What is code in Arduino to turn led on and off

Answer:

In an electric motor, the stator provides a magnetic field that drives the rotating armature; in a generator, the stator converts the rotating magnetic field to electric current. In fluid powered devices, the stator guides the flow of fluid to or from the rotating part of the system.

Answer:

[tex]\mu_{max}=200Mpa[/tex]

Explanation:

From the question we are told that:

Internally pressurized [tex]P_i=100MPa[/tex]

Tangential Stress [tex]P_t=400mpa[/tex]

Axial stress [tex]P_a=200mpa[/tex]

Generally the equation for maximum normal and shear stresses are experienced by the outer surface is mathematically given by

 [tex]\mu_{max}=|\frac{P_t-P_a}{2}|,|\frac{P_t}{2}|,|\frac{P_t}{2}|[/tex]

Therefore

 [tex]\mu_{max}=|\frac{400-200}{2}|,|\frac{400}{2}|,|\frac{200}{2}|[/tex]

 [tex]\mu_{max}=200Mpa[/tex]

Answer:

a)  [tex]Tb=1845.05K[/tex]

b)  [tex]Q=1000.25KJ[/tex]

c)  [tex]\mu=0.59[/tex]

Explanation:

From the question we are told that:

Temperature x [tex]Tx=450c=>723K[/tex]

Pressure x [tex]Px=2.5MPa[/tex]

Temperature y [tex]Ty=600c=>873K[/tex]

Pressure y [tex]Py=0.45MPa[/tex]

Let

Air atmospheric temperature be [tex]25c[/tex]

Therefore

Temperature [tex]Ta=25+273=298k[/tex]

Generally the equation for Otto cycle is mathematically given by

 [tex]\frac{Tb}{Tx}=\frac{Ty}{Ta}[/tex]

 [tex]Tb=\frac{873*723}{298}[/tex]

 [tex]Tb=2118.05[/tex]

Therefore the peak cycle temperature (°C)

 [tex]Tb=2118.05k[/tex]

 [tex]Tb=2118.05-273[/tex]

 [tex]Tb=1845.05K[/tex]

Generally the equation for Heat addition is mathematically given by

 [tex]Q=Cv(Tb-Tx)[/tex]

 [tex]Q=Cv(2118.05-723)[/tex]

 [tex]Q=1000.25KJ[/tex]

Generally the equation for Thermal  efficiency is mathematically given by

 [tex]\mu=1-\frac{Ta}{Tx}[/tex]

 [tex]\mu=1-\frac{298}{723}[/tex]

 [tex]\mu=0.59[/tex]

Answer:

the factor of safety was used in the design of the cable is 2.6146

Explanation:

Given the data in the question;

Load on the main capable [tex]P_{initial[/tex] = 2600000 lb

number of parallel wires n = 1470

Diameter d = 0.16 in

average ultimate strength [tex]S_{ultimate[/tex] = 230000 psi

First we calculate the Load acting on each cable;

[tex]P_{initial[/tex] = P × n

P = [tex]P_{initial[/tex] / n

we substitute

P = 2600000 lb / 1470

P = 1768.70748 lb

Next we determine the working stress acting in a member;

[tex]S_{working[/tex] = P/A

{ Area A = [tex]\frac{\pi }{4}[/tex]d² }

[tex]S_{working[/tex] = P / [tex]\frac{\pi }{4}[/tex]d²

we substitute

[tex]S_{working[/tex] = 1768.70748 / [tex]\frac{\pi }{4}[/tex](0.16)²

[tex]S_{working[/tex] = 1768.70748 / 0.02010619298

[tex]S_{working[/tex] = 87968.29 psi

Now we calculate the factor of safety F.S

F.S = [tex]S_{ultimate[/tex] / [tex]S_{working[/tex]

we substitute

F.S = 230000 psi / 87968.29 psi

F.S = 2.6145785 ≈ 2.6146

Therefore, the factor of safety was used in the design of the cable is 2.6146

Complete Question

Analyze the boundary work done during the process having a rigid tank contains air at 500 kPa and 150°C. As a result of heat transfer to the surroundings, the temperature and pressure inside the tank drop to 65°C and 400 kPa, respectively.

