Shorter lines are faster than longer lines is an example of an algorithm.

Answers

Answer 1

Answer:

Programmers count the number of lines of code in an algorithm.

Answer 2

Programmers count the number of lines of code in an algorithm. Thus, shorter lines are faster than longer lines is an example of an algorithm.

What do you mean by algorithm?

An algorithm is a finite sequence of exact instructions that is used in mathematics and computer science to solve a class of particular problems or carry out a computation.

For performing calculations and processing data, algorithms are employed as specifications. Conditionals can be used by more sophisticated algorithms to divert code execution along several paths and draw reliable inferences, ultimately leading to automation.

Alan Turing was the first to use terminology like "memory," "search," and "stimulus" to describe human traits as metaphorical descriptions of machines.

A heuristic, on the other hand, is a method for addressing problems that may not be fully articulated or may not provide accurate or ideal solutions, particularly in problem domains where there isn't a clearly defined proper or ideal conclusion.

Learn more about algorithm, here

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Related Questions

A cylindrical specimen of some metal alloy 10 mm in diameter and 150 mm long has a modulus of elasticity of 100 GPa. Does it seem reasonable to expect a tensile stress of 200 MPa to produce a reduction in specimen diameter of 0.08 mm

Answers

Answer:

N0

Explanation:

It does not seem reasonable to expect a tensile stress of 200 MPa to produce a reduction in specimen diameter of 0.08 mm

Given data :

Diameter ( d ) = 10 mm

length ( l ) = 150 mm

elasticity ( ∈ ) = 100 GPa

longitudinal strain ( б ) 200 MPa

Poisson ratio ( μ )  ( assumed ) =0.3

Assumption : deformation totally elastic

attached below is the detailed solution to why it is not reasonable .

The Sd value = 0.08 > the calculated Sd value ( 6*10^-3 ) hence it is not reasonable to expect a tensile stress of 200 MPa to produce a reduction in specimen

Match the example to the model type it represents.

1. The client complains about the way the keyboard feels
1.mock-up
2. The engineering team tests how the tire treads on a new SUV perform on 2.various road conditions
preproduction model
3. The engineering team performs tests on the efficiency of the manufacturing process used for a recumbent bicycle
3.presentation model

Answers

Answer:

represnt

Explanation:

Determine the maximum shear stress in the beam and the minimum yield strength that should be considered to obtain a minimum factor of safety of 2 based on the distortion-energy theory. The maximum shear stress in the beam is 1875 Numeric ResponseEdit Unavailable. 1875 correct.psi. The minimum yield strength based on the distortion-energy theory is

Answers

Answer: hello some parts of your question is missing attached below is the missing part

max shear stress = 1875 psi

minimum yield strength = 14911.76 psi

Explanation:

Since The maximum shear stress in the Beam is already given as 1875 psi , I will calculate for the minimum yield strength

Determine minimum yield strength

attached below is the detailed solution

minimum yield strength = 14911.76 psi

Write a python program to get the following output. 1-----99 2-----98 3-----97 . . . . . . 98-----2 99-----1

Answers

Answer:

i dont know th answer can u help ?

Explanation:

Which option identifies the section of a project charter represented in the following scenario?
Updated POS terminals will be available to the following five departments by July 31, 2015.
O project assumptions
O project deliverables
O project constraints
O project requirements

Answers

it’s b! i know this because i did the test!
Answer B , Just took the test and got 100%

An infinite cylindrical rod falls down in the middle of an infinite tube filled with fluidat a constant speed V (terminal velocity). The density of the rod and the fluid are different.Assume that the pressure field is hydrostatic.(a)[5pts] Solve for the velocity profileas a function of rin terms of V and the other variables.(b)[2pts] Calculate upward force per unit length of the rod from the fluid wall shear stress on the rod.(c)[2 pts] Calculate upward force per unit length of the rod from bouyancy.(d)[1pts] Calculate V.VR1

Answers

Answer:

the speed of your poop

Explanation:

A water jet pump involves a jet cross-sectional area of 0.01 m^2, and a jet velocity of 30 m/s. The jet is surrounded by entrained water. The total cross-sectional area associated with the jet and entrained streams is 0.075 m^2. These two fluid streams leave the pump thoroughly mixed with an average velocity of 6 m/s through a cross-sectional area of 0.075 m^2. Determine the pumping rate (i.e., the entrained fluid flowrate) involved in liters/s.

Answers

Answer:

the entrained fluid flowrate is 150 liters/s

Explanation:

Given the data in the question;

we determine the flow rate of water though the jet by using the following expression;

Q₂ = A₂ × V₂

where Q₂  is the flow rate of water though the jet, A₂ is the cross sectional area of the jet( 0.01 m² ) and V₂ is the jet velocity( 30 m/s )

so we substitute

Q₂ = 0.01 m² × 30 m/s

Q₂ = 0.3 m³/s

Next we determine the flow rate of water through the pump by using the following expression

Q₃ = A₃ × V₃

where Q₃  is the flow rate of water though the pump, A₃ is the cross sectional area of the pump( 0.075 m² ) and V₃ is the average velocity of mixing( 6 m/s )

so we substitute

Q₃ = 0.075 m² × 6 m/s

Q₃ = 0.45 m³/s

so to calculate the flow pumping rate of water into the water jet pump, we use the expression;

Q₁ + Q₂ = Q₃

we substitute

Q₁ + 0.3 m³/s = 0.45 m³/s

Q₁ = 0.45 m³/s - 0.3 m³/s

Q₁ = 0.15 m³/s

we know that 1 m³/s = 1000 Liter/second

so

Q₁ = 0.15 × 1000 Liter/seconds

Q₁ = 150 liters/s

Therefore, the entrained fluid flowrate is 150 liters/s

Determine the convection heat transfer coefficient, thermal resistance for convection, and the convection heat transfer rate that are associated with air at atmospheric pressure in cross flow over a cylinder of diameter D = 100 mm and length L = 2 m. The cylinder temperature is Ts = ° 70 C while the air velocity and temperature are V = 3 m/s and T[infinity] = 20°C, respectively. Plot the convection heat transfer coefficient and the heat transfer rate from the cylinder over the range 0.05 m ≤ D ≤ 0.5 m.

