A circular pipe with diameter 1.50( m) is filled with sand, the flow rate through the pipe is 0.22( m3/ day), the length of the pipe is 5 m and the hydraulic gradient is 0.03, calculate the hydraulic conductivity K in m/day for the sand.

Answer:

SECTION LEARNING OBJECTIVES

By the end of this section, you will be able to do the following:

Distinguish between static friction and kinetic friction

Solve problems involving inclined planes

Section Key Terms

kinetic friction static friction

Static Friction and Kinetic Friction

Recall from the previous chapter that friction is a force that opposes motion, and is around us all the time. Friction allows us to move, which you have discovered if you have ever tried to walk on ice.

There are different types of friction—kinetic and static. Kinetic friction acts on an object in motion, while static friction acts on an object or system at rest. The maximum static friction is usually greater than the kinetic friction between the objects.

Imagine, for example, trying to slide a heavy crate across a concrete floor. You may push harder and harder on the crate and not move it at all. This means that the static friction responds to what you do—it increases to be equal to and in the opposite direction of your push. But if you finally push hard enough, the crate seems to slip suddenly and starts to move. Once in motion, it is easier to keep it in motion than it was to get it started because the kinetic friction force is less than the static friction force. If you were to add mass to the crate, (for example, by placing a box on top of it) you would need to push even harder to get it started and also to keep it moving. If, on the other hand, you oiled the concrete you would find it easier to get the crate started and keep it going.

Figure 5.33 shows how friction occurs at the interface between two objects. Magnifying these surfaces shows that they are rough on the microscopic level. So when you push to get an object moving (in this case, a crate), you must raise the object until it can skip along with just the tips of the surface hitting, break off the points, or do both. The harder the surfaces are pushed together (such as if another box is placed on the crate), the more force is needed to move them.

Answer:

The number of vehicles added to this highway before the capacity is reached is 1,511 vehicles.

Explanation:

see attached image

Answer:

The thrust of the engine calculated using the cold air is 34227.35 N

Explanation:

For the turbofan engine, firstly the overall mass flow rate is considered. The mass flow rate is given as

\(\dot{m}=\rho AV_a\)

Here

So the equation becomes

\(\dot{m}=\rho AV_a\\\dot{m}=\dfrac{P}{RT} AV_a\\\dot{m}=\dfrac{50506.80}{286.9\times 253} \times (\dfrac{\pi}{4}\times 2^2)\times 250\\\dot{m}=546.4981\ kgs^{-1}\)

Now in order to find the flow from the fan, the Bypass ratio is used.

\(\dot{m}_f=\dfrac{BPR}{BPR+1}\times \dot{m}\)

Here BPR is given as 8 so the equation becomes

\(\dot{m}_f=\dfrac{BPR}{BPR+1}\times \dot{m}\\\dot{m}_f=\dfrac{8}{8+1}\times 546.50\\\dot{m}_f=485.77\ kgs^{-1}\)

Now the exit velocity is calculated using the total energy balance which is given as below:

\(h_4+\dfrac{1}{2}V_a^2=h_5+\dfrac{1}{2}V_e^2\)

Here

So the equation becomes

\(h_4+\dfrac{1}{2}V_a^2=h_5+\dfrac{1}{2}V_e^2\\c_pT_4+\dfrac{1}{2}V_a^2=c_pT_5+\dfrac{1}{2}V_e^2\)

Rearranging the equation gives

\(\dfrac{1}{2}V_e^2=c_pT_4-c_pT_5+\dfrac{1}{2}V_a^2\\\dfrac{1}{2}V_e^2=c_p(T_4-T_5)+\dfrac{1}{2}V_a^2\\V_e^2=2c_p(T_4-T_5)+V_a^2\\V_e=\sqrt{2c_p(T_4-T_5)+V_a^2}\\V_e=\sqrt{2\times 1005\times (253-233)+(250)^2}\\V_e=320.46 m/s\)

Now using  the cold air approach, the thrust is given as follows

\(T=\dot{m}_f(V_e-V_a)\\T=485.77\times (320.46-250)\\T=34227.35\ N\)

So the thrust of the engine calculated using the cold air is 34227.35 N

Answer:

Perspective view drawing.

Answer: It does make sense, because I've been involved in these careers and have a long family line of them. And other questions?

Explanation:

A high-performance graphics card (GPU) is a type of device that would use a PCIe 6/8-pin connector.

A PCIe 6/8-pin connector, also known as a PCI Express power connector, is designed to provide additional power to high-end graphics cards. These connectors are commonly found on modern power supplies and are used to supply the necessary power directly to the graphics card.

High-performance graphics cards, particularly those intended for gaming or intensive graphical applications, require a significant amount of power to operate efficiently. The PCIe 6/8-pin connector ensures a stable and dedicated power supply to the graphics card, allowing it to deliver optimal performance without relying solely on the power provided by the motherboard.

