N elements are inserted from a min-heap with N elements. The total running time is:
a) O(N2) worst case
b) O(logN) worst case
c) O(N) worst case
d) None of these

Answer:

- the process should operate each day for 7.9286 hours

- the amount of brine that must be fed to the evaporator is 94.594 kg/hr

Explanation:

Given that;

concentration of brine = 26%

so water concentration will be 100% - 26% = 74%

for evaporation of 70kg water per hour, residual is pure salt( 0% moisture)

so mass flow rate of brine = 70/(74%) = 70/0.74 = 94.5945 kg/hr

Amount of pure dry salt produced(0%) = 94.5945 - 70 = 24.5945 kg/hr

Now for production of 195kg of pure dry salt, number of hours required will be

T = 195 / 24.5945 = 7.9286 hrs

Therefore the process should operate each day for 7.9286 hours.

Total brine solution required ( 26% salt conc.) = 195/0.26 = 750 kg

Feed rate of brine solution ( 26% salt conc.) = 750 / 7.9286 = 94.594 kg/hr

Therefore the amount of brine that must be fed to the evaporator is 94.594 kg/hr

Answer:

C) a winding connected in series with the armature.

Explanation:

In a series motor, an electromagnet is used as a stator to generate its magnetic field. The field coil of this stator are connected through a commutator in series with the rotor windings. This stator which is the armature windings will conduct AC even on a DC machine, due to the  periodically reverses current direction (commutation) or due to electronic commutation (as in brushless DC motors).

Aprendizajes del buen vivir ya que al compartir con la comunidad y te desarrollas, tu desarrollas otra manera de pensar y adquieres el aprendizaje del buen vivir con todos sin excepción alguna

Además el aprendizaje de experimentar compartir con otras personas que no son tu familia, eso hará que desarrolles el aprendizaje de tus cualidades y en este caso sería la cualidad de la imparcialidad

It is to be noted that "All of the above" (Option G) Belt Levels must be completed before a Six Sigma Certification is awarded.

Six Sigma is a certification program that teaches employees how to analyze processes and results in the objective of eliminating waste and faults. There are numerous levels of certification, ranging from main end user to master Six Sigma user.

Six Sigma is a quality control certification that teaches employees how to analyze processes and outcomes in order to eliminate waste and faults. Certification levels range from primary end user to master Six Sigma user who serves as a senior quality control member.

It is to be noted that the ASQ Six Sigma Black Belt exam is a demanding and tough exam. The disciplined study, experience, and a solid preparation and test-taking plan are required. Many students study for months and then fail.

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The following should be done by the team member:

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\(\boxed{ℒ_t[f(t)] =\frac{30}{{s}^{4} } - \frac{9}{s} + \frac{6}{s + 5} + \frac{1}{s - 3}} \)

This holds true because:

\(\lim_{s\to\infty} [\frac{30}{{s}^{4} } - \frac{9}{s} + \frac{6}{s + 5} + \frac{1}{s - 3}] =0 \)

[The answer is throughly verified, hence you can trust this :)]

A) Sewage Factor: It represents the proportion of water inflow that is expected to be discharged as sewage. It is used in the design of sewer systems to estimate the quantity of sewage flow that will be generated.

B) Connection Factor: It represents the percentage of buildings or properties that are expected to connect to a sewer system. It is used in the design of sewer systems to estimate the total number of connections that will be made to the system.

C) Infiltration Coefficient: It represents the rate at which water enters a sewer system through cracks, joints, and other defects in pipes. It is used in the design of sewer systems to estimate the volume of infiltration that will occur during wet weather conditions.

D) Design Period: It is the length of time for which a particular engineering project is designed to function effectively. For example, in the case of water supply systems, the design period may be 20-30 years, during which the system is expected to meet the water demand requirements of the users.

E) Drop Manhole: It is a type of manhole that is constructed at a location where the sewer pipe changes direction from a horizontal to a vertical alignment. The purpose of a drop manhole is to reduce the velocity of the sewage flow and prevent damage to the downstream sewer structures.

