F_{BC}=\frac{5}{4}P\\\\F_{BC}=1.25P\)

∑ F(x)= 0 (Towards Right is positive)

\(N_{a-a} - F_{BC}cos\theta=0\\N_{a-a}-1.25P(\frac{3}{5})=0\\N_{a-a}=0.75P\)

∑ F(y)= 0 (Upwards is positive)

\(F_{BC}sin\theta- V_{a-a}= 0\\(\frac{4}{5})1.25P-V_{a-a}=0\\V_{a-a}=P\)

The cross sectional area of a-a:

\(A_{a-a}= \frac{(0.026)(0.026)}{3/5}\\A_{a-a}= 1.127\cdot10^{-3}\)

σ = N(a-a)/A(a-a)

Substitute values:

\(160\cdot10^6=\frac{0.75P}{1.127\cdot10^{-3}}\\P=240.42\cdot10^3 N\\P=240.42 kN\)

Т= V(a-a)/A(a-a)

Substitute values:

\(60\cdot10^6=\frac{P}{1.127\cdot10^{-3}}\\P=67.62\cdot10^{3}N\\P=67.62kN\)

To satisfy both the condition, we have to choose the lower value of P.

P=67.62 kN

Answer:

12ma²

Explanation:

The moment of inertia I = ∑mr² where m = mass and r = distance from shaft

Since we have three masses,

So, I = m₁r₁² + m₂r₂² + m₃r₃² where m₁ = first mass = m, m₂ = second mass = 3m and m₃ = third mass = 2m. Also, r₁ = distance of first mass from shaft = a, r₂ = distance of second mass from shaft = a and r₃ = distance of third mass from shaft = 2a.

I = m₁r₁² + m₂r₂² + m₃r₃²

I = ma² + 3ma² + 2m(2a)²

I = ma² + 3ma² + 2m(4a²)

I = ma² + 3ma² + 8ma²

I = 12ma²

The following have an effect on the present day sporting ability of an electrical conductor is Voltage, Insulation type, Power.

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The electricity cost of the refrigerator per month, running only one quarter of the time, is : option (a) $4.9.

The power consumed by the refrigerator, P = 320 W

Unit cost of electricity, C = $0.13/kWh

Let the time for which the refrigerator runs in a day = t.

The total time taken by the refrigerator for running in 30 days = 30t.

We know that,Energy consumed = power x time. Therefore,The energy consumed by the refrigerator in

30 days = P × 30t = 320 × 30t = 9600t Wh

The electricity cost for 1 kWh = $0.13

Therefore, the electricity cost for 9600t Wh = $0.13 × 9600t = 1248t cents

We know that, 100 cents = $1

Therefore, the electricity cost for 9600t Wh = 12.48t $For 24 hours, the refrigerator runs 1/4th of the time.So,

t = (1/4) × 24 = 6 hours = 0.25 days

Therefore, The energy consumed by the refrigerator in 30 days = 9600t Wh = 9600 × 0.25 Wh = 2400 Wh = 2.4 kWh

The electricity cost for 2.4 kWh = 2.4 × 0.13 = $0.312Therefore, the main answer is option (a) $4.9.

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The pressure at Section 2 in the sewage system is the same as the pressure at Section 1, which is 50 psi.

To determine the pressure at Section 2, we can use the principle of conservation of energy in fluid flow, specifically Bernoulli's equation. Let's follow these steps:

Step 1: Understand the given information and variables:

- Diameter at Section 1: D1 = 16 inches

- Diameter at Section 2: D2 = 8 inches

- Flow rate through the pipe: Q = 6 cfs (cubic feet per second)

- Pressure at Section 1: P1 = 50 psi (pounds per square inch)

- Pressure at Section 2: P2 (to be determined)

Step 2: Convert the diameters from inches to feet:

Since the flow rate is given in cubic feet per second, it is essential to have all dimensions in consistent units. Convert the diameters from inches to feet by dividing them by 12.

D1 = 16 inches / 12 = 1.33 feet

D2 = 8 inches / 12 = 0.67 feet

Step 3: Calculate the areas of the pipe sections:

The cross-sectional area of a circular pipe can be calculated using the formula: A = (π/4) * D^2.

A1 = (π/4) * (1.33 feet)^2

A2 = (π/4) * (0.67 feet)^2

Step 4: Apply Bernoulli's equation:

Bernoulli's equation states that the total energy per unit weight of a fluid flowing in a pipe is constant along a streamline. It can be expressed as:

P1 + (ρg/144) + (V1^2/2g) = P2 + (ρg/144) + (V2^2/2g)

where:

P1 and P2 are the pressures at Sections 1 and 2, respectively

ρ is the fluid density (assumed to be constant)

g is the acceleration due to gravity

V1 and V2 are the velocities of the fluid at Sections 1 and 2, respectively

Since we are interested in comparing pressures, we can disregard the terms involving density and velocity. Therefore, the equation becomes:

P1 = P2

Hence, the pressure at Section 2 is equal to the pressure at Section 1, which is 50 psi.

To summarize, the pressure at Section 2 in the given sewage system is the same as the pressure at Section 1, which is 50 psi.

