please help me make a lesson plan. the topic is Zigzag line. and heres the format.
A. Objective
B. Subject matter
C. Learning activities.
D. Assessment.
E. Reinforcement​

Answer: The 8-core machine saves  87.5% of the dynamic power.

Explanation:

Let Fold = f , Vold = V , Cold = Capacitance

so

Old Dynamic power = Cold × (Vold × Vold) × f

therefore for the 8-core machine

 Fnew / Fold = 1/4

Fnew = Fold/4

we were told that Voltage decreases proportional to frequency,

so

Vnew / Vold = 1/4

Vnew = V / 4

So New Capacitance will be;

Cnew = Cold

Thus, New Dynamic power = 8 × Cnew × ( Vnew × Vnew ) ×  Fnew

= 8 × Cold × (Vold × Vold/16) × ( f/4 )

=  8 × ( Cold ) × ( Vold × Vold ) × ( f ) / 64

= (Old Dynamic Power) / 8

therefore

Old Dynamic Power / New Dynamic Power = 8

Thus, Percentage of power saved will be;

Percentage power saved = 100 × ( Old Dynamic Power - New Dynamic Power ) / Old Dynamic Power

=   100 × (8-1) / 8

= 87.5 %

Therefore The 8-core machine saves  87.5% of the dynamic power.

Answer:

(a)This portion of the problem asks that we calculate, for a Pb-Mg alloy, the mass of lead in 7.5 kg of thesolidphase at 300C just below the solubility limit.From Figure 9.20, the solubility limit for thephase at

Explanation:

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Answer:the critical radius for the homogeneous nucleating ice is  0.986 nm

Explanation:

Using the formulae below to calculate the  critical radius for homogeneous nucleation,

We Have that

Critical radius ( r *) =  [2γ Tm/ΔHf]{ 1/ Tm- T]

where γ = surface free energy =25 × 10^-3 J/m^2

Tm= solidification temperature at equilibrium =273K

Hf= latent heat of fusion = -3.1 × 10^8 J/m3

Temperature , T = -44.6°C

Critical radius ( r *) = (2 X 25 × 10^-3 J/m^2 x 273)/ (-3.1 × 10^8 J/m3 ) X ( 1/ 273 - ( -44.6 +273)

4.40x 10^-8X ( 1/273-228.4)

4.40x 10^-8  X 1/44.6

4.40x 10^-8  x 0.0224

9.86x 10 ^-10m =0.986 x 10 ^-9m = 0.986nm

Answer:

a and c

Explanation:

Answer:

The answer for the question is true

Explanation:

If you get a virus or get hacked you will still have it saved

Answer:

stagflation

Explanation:

it was used in the article

A word which was first used during the 1970s economic crisis is stagflation.

In the 1970s, an energy crisis took place in the United States of America due to the oil embargo that was imposed on it by OPEC. This oil embargo was imposed on the United States of America by the Organization of Petroleum Exporting Countries (OPEC) in 1973 because of its role in the Arab-Israeli War.

Consequently, the economy of the United States of America experienced stagflation in the following ways:

Read more on stagflation here: https://brainly.com/question/25505087

Answer: the mole fraction of D in the mixture is 0.2222

Explanation:

Given that;

mixture analysis shows 2 kmol A, 1 kmol B, 4 kmol C and some unknown kmol of D was present.

4A+3B→2C+D

As from reaction stoichiometry, for every 2 kmol of C produced, kmol of D produced = 1 kmol

so, for 4 kmol C, kmol of D produced = 4/2 × 1 kmol = 2 kmol

Now our mixture has 2 kmol A, 1 kmol B, 4 kmol C and also 2 kmol of D

so, total moles in mixture, we have (2 + 1 + 4 + 2) kmol = 9 kmol

mole fraction of D in mixture  will be;

( Kmol of D) / (total moles in mixture) = 2 / 9 = 0.2222

Therefore the mole fraction of D in the mixture is 0.2222

Answer: the critical stress for the sodalime glass = 5.7MPa

Explanation:

Ist Step

we Calculate  half length   of internal crack as

2a =0.8 mm

a = 0.8/2 = 0.4 mm

Changing  to meters becomes = 0.4 / 1000 =0.0004m

2nd Step

Now  the critical stress required for the propagation of the internal crack can be calculated using the formulae

Critical Stress (σc) =      (2 E γs/ πa) 1/2

where E= modulus of elasticity

γs= specific surface energy for soda–lime glass

a= Length

= (2 x 69 x 10 ^9 x 0.30/ π x 0.0004)1/2

=[tex]\sqrt{ 32,940,802,036,919}[/tex]

= 5,739,407.8= 5.7 x 10^6 N/m^2

= 5.7MPa

Answer:

406.140 KHz

Explanation:

Given data:

Rsig = 100 kΩ

Rin = 100kΩ

Cgs = 1 pF,

Cgd = 0.2 pF,  and   etc.