By connecting the PCIe 6/8-pin power cables from the power supply to the corresponding connectors on the graphics card, users can ensure that their high-end GPU receives the necessary power to operate reliably and perform at its best.

In summary, a device that would use a PCIe 6/8-pin connector is a high-performance graphics card (GPU) that requires additional power beyond what the motherboard can provide.

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Answer:

2.47  ft

Explanation:

Given that:

The initial temperature of air = 80°F

Diameter of the pipe = 72 ft

average velocity \(v_{air}\) of the air flow through the pipe =  34 ft/s

The objective is to determine the diameter of the  pipe to  be used to move water at:

At a temperature = 60°F   &

An average velocity \(v_{water}\) of 71 ft/s

Assuming Reynolds number similarity is enforced;

where :

kinematic viscosity (V_air) of air at 80 °F  (V_air)  = 1.69 × 10⁻⁴ ft²/s

kinematic viscosity of water  at 60 °F (V_water) = 1.21 × 10⁻⁵ ft²/s

The diameter of the pipe can be calculated by using the expression:

\(D_{water} = \dfrac{V_{water}}{V_{air}}*\dfrac{v_{air}}{v_{water}}* D_{air}\)

\(D_{water} = \dfrac{1.21*10^{-5} \ ft^2/s}{1.69*10^{-4} \ ft^2/s}*\dfrac{34 \ ft/s}{71 \ ft/s}* 72 \ ft\)

\(D_{water} =\) 2.4686  ft

\(D_{water} =\) 2.47 ft   ( to two decimal places)

Thus; diameter pipe to be use to move water at the given temperature and average velocity is 2.47 ft

Answer:

2.47  ft

Explanation:

Given that:

The initial temperature of air = 80°F

Diameter of the pipe = 72 ft

average velocity  of the air flow through the pipe =  34 ft/s

The objective is to determine the diameter of the  pipe to  be used to move water at:

At a temperature = 60°F   &

An average velocity  of 71 ft/s

Assuming Reynolds number similarity is enforced;

where :

kinematic viscosity (V_air) of air at 80 °F  (V_air)  = 1.69 × 10⁻⁴ ft²/s

kinematic viscosity of water  at 60 °F (V_water) = 1.21 × 10⁻⁵ ft²/s

The diameter of the pipe can be calculated by using the expression:

2.4686  ft

2.47 ft   ( to two decimal places)

Thus; diameter pipe to be use to move water at the given temperature and average velocity is 2.47 ft

The peak frequency deviation, minimum bandwidth, and baud for a binary FSK signal for the given frequencies are respectively;

a) 0.5 kHz

b) 9 kHz

c) 4000

1) The peak frequency deviation is gotten from the formula;

∆f = |f_m - f_s|/f_b

where;

Thus;

∆f = |38 - 40|/4

∆f = 0.5 kHz

2) The minimum bandwidth is given by the formula;

B = 2(∆f + f_b)

B = 2(0.5 + 4)

B = 9 kHz

3) For FSK signal, N = 1, and the baud is gotten from the Equation;

baud = f_b/1

f_b = 4 kbps = 4000 bps

Thus; baud = 4000/1 = 4000

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Answer:

yes answer d is correct

Communication Process are actions that create physical solutions to problems. The correct option is d.

Human existence and organisational survival both depend on effective communication. It is a process of generating and disseminating thoughts, facts, opinions, and sentiments from one place, individual, or group to another. The Management function of Directing depends on effective communication.

The sending party, message encoding, channel selection, message receipt by the recipient, and message decoding are all aspects of the communication process.

Feedback is when the recipient communicates something back to the original sender. These procedures are actions that result in tangible fixes for issues.

Thus, the correct option is d.

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If an engine fails during rollout or just before takeoff, immediately shut both throttles and land the aircraft safely. Before reaching a safe single engine speed right away after takeoff, drop your nose to increase velocity.

Closing the throttle, turning off the fuel pump, setting the mixture control to idle cutoff, and simply cranking the engine is the most reliable hot start method I've found.

If an engine fails during rollout or just before takeoff, immediately shut both throttles and land the aircraft safely. Before reaching a safe single engine speed right away after takeoff, drop your nose to increase velocity. If you are unable to climb, close both throttles and land straight ahead.

The typical practice for the majority of aircraft would be to abandon takeoff if an engine failed during takeoff. In small aircraft, the pilot should turn the throttles down to idle, activate the speed brakes (if provided), and apply the brakes as needed if the engine fails before VR (Rotation Speed).

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The three main types of waves are Transverse Waves, Longitudinal Waves, and Surface Waves.

