Question 2: EOQ, varying t
(a) In class we showed that the average inventory level under the EOQ model was Q/2 when we look over a time period that is a multiple of T. What is this average inventory level over the period of time from 0 to t for general t? Provide an exact expression for this.
(b) Using your expression above, plot the average inventory (calculated exactly using your expression from part a) and the approximation Q/2 versus Q over the range of 1 to 30. Use t=100 and λ=2.
Note that lambda is a keyword in python and using it as a variable name will cause problems. Pick a different variable name, like demand_rate.
You should see that the approximation is quite accurate for large t, like 100, and is less accurate for small t.

Answer:

Step On: Your foot forces the clutch pedal down and then causes it to take up the slack. This, in turn, causes the clutch friction disk to slip, creating heat and ultimately wearing your clutch out.

Step Off: When the clutch pedal is released, the springs of the pressure plate push the slave cylinder's pushrod back, which forces the hydraulic fluid back into the master cylinder.

Answer: A. High

Explanation:

In order to have an obstructed view from your seat, you should always adjust your seat as high as possible, while staying comfortable.

Answer:

A

Explanation:

I put high and got it right.

There are several types of firewall screening technologies available, each with its own unique approach to protecting networks from malicious traffic.

Packet filtering: This is a basic type of firewall screening that filters incoming and outgoing network traffic based on specific criteria, such as IP addresses, ports, and protocols. Packet filtering is a simple but effective way to block known threats and unwanted traffic from entering or leaving the network.

Stateful inspection: This type of firewall screening technology examines the state of the connection to ensure that only legitimate traffic is allowed through. Stateful inspection keeps track of the state of each network connection and allows only packets that belong to established sessions to pass through.

Proxy servers: A proxy server acts as an intermediary between the network and the internet, intercepting all incoming and outgoing traffic and filtering it for malicious content. Proxy servers can also be used to enforce security policies and perform other network functions.

Deep packet inspection: This is an advanced type of firewall screening that analyzes the entire packet, including the data payload, to detect and block malicious content. Deep packet inspection can detect complex threats such as malware and viruses that may be hidden within seemingly harmless network traffic.

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Answer:

Massive or linteled

Explanation:

Pyramid is a type of massive or linteled structure.

These structures do no have not much internal spaces and they are huge edifices.

A pyramid is a solid body with outer triangular faces that converges on top. To construct a pyramid, large amounts of materials are usually involved. Pyramids were more prominent in times past before this present civilization.

In snooker, the maximum number of points that can be scored during a single frame is 147. This is achieved by potting all 15 red balls, each worth 1 point, and subsequently potting the colored balls in sequence: yellow (2 points), green (3 points), brown (4 points), blue (5 points), pink (6 points), and black (7 points). Each color is then returned to its original position and can be potted again for additional points.

The angle and wind speed mentioned in your question are not relevant to determining the maximum number of points in a snooker frame. The speed and direction of the balls are controlled by the player's shots, and the rug coating on the table doesn't affect the scoring.

So, regardless of the angle, wind speed, or table surface, the maximum number of points that can be scored in a frame of snooker is 147.

Answer:

Gasoline is injected directly into the cylinder.

Explanation:

In a direct injection system, the air and gasoline are not pre-mixed. Rather, air comes in using the intake manifold, while the gasoline is injected directly into the cylinder.

Two inmates at a Virginia jail used a metal bar and a piece of wire to dig their way out.

The pair had been held at the facility awaiting trial on charges including burglary, grand larceny, and other offenses.

They managed to create a hole in the wall of their cell and then burrowed their way out of the jail's perimeter fence. The escape was discovered during a routine security check.

Law enforcement officials launched a manhunt and eventually located the escapees, who were returned to custody.

The jail is now reviewing its security protocols and taking steps to prevent similar incidents in the future.

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A stall happens when the smooth airflow over the unmanned airplane`s wing is disrupted, and the lift degenerates rapidly. This is caused when the wing "exceeds its critical angle of attack."

The angle of attack refers to the angle at which the airplane's wing meets the air that is flowing over it. When an airplane actually is taking off, it is lifting the nose up into the air. And, if that nose continues to rise ultimately it reaches a point where the air is not able to smoothly flow over the wing, causing the airplane to drop. And, this is the point when the airplane exceeds its critical angle of attack.

Exceeding the critical angle of attack is known to be a stall. This has no concern with the engine stalling, it just concerns with the wings not producing enough lift to keep the airplane in the air.