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The magnitude and sense of the horizontal force applied  is 333.60 N

Force exerted on an object by another object. An example of applied force is a person pushing a barrel. When the individual pushes the barrel, a force is applied to the barrel.

A contact force applied to an object by external means is referred to as an applied force. The object moves or deforms as a result of applied force.

Since C is horizontal C i =C

r = DC = (0.2m )i - (0.125 m )j

MD = r × Ci = C = (0.125 m )k

41.7 N  m = (0.125 m )(C )

C = 333. 60 N

(c) The smallest force C must be perpendicular to DC; thus, it forms a with the vertical

tan α = 0.125m/0.2m

α = 32.0²

MD = C(DC);DC = √[(0.2m)²+(0.125m)²]

= 0.23585m

41.70 N = C (0.23585 m )

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Answer:

5,465.4165939453

Explanation:

formula

A=P(1+r/n)^n(t)

p=1000

r=0.09

n=4

t=5

The waveforms delta (0.5 to 4Hz), theta (4 to 7Hz), alpha (8 to 12Hz), sigma (12 to 16Hz), and beta are among the most frequently researched waveforms (13 to 30Hz).

EEG frequency patterns can be divided into four categories: beta (14–30 Hz), alpha (8–13 Hz), theta (4–7 Hz), and delta (1-3 Hz). In general, when the frequency drops, the EEG's amplitude rises. Each of the four EEG frequencies corresponds to a distinct degree of cerebral cortex arousal.

The frequency is then calculated by taking I and dividing it by the duration, which in this case results in a 5Hz wave. the amount of analogue millimetres in a second.

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Answer:

Applied pressure on piston (P) = 50 psi. We know that work done is given by W = Force x displacement. S Force = Pressure x Areas > W = ( PXA ) X L W = PX ( AL ) W = PX AV = 50x 43.7088. W= 2185.

Please give me brainliest -  you get 25% as well! I swear!

The total heat lost per meter of length can be calculated by adding the heat transfers through the pipe, the first layer of insulation, and the second layer of insulation: Heat lost per meter of length = Q1 + Q2 + Q3.

Heat transfer through the pipe:The heat transfer through the pipe can be calculated using the formula:  Q1 = (2 * π * k1 * L * (T1 - T2)) / ln(r2 / r1) Q1 is the heat transfer through the pipe per meter of length.π is a mathematical constant (approximately 3.14).k1 is the thermal conductivity of the pipe material (47 W/m·°C).L is the length of the pipe.T1 is the inside surface temperature of the pipe (250 °C).T2 is the outside surface temperature of the pipe (20 °C).r1 is the inside radius of the pipe (half of the inside diameter, 8 cm).r2 is the outside radius of the pipe (r1 + wall thickness, 8 cm + 0.055 cm).

Heat transfer through the first layer of insulation:The heat transfer through the first layer of insulation can be calculated using the formula: Q2 = (2 * π * k2 * L * (T2 - T3)) / ln(r3 / r2)
Where: Q2 is the heat transfer through the first layer of insulation per meter of length.k2 is the thermal conductivity of the first layer of insulation (0.5 W/m·°C).T3 is the outside surface temperature of the first layer of insulation (20 °C). r3 is the outside radius of the first layer of insulation (r2 + 9 cm).The total heat lost per meter of length can be calculated by adding the heat transfers through the pipe, the first layer of insulation, and the second layer of insulation:Heat lost per meter of length = Q1 + Q2 + Q3.

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Answer:

Explanation:

The critical path is the longest path through a project network diagram, which includes all the activities that are critical to the completion of the project. In order to calculate the critical path, you will need to follow these steps:

Identify all the activities that need to be completed for the project.

Determine the duration of each activity. This can be estimated based on past experience or expert opinion, or by conducting time studies.

Create a network diagram that shows the sequence of activities and their dependencies. This can be done using software such as Microsoft Project or by drawing a diagram manually.

Identify the earliest start and finish times for each activity, based on the dependencies and durations.

Identify the latest start and finish times for each activity, based on the project deadline.

Calculate the total float or slack for each activity, which is the amount of time that an activity can be delayed without affecting the overall project completion date.

Identify the critical path, which is the longest path through the project network diagram that has zero total float or slack. This means that any delay in one of the activities on the critical path will cause a delay in the overall project completion date.

Once you have identified the critical path, you can focus your attention on managing and monitoring the activities on that path to ensure that the project stays on schedule.

In any mechanical project, calculating the measurements of the equipment is an essential task.

As a result, it is paramount to reference the numerical details outlined in either the drawings or specifications related to that endeavor.

Fundamentally, these documents comprise height, width, and length. If precise dimensions are not available, collaborating with mechanical engineers or equipment manufacturers could be beneficial.

Moreover, guarantee that the measurements harmonize with the conception and capacity of the project to keep away from probable superfluous installation or usefulness snags.

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Answer:

Attached below is the Radial power circuit arrangement

Explanation:

Radial power circuit arrangement is done in a way that a single cable starts from the fuse box and connects to all the outlet socket contained in the circuit also the cable contains wires ( live , neutral and earth )

The advantage of a radial power circuit arrangement is that it enables easy identification of electrical faults on the circuit.