F) Self Cleansing Velocity: It is the minimum velocity required in a sewer pipe to prevent the deposition of solids and ensure the self-cleansing of the pipe. A value of twice the average velocity is commonly used as the self-cleansing velocity.

The factors that affect water demand include population size, economic activity, climate, lifestyle, and water pricing policies. Changes in any of these factors can influence the level of water demand in a given area. For example, an increase in population size or economic activity can lead to a higher demand for water, while the implementation of water conservation measures can reduce water demand.

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Answer:

1279

Explanation:

We have the mean u = 1497

Standard deviation sd = 322

We find the x distribution using 25%

P(Z<z) = 0.25

Z = -0.675

From here we use the formula for z score

X = z(sd) + u

X = -0.675*322 + 1497

X = -217.35 + 1497

X = 1279.6

Which is approximately 1279

So we conclude that the maximum sat scores in year 2014that meets with the requirements of this course is 1279

Answer:

1831

Explanation:

When pulling a trailer down a long steep hill, it is important to take certain precautions to ensure your safety and the safety of other drivers on the road. One of the most important things to remember is to never increase your speed.

Instead, it is recommended that you drive in a lower gear, which will help to control your speed and reduce the risk of losing control of your trailer. By driving in a lower gear, you can also use your vehicle's engine to slow down, rather than relying solely on your brakes. This is important because overheating your brakes can cause them to fail, which can be dangerous when you are towing a heavy load.

In addition to driving in a lower gear and avoiding speeding, it is also important to maintain a safe following distance between your vehicle and other vehicles on the road. This will help to ensure that you have enough time to react if the vehicle in front of you suddenly stops or slows down. It is also a good idea to keep to the right lane when going down a steep hill, as this can help to reduce the risk of collisions with other vehicles.
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Answer:

4. Competent person

Explanation:

A competent person should inspect the crane

OSHA regulations only require that such equipment be inspected during initial use and annually thereafter by a "competent person"

The company should implement SSL/TLS to prevent this type of attack from occurring in the future.

SSL/TLS (Secure Sockets Layer/Transport Layer Security) is a protocol used to provide secure communication over the internet. It ensures the confidentiality and integrity of the data being transmitted between two systems, such as a web server and a client's web browser.

In this case, SSL/TLS can prevent the attack by ensuring that the communication between the company's web server and a client's web browser is encrypted and authenticated. With SSL/TLS, the attacker cannot redirect the client's web browser to their own server without being detected, as the encryption ensures that the client's browser can only communicate with the legitimate web server.

IPSec (Internet Protocol Security) is another protocol that provides secure communication over the internet. However, it is more commonly used for site-to-site VPN (Virtual Private Network) connections rather than web traffic.

DNSSEC (Domain Name System Security Extensions) is a protocol that adds security to the domain name system (DNS) by digitally signing DNS records. It does not prevent attacks on the company's website.

S/MIME (Secure/Multipurpose Internet Mail Extensions) is a protocol used for securing email communication. It is not relevant to the company's website and does not prevent attacks on it. Thus, SSL/TLS is the right answer.

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Answer:

When a force is exerted on an object it can change the object's speed, direction of movement or shape. Pressure is a measure of how much force is acting upon an area. Pressure can be found using the equation pressure = force / area. Therefore, a force acting over a smaller area will create more pressure

Explanation:

hope it will become helpful to you ☺️☺️

The number of gold atoms per cubic centimeter in the silver-gold alloy is 1.44 x 10^22 atoms/cm3.

Determine the alloy's density. Determine the molar composition of the alloy . Determine the number of atoms of gold per cubic centimetre. Find out the alloy's molar composition. We must convert the mass fractions of gold and silver to molar fractions in order to estimate the alloy's molar composition. Silver's molar fraction is equal to 0.930 by multiplying (0.810 x (196.97 g/mol + 107.87 g/mol)) by (107.87 g/mol x 12.05 g/cm3). Determine the amount of atoms of gold per cubic centimetre . Ultimately, we can determine how many gold atoms there are in every cubic centimetre. As a result, the silver-gold alloy has 1.44 x 1022 gold atoms per cubic centimetre.