Answers

Answer:

attached below

Explanation:

Attached below is  a detailed solution to the question above

Step 1 : determine the Reynolds number using the characteristics of Air at 45°c

Step 2 : calculate the Nusselt's number

Step 3 : determine heat transfer coefficient

Step 4 : calculate heat transfer ratio and thermal resistance

Repeat steps 1 - 4 for each value of diameter from 0.05 to 0.5 m

attached below is a detailed solution

What's the ampacity of a No. 10 type TW copper Wire in a raceway containing six wires and located in an area where the ambient Temperature is 50 C

A. 15.1 A
B. 8.6 A
C. 13.9 A
D. 10.8 A

Answers

I would say B 8.6A!!!!

Find the minimum stages at total reflux for separating ethanol-acetone mixture to achieve 95 mol% purity of acetone in the distillate and 95 mol% purity of ethanol in the bottoms product. Use equilibrium data from problem 4D7. x y Equilibrium data for acetone (A) and ethanol at 1 atm (Perry et al., 1963, pp. 13-4) are
0.100.150.200.250.300.350.40 0.50 0.60 0.70 0.80 0.90 0.262 0.348 0.417 0.478 0.524 0.566 0.605 0.674 0.739 0.80 20.865 0.929

Answers

Answer:

minimum number of stages ≈ 9 ( from Image number 1 )

Explanation:

Attached below is a detailed solution to the question

Total reflux = R Tending to infinity

we can determine the operation equation by stripping and rectifying section in total reflux condition

minimum number of stages ≈ 9 ( from Image number 1 )

Choose two other elements from the periodic table that you predict should react to form something like table salt


Please pleassssss helppp

I give branlistttttt

Answers

sodium (Na) and chloride (Cl) make table salt if that’s what you’re asking

A utility generates electricity with a 36% efficient coal-fired power plant emitting the legal limit of 0.6 lb of SO2 per million Btus of heat into the plant. Suppose the utility encourages its customers to replace their 75-W incandescents with 18-W compact fluorescent lamps (CFLs) that produce the same amount of light. Over the 10,000-hr lifetime of a single CFL.

Required:
a. How many kilowatt-hours of electricity would be saved?
b. How many 2,000-lb tons of SO2 would not be emitted?
c. If the utility can sell its rights to emit SO2 at $800 per ton, how much money could the utility earn by selling the SO2 saved by a single CFL?

Answers

Answer:

a) 570 kWh of electricity will be saved

b) the amount of  SO₂ not be emitted or heat of electricity saved is 0.00162 ton/CLF

c) $1.296 can be earned by selling the SO₂ saved by a single CFL

Explanation:

Given the data in the question;

a) How many kilowatt-hours of electricity would be saved?

first, we determine the total power consumption by the incandescent lamp

[tex]P_{incandescent}[/tex] = 75 w × 10,000-hr = 750000 wh = 750 kWh

next, we also find  the total power consumption by the fluorescent lamp

[tex]P_{fluorescent}[/tex] = 18 × 10000 = 180000 = 180 kWh

So the value of power saved will be;

[tex]P_{saved}[/tex] = [tex]P_{incandescent}[/tex]  - [tex]P_{fluorescent}[/tex]

[tex]P_{saved}[/tex] = 750 - 180

[tex]P_{saved}[/tex]  = 570 kWh

Therefore, 570 kWh of electricity will be saved.

now lets find the heat of electricity saved in Bituminous

heat saved = energy saved per CLF / efficiency of plant

given that; the utility has 36% efficiency

we substitute

heat saved =  570 kWh/CLF / 36%

we know that; 1 kilowatt (kWh) = 3,412 btu per hour (btu/h)

so

heat saved =  570 kWh/CLF / 0.36 × (3412 Btu / kW-hr (

heat saved = 5.4 × 10⁶ Btu/CLF

i.e eat of electricity saved per CLF is 5.4 × 10⁶

b) How many 2,000-lb tons of SO₂ would not be emitted

2000 lb/tons = 5.4 × 10⁶ Btu/CLF

0.6 lb SO₂ / million Btu = x

so

x = [( 5.4 × 10⁶ Btu/CLF ) × ( 0.6 lb SO₂ /  million Btu )] / 2000 lb/tons

x = [( 5.4 × 10⁶ Btu/CLF ) × ( 0.6 lb SO₂ )] / [ ( 10⁶) × ( 2000 lb/ton) ]

x = 3.24 × 10⁶ / 2 × 10⁹

x = 0.00162 ton/CLF

Therefore, the amount of  SO₂ not be emitted or heat of electricity saved is 0.00162 ton/CLF

c)  If the utility can sell its rights to emit SO2 at $800 per ton, how much money could the utility earn by selling the SO2 saved by a single CFL?

Amount = ( SO₂ saved per CLF ) × ( rate per CFL )

we substitute

Amount = 0.00162 ton/CLF × $800

= $1.296

Therefore; $1.296 can be earned by selling the SO₂ saved by a single CFL.

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