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At time t=2 min, the concentration inside the tank is 0.042 kg/kg and the concentration outside the tank is 0.476 kg/kg.

To determine the concentration inside and outside the tank at 2 minutes, we need to use the mass balance equation:

mass in - mass out + mass generated = change in mass in the tank

At time t=0, the mass in the tank is 8 kg, and there is no mass generated. Thus, we have:

(2.2 + 0.2) kg/min - 2 kg/min + 0 kg/min = dM/dt

Simplifying, we get:

dM/dt = 0.4 kg/min

Integrating both sides, we get:

M(t) = M(0) + (dM/dt) * t

Plugging in the values, we get:

M(2) = 8 kg + (0.4 kg/min) * 2 min = 9.6 kg

Now, we can calculate the concentration inside and outside the tank:

Concentration inside the tank:

At time t=0, the concentration inside the tank is:

Cin(0) = 0 kg/kg (since there is no salt in the tank initially)

At time t=2 min, the amount of salt in the tank is:

Min(2) = 0.2 kg/min * 2 min = 0.4 kg

Thus, the concentration inside the tank is:

Cin(2) = Min(2) / M(2) = 0.4 kg / 9.6 kg = 0.042 kg/kg

Concentration outside the tank:

The rate of salt leaving the tank is 2 kg/min, and at time t=2 min, the total amount of salt leaving the tank is:

Mout(2) = 2 kg/min * 2 min = 4 kg

Thus, the amount of salt remaining in the tank is:

Mrem(2) = M(0) + Min(2) - Mout(2) = 8 kg + 0.4 kg - 4 kg = 4.4 kg

The total mass leaving the tank is 2 kg/min * 2 min = 4 kg. Thus, the concentration outside the tank is:

Cout(2) = Mout(2) / (M(0) + Min(2)) = 4 kg / 8.4 kg = 0.476 kg/kg

Therefore, at time t=2 min, the concentration inside the tank is 0.042 kg/kg and the concentration outside the tank is 0.476 kg/kg.

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Answer:

Precision

Explanation

You know the answer is precision because it uses the word range and plus because I just answered this question and got it correct.

.

Answer:

E = 132.69 sin(ωt -11.56)

i(t) = 6.64 sin (ωt +48.44) A

Explanation:

given data

e1 = 80 sin ωt volts                       80 < 0

e2 = 60 sin (ωt + π/2) volts          60 < 90

e3 = 100 sin (ωt – π/3) volts        100 < -60

solution

resultant will be  = e2 + e2 + e3

E =  80 < 0 + 60 < 90 + 100 < -60

\(\bar E\) = 80 + j60 + 50 - j50\(\sqrt{3}\)

\(\bar E\) = 130 + (-j26.60)

\(\bar E\) = 132.69  that is less than -11.56

so

E = 132.69 sin(ωt -11.56)

and

as we have given the impedance

z = (10-j17.3)Ω

z = 19.982 < -60

and

i(t) = \(\frac{132.69}{19.982}\) sin(ωt -11.56 + 60)  

i(t) = 6.64 sin (ωt +48.44) A

The Xtronic CVT (Continuously Variable Transmission) is a transmission system that is capable of continuously changing gear ratios when shifting. Unlike a standard automatic transmission that uses planetary gears to shift gears, the Xtronic CVT uses two adjustable pulleys with a belt running between them. These pulleys are capable of adjusting to any desired diameter, providing an infinite number of gear ratios.

As the driver of a vehicle with a standard automatic transmission shifts gears, the planetary gears mechanically connect and disconnect as they shift to the next gear. This results in a short pause that can be felt by the driver as the gears mesh together. In contrast, the Xtronic CVT avoids all of these issues since it doesn't have gears like a conventional automatic transmission.

Since there are no gears to engage or disengage, the Xtronic CVT is a lot more fluid and gentle than a typical automatic transmission. This makes for a smoother driving experience since the driver can't feel anything happening when the vehicle changes gears. Overall, the Xtronic CVT is a great option for those who want a more comfortable and smooth driving experience.

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Answer:

TRUE

Explanation: I think

protecting the ecosystem saves the species is true because a lot of species lives in the ecosystem and is there habitat / i think

Answer:

13.335 CM (1 ft, 1.335 cm)

I am 80% sure this is the answer, but i am not too keen on math so if i am wrong let me know and i will try my best to fix it!

I hope this helped! Have a good day :]

Answer:

Explanation:

The probability of n bits being in error is 10Cn × p^n, where ...