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Answer:

Explanation:

Considering the flow of mercury in a tube:

When it comes to laminar flow of mercury, the thermal entry length is quite smaller than the hydrodynamic entry length.

Also, the hydrodynamic and thermal entry lengths which is given as DLhRe05.0= for the case of laminar flow. It should be noted however, that Pr << 1 for liquid metals, and thus making the thermal entry length is smaller than the hydrodynamic entry length in laminar flow, like I'd stated in the previous paragraph

This statement is generally true.Cache memory is a type of high-speed memory that is used to improve the performance of a computer by reducing the time it takes to access data from main memory.

Cache memory is typically available in two or more levels, with each level providing a progressively larger and slower cache.Level 1 (L1) cache is the smallest and fastest cache, and it is typically located directly on the CPU chip. This allows for extremely fast access to frequently used data and instructions, which can significantly improve performance.Level 2 (L2) cache is typically larger than L1 cache and is located on a separate chip that is connected to the CPU through a high-speed dedicated interface, such as a backside bus. This allows for faster access to frequently used data than main memory, but it is slower than L1 cache.

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In an orthogonal cutting, a cylinder is turned to reduce the diameter with the following processing conditions Initial The spindle horse power is 11.78 kW.

Shear and normal stresses on the chip-tool interface To determine the shear stress (τ) and normal stress (σ) on the chip-tool interface, the following formula will be used:τ = the thrust force, t is the chip-tool contact length, is the width of the chip.t = 0.5 mm w = 0.1 mm/rev * 2 mm = 0.2 mmτ = 450 N / (0.5 mm * 0.2 mm) = 45000 N/m²σ = 150 N / (0.5 mm * 0.2 mm) = 15000 N/m²Therefore, the shear stress on the chip-tool interface is 45000 N/m², and the normal stress is 15000 N/m².

Shear angle using the Lee and Shaffer's model Lee and Shaffer's model can be used to calculate the shear angle (ϕ) using the formula:ϕ = (1 / tan α_r) * [(1 + sin ψ) / (cos ψ)]whereα_r is the rake angle andψ is the clearance angle.α_r = 10°ψ = 90° - 10° = 80°ϕ = (1 / tan 10°) * [(1 + sin 80°) / cos 80°] = 18.19°Therefore, the shear angle is 18.19°.

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Shear stress on chip-tool interface = 300 MPa ;Normal stress on chip-tool interface = 52.17 MPa ; Shear angle = 5.74°Chip thickness = 0.57 mm ; Shear stress on shear plane = 263.16 MPa ; Normal stress on shear plane = 45.79 MPa ; Specific cutting energy = 113398.2 N/m ; Spindle horse power = 8.44 hp.

Given data:

Initial diameter = 100 mm

Depth of cut = 2 mm

Feed = 0.1 mm/rev

Rake angle = 10°

Chip-tool contact length = 0.5 mm

Cutting force = 450 N

Thrust force = 150 N

Spindle RPM = 60

Formula used:

Shear force = Cutting force - Thrust force

Chisel angle = Tan-1(1/ Tan Φ - Tan Φ / Tan λ)

Shear angle = Tan-1(Tan Φ / (Cos λ - Sin Φ Sin λ))

Chip thickness = Feed / Sin λa)

Shear and normal stresses on chip-tool interface

Chip-tool contact length, l = 0.5 mm

Shear force, Fs = Cutting force - Thrust force= 450 - 150= 300 N

Area of contact, Ac = t × l= 2 × 0.5= 1 mm2

Shear stress, τ = Fs / Ac= 300 / 1= 300 MPa

Normal force, Fn = Fs Tan λ= 300 × Tan 10°= 52.17 N

Normal stress, σ = Fn / Ac= 52.17 / 1= 52.17 MPab)