Determine the expected 3-dB cutoff frequency

first find the CM miller capacitance

CM = ( 1 + gm*ro || RL )( Cgd )

     = ( 1 + 5*10^-3 * 25 || 20 ) ( 0.2 )

     = ( 11.311 ) pF

now we apply open time constant method to determine the cutoff frequency

Th = 1 / Fh

hence : Fh = 1 / Th = [tex]\frac{1}{(Rsig +Rin) (Cm + Cgs )}[/tex]

                               = [tex]\frac{1}{( 200*10^3 ) ( 12.311 * 10^{-12} )}[/tex] =  406.140 KHz

Answer:

hi tommoro i have phisics exam i needd help only for 20 min how wants to help messege me instgram  meeraalk99

Explanation:

please

Answer:

284.4 m/s

Explanation:

At the inlet of the nozzle P =1 atm.

Temperature T = 250° C

Velocity of the steam at the inlet V_1 = 30 m/s

Change in enthalpy Δh = 40 KJ/kg

let V_2 be the final velocity

then

[tex]V_2 =\sqrt{2\Delta h+V_1^2} \\=\sqrt{2\times40+30^2}\\= 284.4 m/s[/tex]

Answer:

The appropriate solution will be "6.4 cm".

Explanation:

The given values are:

Length,

l = 0.2 m

Thermal conductivity,

K₁ = 1.82 W/m-K

K₂ = 0.095 W/m-K

Temperature,

T = 950 K

T = 300 K

Heat transfer rate,

Q = 830 W/m²

Now,

⇒  [tex]Q = \frac{\Delta T}{\frac{L_1}{K_1 A} +\frac{L_2}{K_2 A} }=\frac{A \Delta T}{\frac{L_1}{K_1 } +\frac{L_2}{K_2 } }[/tex]

⇒ [tex]\frac{Q}{A} =\frac{\Delta T}{\frac{L_1}{K_1} +\frac{L_2}{K_2} }[/tex]

On substituting the above given values in the equation, we get

⇒ [tex]830=\frac{(980-300)}{\frac{0.2}{1.82} +\frac{x}{0.095} }[/tex]

On applying cross-multiplication, we get

⇒ [tex]\frac{0.2}{1.82} +\frac{x}{0.095} =\frac{950-300}{830}[/tex]

⇒ [tex]\frac{0.2}{1.82} +\frac{x}{0.095} =\frac{650}{830}[/tex]

⇒                [tex]x =0.639 \ m[/tex]

⇒                [tex]x=6.345 \ i.e., 6.4 \ m[/tex]  

Answer:

There are several similarities between the nucleus and a liquid droplet.

Explanation:

A droplet of liquid simply is is very small or tiny drop of liquid. It is also considered as a tiny column of liquid that is surrounded by surfaces that have zero shear stress.

A nucleus on the other hand is an assembly between protons and neutrons. The latter is electrically charged whilst the former is positively charged. The number of protons present in an element is very crucial to the qualities of an element.

The main similarities between a nucleus and a liquid droplet are:

1. a nucleus consists of a large amount of neutrons and protons in the same volume as would a liquid which contains large numbers of molecules in the same volume;

2. both the nucleus and the droplet are similar for their homogeneity in electric charge and density;

3. the molecules exert the same amount for forces towards one another as would the nuclear forces in the nucleons.

4. both of them cannot be compressed

5. both molecules and nucleus are can be subject to nuclear fission which simply mean the breaking apart into smaller units (in the case of the nucleus) or the breaking apart into smaller droplets in the case of the liquid molecule.

6. There are two types of phenomena which occurs in both the liquid droplet and the nucleus which are similar to one another. They are:

Evaporation (in the case of the liquid molecule) and reaction emission (in the case of the nucleus). In evaporation, particles are lost, in Atomic transmutation, particles are lost as well.

B)  the forces which determine the stability of droplets are surface tension and gravitation. The smaller the area, the stronger the surface tension available to keep the drops from going out of shape.