  10Cn = 10!/(n!(10-n!))

  10C1 = 10

  10C2 = 45

  10C3 = 120

This is the product of the probability that n bits can be in error and the number of ways that n bits can be chosen from the 10 in the word.

  n = 1: 10p

  n = 2: 45p²

  n = 3: 120p³

Answer:

a. The time required for the tank to empty halfway is presented as follows;

\(t_1 = \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right)\)

b. The time it takes for the tank to empty the remaining half is presented as follows;

\(t_2 = { \dfrac{ D_0^2 }{D} \cdot\sqrt{\dfrac{H}{g} }\)

The total time 't', is presented as follows;

\(t = \sqrt{2} \cdot \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} }\)

Explanation:

a. The diameter of the tank = D₀

The height of the tank = H

The diameter of the orifice at the bottom = D

The equation for the flow through an orifice is given as follows;

v = √(2·g·h)

Therefore, we have;

\(\dfrac{P_1}{\gamma} + z_1 + \dfrac{v_1}{2 \cdot g} = \dfrac{P_2}{\gamma} + z_2 + \dfrac{v_2}{2 \cdot g}\)

\(\left( \dfrac{P_1}{\gamma} -\dfrac{P_2}{\gamma} \right) + (z_1 - z_2) + \dfrac{v_1}{2 \cdot g} = \dfrac{v_2}{2 \cdot g}\)

Where;

P₁ = P₂ = The atmospheric pressure

z₁ - z₂ = dh (The height of eater in the tank)

A₁·v₁ = A₂·v₂

v₂ = (A₁/A₂)·v₁

A₁ = π·D₀²/4

A₂ = π·D²/4

A₁/A₂ = D₀²/(D²) = v₂/v₁

v₂ = (D₀²/(D²))·v₁ = √(2·g·h)

The time, 'dt', it takes for the water to drop by a level, dh, is given as follows;

dt = dh/v₁ = (v₂/v₁)/v₂·dh = (D₀²/(D²))/v₂·dh = (D₀²/(D²))/√(2·g·h)·dh

We have;

\(dt = \dfrac{D_0^2}{D} \cdot\dfrac{1}{\sqrt{2\cdot g \cdot h} } dh\)

The time for the tank to drop halfway is given as follows;

\(\int\limits^{t_1}_0 {} \, dt = \int\limits^h_{\frac{h}{2} } { \dfrac{D_0^2}{D} \cdot\dfrac{1}{\sqrt{2\cdot g \cdot h} } } \, dh\)

\(t_1 =\left[{ \dfrac{D_0^2}{D\cdot \sqrt{2\cdot g} } \cdot\dfrac{h^{-\frac{1}{2} +1}}{-\frac{1}{2} +1 } \right]_{\frac{H}{2} }^{H} =\left[ { \dfrac{D_0^2 \cdot 2\cdot \sqrt{h} }{D\cdot \sqrt{2\cdot g} } \right]_{\frac{H}{2} }^{H} = { \dfrac{2 \cdot D_0^2 }{D\cdot \sqrt{2\cdot g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right)\)

\(t_1 = { \dfrac{2 \cdot D_0^2 }{D^2\cdot \sqrt{2\cdot g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right) = { \dfrac{\sqrt{2} \cdot D_0^2 }{D^2\cdot \sqrt{ g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right)\)

\(t_1 = { \dfrac{\sqrt{2} \cdot D_0^2 }{D^2\cdot \sqrt{ g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right) = { \dfrac{D_0^2 }{D^2\cdot \sqrt{ g} } \cdot \left(\sqrt{2 \cdot H} - \sqrt{{H} } \right) =\dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right)\)The time required for the tank to empty halfway, t₁, is given as follows;

\(t_1 = \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right)\)

(b) The time it takes for the tank to empty completely, t₂, is given as follows;

\(\int\limits^{t_2}_0 {} \, dt = \int\limits^{\frac{h}{2} }_{0 } { \dfrac{D_0^2}{D} \cdot\dfrac{1}{\sqrt{2\cdot g \cdot h} } } \, dh\)

\(t_2 =\left[{ \dfrac{D_0^2}{D\cdot \sqrt{2\cdot g} } \cdot\dfrac{h^{-\frac{1}{2} +1}}{-\frac{1}{2} +1 } \right]_{0}^{\frac{H}{2} } =\left[ { \dfrac{D_0^2 \cdot 2\cdot \sqrt{h} }{D\cdot \sqrt{2\cdot g} } \right]_{0 }^{\frac{H}{2} } = { \dfrac{2 \cdot D_0^2 }{D\cdot \sqrt{2\cdot g} } \cdot \left( \sqrt{\dfrac{H}{2} } -0\right)\)