Determine the boundary work done during this process and heat Lose

Answer:

a)  [tex]W=0[/tex]

b)  [tex]dQ=-61.03KJ/kg[/tex]

Explanation:

From the question we are told that:

Pressure of air [tex]P_1=500kpa[/tex]

Temperature of Air [tex]T_2=150°C[/tex]

Pressure drop [tex]P_2=400kpa[/tex]

Temperature of drop [tex]T_2=65 \textdegree C[/tex]

Generally the Constant Volume Process  is mathematically given by

 [tex]V_1=V_2=V[/tex]

Therefore

a)

Generally the equation for  boundary work w is mathematically given by

 [tex]W=pdv[/tex]

 [tex]W=P(V_2-V_1)[/tex]

 [tex]W=P(V_V)[/tex]

 [tex]W=0KJ[/tex]

b)

Generally the equation for Heat Change is mathematically given by

 [tex]dQ=dU+dW[/tex]

 [tex]dQ=dU[/tex]

 [tex]dQ=C_v(T_2-T_1)[/tex]

Where

   C_v=Specific Heat capacity of Air

  [tex]C_v=0.718 kJ/kg K[/tex]

 [tex]dQ=0.718(338-423)[/tex]

 [tex]dQ=-61.03KJ/kg[/tex]

Answer:

Resistance and voltage drop will still continue to rise, although at a slower pace than on the desired axis.

Explanation:

Pulling the strain gauge in the long axis causes the wires to elongate/thin, the effect of this is that it will lead to an increase in resistance and voltage drop (V = I*R).

As a result of the resultant effect, resistance and voltage drop will still continue to rise, although at a slower pace than on the desired axis, such as the long axis.

This question is incomplete, the missing diagram is uploaded along this answer below;

Answer:

the speed Vc of the current and its direction measured clockwise from the north is 0.71 m/s and 231.02° respectively

Explanation:

Given the data in the question and as illustrated in the diagram below;

The absolute velocity of the ship Vs is 6 Knots due east

so we convert to meter per seconds

Vs = 6 Knots × [tex]\frac{0.51444 m/s}{1 Knots}[/tex] = 3.0866 m/s

Next we determine the relative velocity of the ship Vs/c

Vs/c = AB / t

given that distance between A to B = 10 nautical miles which requires 2 hours

so we substitute

Vs/c = 10 nautical miles / 2 hrs

Vs/c = [10 nautical miles × [tex]\frac{1852 m}{1 nautical-miles}[/tex] ] / [ 2 hrs × [tex]\frac{3600s}{1hr}[/tex] ]

Vs/c = 18520 / 7200

Vs/c = 2.572 m/s

Now, from the second diagram below, { showing the relative velocity polygon }

Now, using COSINE RULE, we calculate the velocity current.

Vc = √( V²s + V²s/c - 2VsSs/ccos10 )

we substitute

Vc = √( (3.0866)² + (2.572)² - (2 × 3.0866 × 2.572 × cos10 ) )

Vc = √( (3.0866)² + (2.572)² - (2 × 3.0866 × 2.572 × 0.9848 ) )

Vc = √( 9.527099 + 6.615184 - 15.6361 )

Vc = √0.506183

Vc = 0.71 m/s

Next, we use the SINE RULE to calculate the direction;

Vc/sin10 = Vs/c / sinθ

we substitute

0.71 / sin10 = 2.572 / sinθ

0.71 / 0.173648 = 2.572 / sinθ

4.0887 = 2.572 / sinθ

sinθ  = 2.572 / 4.0887

sinθ = 0.62905

θ = sin⁻¹( 0.62905 )

θ = 38.98°

So, angle measured clock-wise will be;

θ = 270° - 38.98°

θ = 231.02°

Therefore, the speed Vc of the current and its direction measured clockwise from the north is 0.71 m/s and 231.02° respectively

Answer:

1+1×1 multiplay then you get the answer

Answer: hello your question is incomplete attached below is the missing detail  

answer :

Complex power = 2.5 ∠ 50°  VA

apparent power = 2.5 VA

average power = 1.6 Watts

reactive power = 1.915 Var

power factor = 0.64 ( leading )

Explanation:

i) complex power

P = Vrms *  Irms

  = 17.67∠40°  * 0.1414∠-10°

  = 2.5∠50° VA

ii) Apparent power

s = Vrms * Irms

  = 17.67 * 0.1414

  = 2.5 VA

iii) Average power absorbed

Absorbed power ( p )  = Vrms * Irms * cos∅

  = 17.67 * 0.1414  * cos ( 50 )

  = 1.6 watt

iv) Reactive power

P =  Vrms * Irms * sin∅

  = 17.67 * 0.1414  * sin ( 50 )

  = 1.915 VAR

v) power factor

P.F = cos ∅ = p /s

                   = 1.6 watt / 2.5 VA  = 0.64.