A four-wire distribution system was created from a three-phase, three-wire ungrounded system. The percent must the overcurrent device be set to trip in case of an overload is 125% by nec section 450.5(a)(2).

Thus, for a four-wire distribution system was created from a three-phase, three-wire ungrounded system. The percent must the overcurrent device be set to trip in case of an overload is 125% by nec section 450.5(a)(2).

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Answer:

false

Explanation:

Answer: rate brainliest

Single-element fuses are particularly suited for the protection of circuit breakers with low interrupting ratings because they are designed to operate quickly and reliably under overload or short circuit conditions. This is important for circuit breakers with low interrupting ratings, as these devices may not be able to handle high currents without damaging themselves or causing other hazards. Single-element fuses are able to quickly interrupt the flow of current in the circuit, which can prevent the circuit breaker from being overloaded and help to protect the electrical system from damage.

Explanation:

A Pb-Sn alloy contains 23% primary alpha phase (α) and 77% eutectic microconstituent immediately after the eutectic reaction.

How to determine the composition?

To determine the composition of the alloy, we will refer to the Pb-Sn phase diagram. The eutectic composition in this system is approximately 61.9% Sn and 38.1% Pb. Since the eutectic microconstituent makes up 77% of the alloy, the primary alpha phase (which is the Pb-rich phase) contributes the remaining 23%. By using a lever rule calculation, you can determine the exact composition of the alloy.

Therefore, the composition of the alloy immediately after the eutectic reaction is 0% alpha and 100% eutectic microconstituent.

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Answer:2%

When the voltages between the three phases are not equal, the current increases dramatically in the motor winding's, and if allowed to continue, the motor will be damaged.

Explanation:

To prepare the hot gas leaving the compressor for condensation, it needs to undergo a process called desuperheating, which involves reducing its temperature.

The hot gas discharged from the compressor is at a significantly higher temperature compared to the condenser line because of the compression process. When the refrigerant is compressed, its pressure and temperature increase. However, the pressure-temperature relationship does not hold true for the discharge line due to the presence of superheated gas. Superheated gas is a state where the refrigerant temperature is higher than its saturation temperature at a given pressure. To facilitate condensation, the hot gas leaving the compressor needs to be desuperheated. Desuperheating involves removing excess heat from the gas to bring its temperature closer to the saturation temperature for a particular pressure.

This is typically achieved by passing the gas through a desuperheater or heat exchanger, where it comes in contact with a cooling medium, such as air or water. The cooling medium absorbs heat from the gas, reducing its temperature. Desuperheating is an important step before condensation because it ensures that the refrigerant is at the appropriate temperature for efficient heat transfer in the condenser. If the hot gas were directly condensed without desuperheating, it could result in inadequate heat rejection, reduced efficiency, and potential damage to the condenser due to excessive heat load. Therefore, desuperheating helps optimize the overall performance and reliability of the refrigeration or air conditioning system.

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The average cost of each trip is $8 per trip while the marginal cost is $0

We are given;

Monthly payment by Frank for Membership fees = $80

Number of times frank goes to the center monthly = 10 times

Thus, average cost is;

Average cost = Monthly payment/Number of visits in a month

Average Cost = $80/10

Average Cost = $8 per trip

The marginal cost is the change in total production cost that comes from making or producing one additional unit. However, in this question, his monthly pay covers an unlimited use of its facilities. Thus, there can be no marginal cost for an additional work-out session. Thus;

Marginal Cost = $0

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Answer:a

Ieieksdjd snsnsnsnsksks

B

a jsdnjwevhfgruewbkuwygru

Answer:

a) diameter available = 0.0384 nm

b)The space  is smaller than the carbon atom which has a  radius of  0.077 nm and this simply means that the carbon atom will not conquer these sites

Explanation:

For BCC iron

From Appendix B given,select the lattice parameter ( a ) as = 0.2866 nm

The BCC iron has 4 atomic radii and therefore the body diagonal length = \(a(3)^\frac{1}{2}\)

expressing the atomic radius of the BCC iron

4r = \(a(3)^\frac{1}{2}\)

insert the value of (a) from appendix B which is = 0.2866 nm

4r = \(0.2866 nm (3)^\frac{1}{2}\)

therefore  r =  0.4964 nm / 4 = 0.1241 nm

Refer again to appendix C given select the atomic radius of the BCC iron as = 0.1241 nm   assuming the atomic radius of the iron are the same

then the radius ratio = 0.62

Refer to the Figure 3.2 given, the amount of space required for an interstitial at the BCC position is between the atoms at the FCC position and also in this space there are two atoms that are equal to a radius of 0.2482 nm

The diameter of the minimum space available

\(d_{a} = a - r_{a}\)

\(r_{a} = atomic radii\) = 0.2482 nm

a = 0.2666 nm

therefore

d\(_{a}\) = 0.2866 nm - 0.2482 nm = 0.0384 nm

comparing this to the diameter of a carbon atom

The space  is smaller than the carbon atom which has a  radius of  0.077 nm and this simply means that the carbon atom will not conquer these sites