\(t_2 = { \dfrac{ D_0^2 }{D} \cdot\sqrt{\dfrac{H}{g} }\)

The time it takes for the tank to empty the remaining half, t₂, is presented as follows;

\(t_2 = { \dfrac{ D_0^2 }{D} \cdot\sqrt{\dfrac{H}{g} }\)

The total time, t, to empty the tank is given as follows;

\(t = t_1 + t_2 = \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right) + t_2 = { \dfrac{ D_0^2 }{D} \cdot\sqrt{\dfrac{H}{g} } = \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \sqrt{2}\)

\(t = \sqrt{2} \cdot \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} }\)

The next 100-hour inspection for the aircraft is due when the tachometer reaches 1349.6 hours.

The 100-hour inspection for an aircraft refers to a maintenance requirement that needs to be performed after every 100 hours of flight time. It is an important safety measure to ensure the aircraft's airworthiness and identify any potential issues or wear and tear that may have occurred during its operation.

In the given scenario, the next 100-hour inspection for the aircraft is scheduled to take place when the tachometer reading reaches 1349.6 hours. The tachometer is a device that measures the running time of the aircraft's engine. It provides an indication of the total number of hours the engine has been in operation.

When the tachometer reading reaches the specified threshold of 1349.6 hours, it serves as a trigger for the aircraft owner or maintenance personnel to schedule and carry out the 100-hour inspection. This inspection involves a comprehensive examination of various aircraft components, systems, and structures to ensure their proper functioning and adherence to safety standards.

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Answer:

yes

Explanation:

ooooooooooooooooooooooooooooooo

Answer:

Air-cooled direct expansion coils operating below the dewpoint may produce condensate or water along with cool air.

The rate of heat transfer by the air conditioner using constant specific heat of 1.004kj/kg.K is 15.06 kW.

Rate of heat transfer is the power rating of the machine.

Work done and changes in potential and kinetic energy are neglected since it is a steady state process.

The specific heat in terms of specific heat capacity and temperature change is given as:

\(q_{out} = Cp(Ti - Te)\)

\(q_{out} = 1.004(30 - 10) = 20.08 kJ/kg \\ \)

The rate of heat transfer, is then determined as follows:

Qout = 0.75 × 20.08 = 15.06 kW

Therefore, the rate of heat transfer by the air conditioner is 15.06 kW.

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The correct option that can replace the missing loop header so the method count Peaks works as intended is option c. p < data. length - 1.

As it contains the correct syntax for the loop header.

Given a problem in which we are to determine which of the loop header is missing from the loop statement so that the method `count Peaks` works as intended.

In order for the `count Peaks()` method to work as intended, the loop header should be as follows:

` for(int p = 1; p < data.length - 1; p++)`

We need to look for the loop header in the provided options that matches the above syntax. The correct option that has the syntax that matches the above loop header syntax is c. `p < data. length - 1`

Therefore, the option C can replace the missing loop header so the method count Peaks works as intended and we can write the loop header as `for(int p = 1; p < data. length - 1; p++)`.

Hence, option (C) is the correct answer.

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The potential solution for Windows hosts that have the Remote Registry Service disabled is to enable the Dissolvable Agent compliance scanning option. This option allows for scanning and evaluation of the host's compliance without relying on the Remote Registry Service.

Enabling the Dissolvable Agent provides an alternative method for conducting compliance scans on Windows hosts that do not have the Remote Registry Service enabled. The Dissolvable Agent acts as a temporary scanning tool that can be deployed on the host, allowing for the collection of necessary data and evaluation of compliance. By utilizing this option, organizations can ensure compliance scanning even on hosts where the Remote Registry Service is disabled, providing a comprehensive solution for maintaining compliance across their Windows infrastructure.

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Answer:

Input power = 465.63 W

current = 1.94 A

Explanation:

we have the following data to answer this question

V = 9130

i = 0.051

the input power = VI

I = 51.0 mA = 0.051

= 9130 * 0.051

= 465.63 watts

the current = 465.63/240

= 1.94A

therefore the input power is 465.63 wwatts

while the current is 1.94A

the input power is the same thing as the output power.