Answer:

Cabinet blower switch ( A )

Explanation:

A major component of a class II BSC ( Biological safety cabinet ) is  Cabinet blower switch  because the Cabinet blower is an integral part of a class II BSC hence the switch is also a major component.

Class II BSC provides protection for the user, environment and sample to be manipulated in the laboratory ( mostly ; Pharmaceutical laboratories, Microbiology laboratories )

Answer:

The current drawn from the outlet is 0.2 A

The number of turns on the input side is 350 turns

Explanation:

Given;

number of turns of the secondary coil, Ns = 35 turns

the output current, [tex]I_s[/tex] = 2 A

power supplied, [tex]P_s[/tex] = 24 W

the standard wall outlet in most homes = 120 V = input voltage

For an ideal transformer; output power = input power

the current drawn from the outlet is calculated;

[tex]I_pV_p = P_s\\\\I_p = \frac{P_s}{V_p} = \frac{24}{120} = 0.2 \ A[/tex]

The number of turns on the input side is calculated as;

[tex]\frac{N_p}{N_s} = \frac{I_s}{I_p} \\\\N_p = \frac{N_sI_s}{I_p} \\\\N_p = \frac{35 \times 2}{0.2} \\\\N_p = 350 \ turns[/tex]

Answer:

It would break I think need to try it out

Explanation:

a.No it will be not an easy matter for Peter to prove that the Er staff caused peter to lose his promotion

Here's why?

1)The boss may think he might not able to stretch himself in emergency situations

2)He can think he can't handle normal pressure

b.What precautions can be taken to avoid giving confidential information to medical personnel who have no need to see it?

1)Peter should not let the medical personnel about his personal life

2)He might have not said extra information about his job promotion

3)Will it be easy matter for Peter to prove that the ER staff caused Peter to lose his promotion? Explain your answer

No, it will not be easy, Peter has to prove in one or in another situation that he can face the difficulties and be able to handle anxiety attacks in the future.

Hence above answers are suitable for the given situation

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Answer:

Gently push the micropipette into the tip box and tag tightly to load the tip.

Explanation:

The recommended practice when putting a tip on a micropipette is ;  Gently push the micropipette into the tip box and tag tightly to load the tip.

Given that it is not advisable to remove tip from rack so as not to contaminate it, if we want to put a tip on a micropipette we should gently push the micropipette into the tip box.

Answer:

Yes

Explanation:

Answer:

true

Explanation:

hehehe

Complete Question

Air at 40C flows over a 2 m long flat plate with a free stream velocity of 7m/s. Assume the width of the plate (into the paper) is 0.5 m. If the plate is at a co temperature of 100C,find:

The total heat transfer rate from the plate to the air

Answer:

[tex]q=1.7845[/tex]

Explanation:

From the question we are told that:

Air Temperature [tex]T_1=40c[/tex]

Length [tex]l=2m[/tex]

Velocity [tex]v=7m/s[/tex]

Width [tex]w=0.5[/tex]

Constant temperature [tex]T_t= 100C[/tex]

Generally the equation for Total heat Transfer is mathematically given by

 [tex]q=hA(T_s-T_\infty)[/tex]

Where

h=Convective heat transfer coefficient

 [tex]h=29.9075w/m^2k[/tex]

Therefore

 [tex]q=h(L*B)(T_s-T_\infty)[/tex]

 [tex]q=29.9075*(2*0.5)(100+273-(40+273))[/tex]

 [tex]q=1794.45w[/tex]

 [tex]q=1.7845[/tex]

This question is incomplete, the complete question is;

An air-standard Diesel cycle engine operates as follows: The temperatures at the beginning and end of the compression stroke are 30 °C and 700 °C, respectively. The net work per cycle is 590.1 kJ/kg, and the heat transfer input per cycle is 925 kJ/kg. Determine the a) compression ratio, b) maximum temperature of the cycle, and c) the cutoff ratio, v3/v2.

Use the cold air standard assumptions.