Shear angle using the Lee and Shaffer's model

Here, λ = 10°

Chisel angle, Φ = Tan-1(1 / Tan λ)= Tan-1(1 / Tan 10°)= 5.71°

Shear angle, α = Tan-1(Tan Φ / (Cos λ - Sin Φ Sin λ))= Tan-1(Tan 5.71° / (Cos 10° - Sin 5.71° Sin 10°))= Tan-1(0.1)= 5.74°c) Chip thicknessHere, λ = 10°Feed, t = 0.1 mm

Chip thickness, h = t / Sin λ= 0.1 / Sin 10°= 0.57 mmd)

Shear and normal stresses on shear plane

Shear force, Fs = 300 N

Shear plane area, As = t × d= 2 × 0.57= 1.14 mm2

Shear stress, τ = Fs / As= 300 / 1.14= 263.16 MPa

Normal stress, σ = Fn / As= 52.17 / 1.14= 45.79 MPae)

Specific cutting energy

Cutting power, Pc = Fs × vc= Fs × πdn/1000= 300 × π × 100 × 60/1000= 5669.91 W

Specific cutting energy, E = Pc / Vt= Pc / (f × Vf)= 5669.91 / (0.1 × 0.5)= 113398.2 N/mmf)

Spindle horse power

Spindle power, Ps = Pc / η= Pc / 0.9= 5669.91 / 0.9= 6299.90 W= 6.2999 kW= 8.44 hp (1 hp = 0.7457 kW)

Therefore,

Shear stress on chip-tool interface = 300 MPa

Normal stress on chip-tool interface = 52.17 MPa

Shear angle = 5.74°Chip thickness = 0.57 mm

Shear stress on shear plane = 263.16 MPa

Normal stress on shear plane = 45.79 MPa

Specific cutting energy = 113398.2 N/m

Spindle horse power = 8.44 hp

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This question requires a thermal analysis of the heat exchanger system, which would involve the calculation of heat transfer rate and the resulting temperature change in both the light oil and the hot oil from the still bottom. This would require an understanding of heat transfer principles and the application of energy balance equations.

To determine the maximum temperature attained by the light oil in the heat exchanger, you would need to consider the flow arrangement of the two fluids, whether they are flowing in counter-current or co-current direction. This would affect the heat transfer rate and the resulting temperature change in each fluid.

To calculate the temperatures, you would need to consider the heat transfer rate between the two fluids, the inlet and outlet temperatures of each fluid, and their specific heat capacities. The equations would need to take into account the heat transfer rate, mass flow rate, and temperature change for each fluid.

Once the temperatures have been calculated, you can then determine the maximum temperature attained by the light oil and the hot oil, which would depend on the flow arrangement and the heat transfer rate between the two fluids.

It is recommended that you consult with a specialist in the field of thermal analysis or heat transfer to accurately and properly determine the maximum temperature attained by the light oil in the heat exchanger.

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Answer:

Option B, qualification-based selection

Explanation:

Whenever bidding is done for seeking professional services, selection of contractor should be done using qualifications-based selection.

While seeking professional service, focus is on selecting the best quality and value matter, hence the contracting agency with best experience and skill for the job is selected.

Answer:

C. 14.55

Explanation:

12 x 10 = 120

120 divded by 10 is 12

so now we do the left side

7 x 3 = 21 divded by 10 is 2

so now we have 14

and the remaning area is 0.55

so 14.55

Answer:

  at frequencies below resonance

Explanation:

Below resonance, circuit impedance decreases as XC decreases with increasing frequency. Above resonance, circuit impedance increases as XL increases with frequency. Hence increasing circuit impedance will only accompany increasing XC below resonance (as frequency decreases).

____

Below resonance, |XL| < |XC|, so XC dominates series impedance. Above resonance, the reverse is true.

A toddler who can recognize herself in a mirror is more likely to be confident in her daycare class. Explanation: A toddler who can recognize herself in a mirror is more likely to be confident in her daycare class.

It shows that she is developing a sense of self-awareness and is more aware of her place in the world. This can lead to increased confidence and self-esteem, which can translate to better social skills and interactions with others in her daycare class.