The maximum allowable power for the silicon chip when it is flush mounted in a substrate with an unheated starting length of 20 mm is 0.136 W.

To find the maximum allowable power for the silicon chip, we need to use the equation for heat transfer from a flat plate in parallel flow:

q'' = h(Ts - T∞)

Where q'' is the heat flux, h is the heat transfer coefficient, Ts is the surface temperature, and T∞ is the free stream temperature.

We can rearrange this equation to find the maximum allowable power:

q'' = h(Ts - T∞)Pmax = q''A = hA(Ts - T∞)

Where Pmax is the maximum allowable power and A is the area of the chip. We are given the values for u∞, T∞, Ts, and A, so we can plug these into the equation to find Pmax:

u∞ = 20 m/s

T∞ = 24°C

Ts = 80°CA

= (10 mm × 10 mm)

= 0.0001 m²

We also need to find the value for h, which we can do using the equation for the Nusselt number for a flat plate in parallel flow:

\(NuL = 0.664Re^{0.5}Pr^0.33\)

Where NuL is the Nusselt number, Re is the Reynolds number, and Pr is the Prandtl number. We can rearrange this equation to find h:

h = (NuLk)/where k is the thermal conductivity and L is the length of the chip. We can find the values for Re and Pr using the equations:

Re = (u∞L)/νPr = (Cpμ)/k

Where ν is the kinematic viscosity, Cp is the specific heat capacity, and μ is the dynamic viscosity. We can find the values for these properties using the given values for u∞ and T∞ and looking up the properties of air at these conditions:

ν = 15.68 × 10^-6 m²/s

Cp = 1007 J/kg·Kμ

= 18.46 × 10^-6 kg/m·sk

= 0.02624 W/m·K

We can now plug these values into the equations to find Re, Pr, NuL, and h:

Re = (20 m/s × 0.01 m)/(15.68 × 10^-6 m²/s)

= 12755.1Pr =

(1007 J/kg·K × 18.46 × 10^-6 kg/m·s)/(0.02624 W/m·K)

= 0.708NuL

= 0.664(12755.1^0.5)(0.708^0.33)

= 59.58h

= (59.58 × 0.02624 W/m·K)/0.01 m

= 155.6 W/m²

We can now plug these values into the equation for Pmax to find the maximum allowable power:

Pmax = (155.6 W/m²·K)(0.0001 m²)(80°C - 24°C) = 0.874 W

We can use the equation for the Nusselt number for a flat plate with an unheated starting length:

NuL = 0.3387(ReLPr)^(1/3)

Where L is the length of the heated portion of the plate.

We can plug in the values for Re, Pr, and L to find NuL:

NuL = 0.3387(12755.1 × 0.01 m × 0.708)^(1/3)

= 9.23

We can now use this value for NuL to find a new value for h:

h = (9.23 × 0.02624 W/m·K)/0.01 m = 24.25 W/m²·

We can now plug this value for h into the equation for Pmax to find the maximum allowable power:

Pmax = (24.25 W/m²·K)(0.0001 m²)(80°C - 24°C)

= 0.136 W

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It is to be noted that all UHRS platforms are optimized for the popular kinds of internet browser applications.

The Universal Human Relevance System (UHRS) is a crowdsourcing platform that allows for data labeling for a variety of AI application situations.

Vendor partners link people referred to as "judges" to offer data labeling at scale for us. All UHRS judges are bound by an NDA, ensuring that data is kept protected.

A browser is a software tool that allows you to see and interact with all of the knowledgeon the World Wide Web. Web sites, movies, and photos are all examples of this.

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When as-built drawings are received after a project is completed, they should contain the Field notes by the contractor during the construction project before they are accepted.

As-built drawings are a blueprint or drawing that indicates how a structure has been constructed and incorporates the modifications made during the building process. It is often used to show how an engineering process has been completed to support future maintenance or modification work. Before accepting the as-built drawings, they must be reviewed to ensure that they are detailed enough and accurately represent the finished product. An as-built drawing is used to verify that a structure has been completed as per the approved plans and drawings. It's crucial to have them on hand for future renovations, repairs, or to show compliance with the building codes.A proper set of as-built drawings should include the following:Drawings for each floor of the building that show the layout of rooms, staircases, doors, and windows.Exterior building drawings showing the layout of the building on the lot and any landscape or hardscape features.The plumbing, electrical, and HVAC systems are illustrated in separate drawings.The construction's structural drawing.In conclusion, as-built drawings should contain field notes by the contractor during the construction project before they are accepted.

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