Answer:

a) The compression ratio is 18.48

b) The maximum temperature of the cycle is 1893.4 K

c) The cutoff ratio, v₃/v₂ is 1.946

Explanation:

Given the data in the question;

Temperature at the start of a compression T₁ = 30°C = (30 + 273) = 303 K

Temperature at the end of a compression T₂ = 700°C = (700 + 273) = 973 K

Net work per cycle [tex]W_{net[/tex] = 590.1 kJ/kg

Heat transfer input per cycle Qs = 925 kJ/kg

a) compression ratio;

As illustrated in the diagram below, 1 - 2 is adiabatic compression;

so,

Tγ[tex]^{Y-1[/tex] = constant { For Air, γ = 1.4 }

hence;

⇒ V₁ / V₂ = [tex]([/tex] T₂ / T₁ [tex])^{\frac{1}{Y-1}[/tex]

so we substitute

⇒ V₁ / V₂ = [tex]([/tex]  973 K / 303 K  [tex])^{\frac{1}{1.4-1}[/tex]

= [tex]([/tex]  3.21122  [tex])^{\frac{1}{0.4}[/tex]

= 18.4788 ≈ 18.48

Therefore, The compression ratio is 18.48

b) maximum temperature of the cycle

We know that for Air, Cp = 1.005 kJ/kgK

Now,

Heat transfer input per cycle Qs = Cp( T₃ - T₂ )

we substitute

925 = 1.005( T₃ - 700 )

( T₃ - 700 ) = 925 / 1.005

( T₃ - 700 ) = 920.398

T₃ = 920.398 + 700

T₃ = 1620.398 °C

T₃ = ( 1620.398 + 273 ) K

T₃ = 1893.396 K ≈ 1893.4 K

Therefore, The maximum temperature of the cycle is 1893.4 K

c)  the cutoff ratio, v₃/v₂;

Since pressure is constant, V ∝ T

So,

cutoff ratio S = v₃ / v₂  = T₃ / T₂

we substitute

cutoff ratio S = 1893.396 K / 973 K

cutoff ratio S = 1.9459 ≈ 1.946

Therefore, the cutoff ratio, v₃/v₂ is 1.946

Answer:

a) the ultimate tensile stress is 66717.8 psi

b) the ductility of the material in terms of percent elongation is 26%

Explanation:

Given the data in the question;

ultimate load P = 13,100 lb

elongation δl = 0.52 in

diameter of specimen d = 0.50 in

gage length l = 2.00 inch

First we determine the cross-sectional area of the specimen

A = [tex]\frac{\pi }{4}[/tex] × d²

we substitute

A = [tex]\frac{\pi }{4}[/tex] × ( 0.50 )²

A = 0.1963495 in²

a) the ultimate tensile stress σ[tex]_u[/tex]

tensile stress σ[tex]_u[/tex] = P / A

we substitute

tensile stress σ[tex]_u[/tex] = 13,100 / 0.1963495

tensile stress σ[tex]_u[/tex] = 66717.766 ≈ 66717.8 psi

Therefore, the ultimate tensile stress is 66717.8 psi

b) ductility of the material in terms of percent elongation;

percentage elongation of specimen = [change in length / original length]100

% = [ δl / l ]100

we substitute

% = [ 0.52 in / 2.00 in ]100

= [ 0.26 ]100

= 26

Therefore, the ductility of the material in terms of percent elongation is 26%

Answer:

The units on the rate of heat transfer are Joule/second, also known as a Watt.

Explanation:

Heat flow is calculated using the rock thermal conductivity multiplied by the temperature gradient. The standard units are mW/m2 = milli Watts per meter squared. Thus, think of a flat plane 1 meter by 1 meter and how much energy is transferred through that plane is the amount of heat flow.

hope it helps .

The rate of heat transfer is measured in Joules per second, also known as Watts.

Heat transfer is a thermal engineering discipline that deals with the generation, use, conversion, and exchange of thermal energy between physical systems.

Heat transfer mechanisms include thermal conduction, thermal convection, thermal radiation, and energy transfer via phase changes.

The rate of heat transfer through a unit thickness of material per unit area per unit temperature difference is defined as thermal conductivity. Thermal conductivity varies with temperature and is measured experimentally.

Heat is typically transferred in a combination of these three types and occurs at random. Heat transfer rate is measured in Joules per second, also known as Watts.

Thus, Joules per second or watts is the unit of rate of heat transfer.

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Answer:

Dark shadow

Explanation:

Shadow is nothing but space when the light is blocked by an opaque object. It is just that part where light does not reach. When you stand in the sun, you are able to see your shadow behind you. ... This is because our body is opaque and does not allow the light to pass through it

Mark brainliest

Answer:

hope it's helpful please like and Follow me

Answer:

Geography is the science that studies and describes the surface of the Earth in its physical, current and natural aspect, or as a place inhabited by humanity.