Transverse Waves: Transverse waves are waves in which the particles of the medium move perpendicular (or at right angles) to the direction in which the wave is moving. Examples of transverse waves include light waves, water waves, and seismic S waves.

Longitudinal Waves: Longitudinal waves are waves in which the particles of the medium move parallel to the direction in which the wave is moving. Examples of longitudinal waves include sound waves, seismic P waves, and waves in a coiled spring.

Surface Waves: Surface waves are waves that occur at the interface between two different media, such as air and water, or rock and soil. Surface waves are a combination of both transverse and longitudinal motion and move along the surface of the medium. Examples of surface waves include ocean waves, seismic Love waves, and Rayleigh waves.

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❎❎❎❎❎❎❎ sorry but that didn't help me that much

Answer:

Diameter of pipe is 0.535 ft

Explanation:

see attachment, its works out 1st half

Both of the Technician A and Technician B are correct.

The ridged foam can be used as a structural component in various parts of a vehicle which includes pillars and frames. It is a lightweight and strong material that can help improve fuel efficiency and reduce noise and vibration.

In addition, the ridged foam can also provide thermal insulation which can be beneficial in areas where heat or cold transfer is a concern. A proper design and testing should be conducted to ensure that the use of ridged foam is safe and effective in a particular application.

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Answer:

Explanation:

It is useful to consider the voltage at X to be the superposition of two voltages divided by the R1/R2 voltage divider. It is also helpful to remember that OUT = -VX·AVOL.

__

  \(V_X=\dfrac{V_1\cdot R_2+V_{out}\cdot R_1}{R_1+R_2}=\dfrac{V_1\cdot R_2-V_X\cdot A_{VOL}\cdot R_1}{R_1+R_2}\\\\V_X(R_1(1+A_{VOL})+R_2)=V_1\cdot R_2\\\\\boxed{V_X=V_1\dfrac{R_2}{R_2+R_1(1+A_{VOL})}}\)

__

Filling in the given numbers, we find VX to be ...

  \(V_X=-0.01V\cdot\dfrac{400000}{400000+4000(1+10000)}=-0.01V\cdot\dfrac{100}{10101}\\\\\boxed{V_X=\dfrac{-1}{10101}V\approx-99.0001\mu V}\)

__

As we said at the beginning, OUT = -VX·AVOL. Multiplying the expression for VX by -AVOL, we get ...

  \(\boxed{V_{out}=-V_1\cdot\dfrac{A_{VOL}\cdot R_2}{R_2+R_1(1+A_{VOL})}}\)

__

As with the expression, the output voltage is found by multiplying VX by -AVOL:

  \(V_{out}=\dfrac{(-1)(-10000)}{10101}V\\\\\boxed{V_{out}=\dfrac{10000}{10101}V\approx 0.990001V}\)

The dynamic viscosity of the oil in poise is 13.625 pois

The kinematic viscosity of the oil in strokes is 14.34

F viscous is given as n*  ΔFr / Δy

where n = F * Δy / A * ΔVn

We have to define the terms of the formula

Δy = 12.5 x 10⁻³

ΔVr = 2.5m /s

A = 60 x 60 cm² = 0.36m

F = 98.1 n

We have to put the values in the formula

98.1 n *  12.5 x 10⁻³ /  0.36m * 2.5m /s

n = 1.3625  ns / m²

The kinematic viscosity of the oil in strokes if the specific gravity of the oil is 0.95

y = n / e

n =  1.3625  

e = 0.95 x 10³

y = 1.3625  / 0.95 x 10³

= 1.434 x 10⁻³

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Answer:

Used to recall the direction a standard screw

Ragman assigns the mission "Database - Part 1" in Escape From Tarkov. You are asked to get the cargo manifests for Goshan, OLI, and IDEA from Interchange. Contents. Database for Escape from Tarkov: Part 1 Quest Information.

You can keep information on a certain topic in an organized way using a database. It's fantastic when you need to keep a computer system's worth of searchable data or information. Many systems in use today employ databases.

A database is used to store user information on nearly all popular websites with a large user base and almost every online store that sells things. Most databases comprise one or more tables, each of which may have a number of unique fields.

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The critical angle of attack remains constant under all circumstances, despite the fact that changes in weight or center of gravity would alter the airspeed at which the aircraft stalls.

Turbulence over the top wing surface substantially reduces lift at an angle of attack of about 18° to 20° (for most wings), making flight impossible and causing the wing to stall.

With greater angles of attack, the center of pressure advances. The low-pressure area starts to advance towards the wing's leading edge (generally activating the stall warning device mounted on the leading edge of the wing if your aircraft happens to have one).