In addition, the ability to recognize oneself in a mirror is an important developmental milestone that typically occurs around 18-24 months of age. This is an important sign of healthy cognitive and social-emotional development, and can have a positive impact on the child's overall well-being.

Overall, the ability to recognize oneself in a mirror is an important aspect of a toddler's development, and can have positive implications for their social, emotional, and cognitive growth.

It is a sign of increasing self-awareness and confidence, and can help children develop healthy relationships with others in their daycare class.

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Answer:

A power supply unit (PSU) converts mains AC to low-voltage regulated DC power for the internal components of a computer. Modern personal computers universally use switched-mode power supplies

Answer:

Volt is the SI (Standard International) unit of electrical potential of the..

Explanation:

Answer:

the third anwser is right on edgeinuity

Explanation:

Answer:

Technology can pull students’ focus away from their tasks.

Explanation:

Yes, the receiver will discard the frame. Let's perform the CRC calculation for the bit sequence 111001000 with the generator 1101. The first bit in the sequence is a 1, so we take the first 4 bits (1110) and XOR them with the generator (1101) to get the remainder 001.

We then drop the first bit (1) and append the remainder (001) to get the new bit sequence 110010001. We now perform the same process with this new sequence and the same generator. Taking the first 4 bits (1100) and XORing with the generator (1101) gives us the remainder 001. Dropping the first bit (1) and appending the remainder (001) gives us the new bit sequence 10010001.
b) The number of collisions after which, in CSMA/CD protocol, the NIC can make a binary exponential delay of at most 1536 bit times is 10. The binary exponential backoff algorithm doubles the delay time after each collision until a maximum delay time of 1023 slots is reached. After the 10th collision, the delay time would be 1024 slots, which is greater than 1536 bit times. Therefore, the NIC would make a delay of at most 1536 bit times.

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Answer:

I don't know the answer but you can download Ga u t h math to answer that

Explanation:

no spacing it is just can't lay aside

The normal stress on the inclined plane ab is σ = 2.0625 ksi and the shear stress on the inclined plane ab is τ = 1.375 ksi.

Determining the normal stress and shear stress acting on the inclined plane ab, use the stress transformation equations.

The stress transformation equations relate the normal stress, σ, and shear stress, τ, on an inclined plane with the corresponding normal and shear stresses acting on a plane perpendicular to the inclined plane. The equations are:

σ = σn cos²θ + σs sin²θ ± 2τnsinθcosθ

τ = (σn - σs) sinθcosθ ± τn cos²θ ± τs sin²θ

where θ is the angle between the inclined plane and the plane of reference, σn and σs are the normal stresses acting on the plane of reference, τn and τs are the shear stresses acting on the plane of reference, and τns is the shear stress acting on the inclined plane.

Given that τ = 5.5 ksi and σ and τ are to be determined on the inclined plane ab. Assume a plane of reference perpendicular to the inclined plane ab with angle θ in between.

The angle θ between the inclined plane and the plane of reference is the complement of the angle of inclination, so θ = 90° - 30° = 60°.

Assuming that there are no normal stresses acting on the plane of reference, so σn = 0. Therefore, the stress transformation equations simplify to:

σ = σs sin²θ ± 2τnsinθcosθ

τ = (σs/2) ± τn cos²θ ± (τ/2)sin²θ

Solve for σs and τn by using the given value of τ and the fact that there are no normal stresses acting on the plane of reference:

τ = τns = 5.5 ksi

σs = τ/2 = 2.75 ksi

τn = 0

Substitute these values into the stress transformation equations to obtain the normal stress and shear stress on the inclined plane ab:

σ = σs sin²θ = 2.75 ksi sin²60° = 2.75 ksi (0.75) = 2.0625 ksi

τ = (σs/2) ± τn cos²θ ± (τ/2)sin²θ = 2.75 ksi/2 = 1.375 ksi

Therefore, the normal stress on the inclined plane ab is σ = 2.0625 ksi and the shear stress on the inclined plane ab is τ = 1.375 ksi.

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