Cheers

Answer:

attached below is the detailed solution

A) 8288.77 cu.ft

B) 4.96 hours

C) Vss = 131.21 IbVss/day

   Tss = 164 IbTss/day

D) attached below

E ) 0.2

F) 287.23 Ib/day

Explanation:

A) Determine the aeration basin volume

Given

∅c = 20 days

Y = 0.8Ib VSS/Ib BOD

Q = 0.3 mgd

So = 120 mg/l

Se = 2 mg/l

X = 5000 mg/l

Kd = 0.04 per day

attached below is the detailed solution

B) Determine the detention time using this relation

t = ( V / Q )* 24

  = ( 0.062 / 0.3 ) * 24  = 4.96 hours

C ) Determine Vss and Tss

we first calculate the excess biomass Px then assuming Vss ratio to be 80% for sludge age of 20 days

Vss = 131.21 IbVss/day

Tss = 164 IbTss/day

D ) determine BOD loading

  Q = 0.3 mgd , BOD = 220mg/l , V = 8288.77 cu.ft

solution attached below

e) food to microorganism ratio

F/M  = 0.2

solution attached below

f) determine the waste solid

waste solid = Q * SS * % removal of suspended solids

where : Q = 0.3 , SS = 220mgl , % = 50 %

waste solids = 0.3 * 230 * 0.5 * 8.34  = 287.23 Ib/day

Answer:

The hardenability increases with increasing austenite grain size, because the grain boundary area is decreasing. This means that the sites for the nucleation of ferrite and pearlite are being reduced in number, with the result that these transformations are slowed down, and the hardenability is therefore increased.

Answer:

0.21

Explanation:

This would have been a fairly easy one, except for that the first part of the question is missing, and as such, I'd assume a value.

We need to use chord, so, I'm assuming the length of the chord to be 0.045 m

The Area is given by the formula

Area = span * chord

Area = 0.245 * 0.045

Area = 0.011 m²

This area gotten, is what we then divide the glider weight by to get our answer.

Lee = area / weight

Lee = 0.011 / 0.0523

Lee = 0.21

Therefore, using the values of the chord I'd assumed, the Lee of the same wing is 0.21

Answer: 0.2108

Explanation:got it correct

Answer:

At L = 0.1 m

h⁻_lam = 11.004K   W/m^1.5

h⁻_turb = 7.8848K   W/m^1.8

At L = 1 m

h⁻_lam = 3.48K   W/m^1.5

h⁻_turb = 4.975K   W/m^1.8

Explanation:

Given that;

h_lam(x)= 1.74 W/m^1.5. Kx^-0.5

h_turb(x)= 3.98 W/m^1.8 Kx^-0.2

conditions for plates of length L = 0.1 m and 1 m

Now

Average heat transfer coefficient is expressed as;

h⁻ = 1/L ₀∫^L hxdx

so for Laminar flow

h_lam(x)= 1.74 . Kx^-0.5  W/m^1.5

from the expression

h⁻_lam = 1/L ₀∫^L 1.74 . Kx^-0.5   dx

= 1.74k / L { [x^(-0.5+1)] / [-0.5 + 1 ]}₀^L

= 1.74k/L = [ (x^0.5)/0.5)]⁰^L

= 1.74K × L^0.5 / L × 0.5

h⁻_lam= 3.48KL^-0.5

For turbulent flow

h_turb(x)= 3.98. Kx^-0.2 W/m^1.8

form the expression

1/L ₀∫^L 3.98 . Kx^-0.2   dx

= 3.98k / L { [x^(-0.2+1)] / [-0.2 + 1 ]}₀^L

= (3.98K/L) × (L^0.8 / 0.8)

h⁻_turb = 4.975KL^-0.2

Now at L = 0.1 m

h⁻_lam = 3.48KL^-0.5  =  3.48K(0.1)^-0.5  W/m^1.5

h⁻_lam = 11.004K   W/m^1.5

h⁻_turb = 4.975KL^-0.2 = 4.975K(0.1)^-0.2

h⁻_turb = 7.8848K   W/m^1.8

At L = 1 m

h⁻_lam = 3.48KL^-0.5  =  3.48K(1)^-0.5  W/m^1.5

h⁻_lam = 3.48K   W/m^1.5

h⁻_turb = 4.975KL^-0.2 = 4.975K(1)^-0.2

h⁻_turb = 4.975K   W/m^1.8

Therefore

At L = 0.1 m

h⁻_lam = 11.004K   W/m^1.5

h⁻_turb = 7.8848K   W/m^1.8

At L = 1 m

h⁻_lam = 3.48K   W/m^1.5

h⁻_turb = 4.975K   W/m^1.8

Answer: 1.137*10^7 Btu/h.

Explanation:

Given data:

Efficiency of the plant = 4.5percent

Net power output of the plant = 150kw

Solution:

The required collection rate

QH = W/n

= 150/0.045 * 0.94782/ 1 /60 */60 Btu/h.

= 3333.333 *3412.152Btu/h.

= 11373840 Btu/h

= 1.137*10^7 Btu/h.