Solution :

Cost

Destination           Destination         Destination                     Maximum supply

Origin 1                       5                          7                                           600

Origin 2                     10                         10                                          800

                         15, for > 200            15, for > 200

         Demand          500                       700

Variables

Destination       1          2

Origin 1             [tex]$X_1$[/tex]        [tex]$$X_2[/tex]

Origin 2            [tex]$X_3$[/tex]        [tex]$$X_4[/tex]

Constraints   :   [tex]$X_1$[/tex], [tex]$$X_2[/tex], [tex]$X_3$[/tex], [tex]$$X_4[/tex]  ≥ 0

Supply : [tex]$X_1$[/tex] + [tex]$$X_2[/tex]  ≤ 600

              [tex]$X_3$[/tex] + [tex]$$X_4[/tex] ≤ 800

Demand : [tex]$X_1$[/tex] + [tex]$$X_3[/tex]  ≥ 500

              [tex]$X_2$[/tex] + [tex]$$X_4[/tex] ≥ 700

Objective function :

Min z = [tex]$5X_1+7X_2+10X_3+10X_4, \ (if \ X_3, X_4 \leq 200)$[/tex]

[tex]$=5X_1+7X_2+(10\times 200)+(X_3-200)15+(10 \times 200)+(X_4-200 )\times 15 , \ \ (\text{else})$[/tex]

Costs :

                  Destination 1       Destination  2

Origin 1         5                             7

Origin 2        10                           10

                     15                            15

Variables :

[tex]$X_1$[/tex]        [tex]$$X_2[/tex]

300    300  

200   400

[tex]$X_3$[/tex]      [tex]$$X_4[/tex]

Objective function : Min z = 10600

Constraints:

Supply    600 ≤ 600

                600 ≤ 800

Demand   500 ≥ 500

                 700 ≥ 500

Therefore, the total cost is 10,600.

Sheryl should implement steps to optimize the packaging in an eco-friendly manner.

Answer:

The parts of DC motor are as follows -

a) Stator

b) Frame

c) Bearings

d) Windlings /shaft

e) Commutator

f) Brush Assembly

Explanation:

The parts of DC motor are as follows -

a) Stator

b) Frame

c) Bearings

d) Windlings /shaft

e) Commutator

f) Brush Assembly

Answer:

1700 W

Explanation:

The heat transfer rate P = kA(T - T')/d where k = thermal conductivity of wall = 1.7 W/m-K, A = area of wall = 0.5 m × 1.2 m = 0.6 m², T = temperature of inner surface = 1400 K, T = temperature of outer surface = 1150 K and d = thickness of wall = 0.15 m

So, P = kA(T - T')/d

substituting the values of the variables into the equation, we have

P = 1.7 W/m-K × 0.6 m²(1400 K - 1150 K)/0.15 m

P = 1.7 W/m-K × 0.6 m² × 250 K/0.15 m

P = 255 Wm/0.15 m

P = 1700 W

So, the heat transfer rate through the wall is 1700 W

Answer: hello there is a Missing information below is the missing information

surface dimensions of  10m * 8m

answer :

a) 640 m^3/hr

b) 1334.66 m^3

Explanation:

a) Determine the flow rate ( m^3/s ) of the filter handle

Va = Q /A[tex]_{f}[/tex]

where ; Va = filtration rate ( 8.00 m/h ) ,  Q = flow rate of filter handle , A[tex]_{f}[/tex] = surface area ( 10 m * 8 m  )

Q = 8 * ( 10m * 8m ) = 640 m^3 / hr

b) Determine the volume of water needed for backwashing plus rinsing the filter in each filter cycle

Лf = 0.96  ( production efficiency )

Vb + Vr = 0.04 Vf

∴ Vf  =  ( Vb + Vr ) / 0.04  ------ ( 1 )

next step ; determine the volume of filtered water making use of effective filtration rate

Ref = ( Vf - Vb - Vr ) / A[tex]_{f}[/tex]T[tex]_{c}[/tex]

therefore : Vb + Vr = Vf - ( 80 * 52 * 7.7 ) ----  ( 2 )

Input equation 1 into 2

Vb + Vr = ( ( Vb + Vr ) / 0.04  )  - 32032  ---- ( 3 )

Resolve equation 3

hence the Volume of water needed for Backwashing  and rinsing the filter in each filter cycle

= 1334.66 m^3