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The correct answer is option (C): 18862.5 J/kg

To determine the specific work done on the system, we can use Bernoulli's equation for steady, incompressible flow along a streamline. Neglecting the gravitational potential energy variation, Bernoulli's equation can be expressed as:

p/ρ + V²/2 = constant

Using Bernoulli's equation between the inlet and outlet ports of the control volume, the equation becomes:

p1/ρ + C1²/2 = p2/ρ + C2²/2 + L

Given the values of inlet pressure (P), inlet velocity (C), outlet pressure (Pout), outlet velocity (Cout), and viscous dissipations (L), we can substitute them into the equation and solve for L:

1/1000 + 5²/2 = 200/1000 + (-10)²/2 + L/1000

Simplifying the equation gives L = 1000 J/kg.

The specific work done on the system is calculated using the formula:

W = (Pout - P)/ρ + (Cout² - C²)/2 - L/ρ

Substituting the given values, we get:

W = (200 - 1)/1000 + ((-10)² - 5²)/2 - 1000/1000

Simplifying further, we find W = -3.5625 kJ/kg. Multiplying this by 1000, we obtain the specific work done on the system as 18862.5 J/kg.

Therefore, the correct option is (C) 18862.5 J/kg.

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In this case, the timebase control is set to 2.5 ms per division, so the frequency of the sine wave is 40 Hz. (Opton A)

Frequency = (Number of divisions * Vertical sensitivity control) / (Timebase control)

Hence,

Frequency = (10 divisions * 0.5 volts per division) / (2.5 ms per division)

= 40 Hz

Time base control refers to the regulation and adjustment of the timing or speed of a process or system to ensure synchronization or accuracy based on a reference time or desired timeframe.

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Explanation:

A business man who really focus on victory achieved through a right procedure will focus on long-term planning.

Let us understand what a short-term and long-term planning is.

Short-term will plan only for two-three years. But a long-term plan will look for future five years income projection, plan of expansion, bigger goals, etc.

A business man is the person who take risks and achieve more. A victory can be achieved in many ways one is taking bigger risks, next is focusing on long-term plans

If this is wrong, give me the answer choices so I know what's right or wrong. I'll edit the question if given to me.

At 1.2mpa pressure and 50c

What is pressure?
By pressing a knife against some fruit, one can see a straightforward illustration of pressure. The surface won't be cut if you press the flat part of the knife against the fruit. The force is dispersed over a wide area (low pressure).

a)Mass flow rate of the refrigerant
Therefore h1= condenser inlet enthalpy =278.28KJ/Kg
saturation temperature at 1.2mpa is 46.29C
Therefore the temperature of the condenser

T2 = 46.29C - 5
T2 = 41.29C

Now,
d)power consumed by compressor W = 3.3KW
Q4 = QL + w = Q4
QL = mR(h1-h2)-W
= 0.0498 x (278.26 - 110.19)-3.3
=5.074KW
Hence refrigerator load is 5.74Kg

(COP)r = 238/53
(Cop) = 4.490

Therefore the above values are the (a) mass flow rate of the refrigerant, b) the refrigerant load, c) the cop, and d) the minimum power input to the compressor for the same refrigeration load.

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The question is incomplete. Here is the complete question.

Two kg of a two phase liquid vapor mixture of carbon dioxide (CO₂) exists at -40°C in a 0.05m³ tank. Determine the quality of the mixture, if the values of specific volume for saturated liquid and saturated vapor CO₂ at -40°C are \(v_{f}\) = 0.896 x 10⁻³m³/kg and \(v_{g}=\) 3.824 x 10⁻²m³/kg, respectively.

Answer: x = 1

Explanation: In a phase change of a pure substance, at determined pressure and temperature, the substance exists in two different phases: saturated liquid and saturated vapor.

Quality (x) is the ratio of saturated vapor in the mixture and can be written as:

\(x=\frac{m_{vapor}}{m_{liquid}+m_{vapor}}\)

It has value between 0 and 1: x = 0 for saturated liquid and x = 1 for saturated vapor.

When related with volumes, quality is rearranged as:

\(x=\frac{v-v_{f}}{v_{g}-v_{f}}\)

Solving for x:

\(x=\frac{0.05-0.896.10^{-3}}{3.824.10^{-2}-0.896.10^{-3}}\)

\(x=\frac{0.049104}{0.037344}\)

x = 1.3

Quality of mixture of carbon dioxide is x = 1, which means it's for saturated vapor.

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

Answer:

C

Explanation:

they are flat black

Answer:

Technician A

Explanation:

Technician A is correct. Technician B is wrong, as different drive vehicles (FWD, RWD, AWD, 4WD) each require a certain rotation for optimal tire wear.

Answer: metal inert gas (MIG) welding

Explanation:

Gas metal arc welding (GMAW), also called metal inert gas (MIG) welding, is an arc welding process in which the heat for melting the metal is generated by an electric arc between a consumable electrode and the metal (Fig.