Answer:

D. Both pull-in and hold-in windings are energized.

Explanation:

The instant the ignition switch is turned to the start position, "Both pull-in and hold-in windings are energized." This is because the moment the ignition switch is turned to the start position, voltage passes through to the S terminal of the solenoid.

The hold-in winding is attached to the case of the solenoid. Similarly, the pull-in winding is also attached to the starter motor. Thereby, the current will move across both windings by getting energized to generate a strong magnetic field.

Answer: Option D) 298 g/mol  is the correct answer

Explanation:

Given that;

Mass of sample m = 13.7 g

pressure P = 2.01 atm

Volume V = 0.750 L

Temperature T = 399 K

Now taking a look at the ideal gas equation

PV = nRT

we solve for n

n = PV/RT

now we substitute

n = (2.01 atm x 0.750 L) / (0.0821 L-atm/mol-K x 399 K )

= 1.5075 / 32.7579

= 0.04601 mol

we know that

molar mass of the compound = mass / moles

so

Molar Mass = 13.7 g / 0.04601 mol

= 297.7 g/mol  ≈ 298 g/mol

Therefore Option D) 298 g/mol  is the correct answer

The pressure at the bottom of an 18 ft deep storage tank for gasoline is 5.625 psi.

Pressure is defined as the force exerted on a surface's unit area. The mass of the air molecules above is what exerts pressure on the atmosphere. The enormous quantities of air molecules that make up the layers of our atmosphere collectively have a tremendous amount of weight, which bears down on whatever is below.

The storage tank's top pressure is 14.7 psi, or p atm, or 2116.8 lb/ft3.

Gasoline has a density of 45 lb/ft3.

Tank depth, h = 18 foot

The storage tank's bottom pressure is 2926.8 pounds per square foot (Pb = 2116.8 + 45 x 18).

As a result, the pressure difference between the tank's bottom and top is as follows: P = Pb - P atm = 20.325 - 14.7 = 5.625 psi

Thus, the pressure at the bottom of an 18 ft deep storage tank for gasoline is 5.625 psi

To learn more about pressure, refer to the link below:

https://brainly.com/question/12971272

#SPJ2

Answer:

Bottom-up Estimation

Explanation:

Bottom-up estimation is a type of project cost estimation that considers the cost of individual project activities and finally sums them up or finds the aggregates. The summation gives an idea of what the entire project will cost.

This is an effective way of estimating the cost of a project as it evaluates the costs on a wholistic basis. It also considers the tiniest details during the estimation process. The process moves from the simpler details to the more complicated details.

Answer:

0.2

Explanation:

Since the span and chord of the rectangular wing is missing, due to it being from the other question, permit me to improvise, or assume them. While you go ahead and substitute the ones from your question to it, as it's both basically the same method.

Let the span of the rectangular wing be 0.225 m

Let the chord of the rectangular wing be 0.045 m.

Then, the area of any rectangular chord is

A = chord * span

A = 0.045 * 0.225

A = 0.010 m²

And using the weight of the glider given to us from the question, we can find the LER for the wing.

LER = Area / weight.

LER = 0.010 / 0.05

LER = 0.2.

Therefore, using the values of the rectangular wing I adopted, and the weight of the glider given, we can see that the LER of the glider is 0.2

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Answer: 0.2025

Explanation: I got it correct

Answer:

a.dont know e

Explanation:

because d q tlga ammu

Answer:

A. 7199.55 volts

B. 120A

Explanation:

In this question we have the

line voltage = VLL = 12470volts

Phase current = Iph = 120 amps

A.)

We are to calculate the line-to-neutral/phase voltage here

VLL = √3VL-N

VL-N = VLL/√3

VL-N = 12470/√3

This gives a line to neutral phase/voltage of 7199.55 volts.

B.

We are to calculate the line current here:

In this connection, the line current and the phase current are equal

ILL = Iph = 120A

Answer:

I can help but I need to know what it looking for

Answer:

uh, i dont understand?

Explanation:

Answer:

Explanation:

A motor is a device that directs current in electrical energy form to mechanical energy, which is known as direct current (DC) motors.

DC motors are of three types: (a) The series motor, (b) The shunt motor, and (c) the compound motor. Our main focus here is the Compound motor, which is further sub-divided into:

i) The cumulative compound motors

ii) The differential compound motors

The difference between these two are:

Cumulative compound motors                  Differential compound motors

In cumulative compound motors,              In differential compound motors,

both the series and shunt windings          both series and shunt are

are connected in a way that,                     connected in a way that the

production of fluxes through them           production of fluxes via them

assist each other i.e. they aid each          always opposes each other i.e.

other in the production of magnetism      they oppose each other in the

                                                                    production of magnetism.