2. A D flip flop can be made to operate in a toggle mode (divide its CLOCK input frequency by two) by adding an external Inverter gate and making the appropriate connections. Draw the schematic for this circuit.

Answer:

Divide the difference in tax by the amount of income from the investment, and you'll get the economic marginal tax rate from investing. Most people refer to marginal tax rates as being identical to tax brackets.

hope this helps

have a good day :)

Explanation:

Answer: hello your question is incomplete attached below is the complete question

answer : Ac = 5° , A[tex]_{f}[/tex] = 2.5°

Explanation:

Bandgap energy for material A = 0.90 eV

Bandgap energy for material B = 1.10 eV

Calculate the ratio of ni for Material B to Material B

Total derivation ( d ) = d1 + d2

 d = A[tex]_{c}[/tex] ( μ[tex]_{c}[/tex] - 1 ) + A[tex]_{f}[/tex] ( μ[tex]_{f}[/tex] - 1 )  ---- ( 1 )

where : d = 1° , μ[tex]_{c}[/tex] = 1.5 , μ[tex]_{f}[/tex] = 1.6

Input values into equation 1 above

1° = 0.5Ac + 0.6Af  ---- ( 2 )

also d = d1 [ 1 - w/ w1 ] ------ ( 3 )

∴ d = Ac ( μ[tex]_{c}[/tex] - 1 ) ( 1 - w/w1 )

  1° = Ac ( 1.5 - 1 ) ( 1 - 0.06/0.1 ) --- ( 4 )

resolving  equation ( 4 )

Ac = 5°

resolving equation ( 2 )

A[tex]_{f}[/tex] = 2.5°

Answer:

106 kips

Explanation:

Determine the service live load for a first floor interior column

following ASCE-7  guidelines

Area = 20 * 25 = 500 ft^2 which is > 400 ft^2

Nominal live load on every floor = 80 psf

Hence the reduced service live load ( L )

L = Lo [ 0.25 + 15 / √AI ]  ------ ( 1 )

Lo = unreduced load

AT = Tributary area = 500 * number of floors = 2000 ft^2

AI ( influence area ) = AT * number of floors = 2000 * 4 = 8000 ft^2

Lo = 1.6 * 80 psf * 2000 ft^2 = 256 kips

Input values into equation 1 above

L = 106 kips

Answer:

Total number of pastries that will be made from at least [tex]3/8[/tex] a pound of dough is [tex]8.[/tex]

Explanation:

It is given that the line plot shows the weight of the dough for each pastry.

From the given box plot it is clear that,

Number of pastries with weight pound

Number of pastries with weight pound

Number of pastries with weight pound

We know that

We need to find the number of pastries that will be made from at least a pound of dough. It means the number of pastries whose weight is

Number of pastries with weight pound Number of pastries with weight pound

[tex]N=3+5[/tex]

[tex]N=8[/tex]

Therefore, the total number of pastries that will be made from at least a pound of dough is

It is the complete question,

Solution :

Given :

External diameter of the hemispherical shell, D = 500 mm

Thickness, t = 20 mm

Internal diameter, d = D - 2t

                                 = 500 - 2(20)

                                 = 460 mm

So, internal radius, r = 230 mm

                                 = 0.23 m

Density of molten metal, ρ = [tex]$7.2 \ g/cm^3$[/tex]

                                                  = [tex]$7200 \ kg/m^3$[/tex]

The height of pouring cavity above parting surface is h = 300 mm

                                                                                                  = 0.3 m

So, the metallostatic thrust on the upper mold at the end of casting is :

[tex]$F=\rho g A h$[/tex]

Area, A [tex]$=2 \pi r^2$[/tex]

            [tex]$=2 \pi (0.23)^2$[/tex]

            [tex]$=0.3324 \ m^2$[/tex]

[tex]$F=\rho g A h$[/tex]

   [tex]$=7200 \times 9.81 \times 0.3324 \times 0.3$[/tex]

     = 7043.42 N

Answer:

Explanation:

The following code was written in Java. It creates classes for each one of the shapes requested and includes all of their variables and the area method, as well as a toString method. Then in the main method, the menu is created which allows the user to enter the shape that they want and if they decide to exit it will print out every shape within the shapes array. Due to technical reasons I have added the code as a txt file below and in the picture you can see the output.

Answer:

The Mechanical Properties include Elasticity, Plasticity, Ductility, Malleability, Hardness, Toughness, Brittleness, Tenacity, Fatigue, Fatigue resistance, Impact Resistance property, Machineability, Strength, Strain Energy, Resilience, Proof Resilience, Modulus of Resilience, Creep, Rupture, and Modulus of Toughness.

Answer:

The composition of an alloy is 0.75%wt

Explanation:

Let alloy is a hypoeutectoid alloy.

So, we can apply lever rule which is shown below.

[tex]W_{a}=[/tex][tex]\frac{C_{0} -C_{a} }{C_{b}-C_{a} }[/tex]

We know that [tex]C_{a}=0.022[/tex] and [tex]C_{b}=6.7[/tex]

Given that [tex]W_{a}=0.109[/tex], we have to find [tex]C_{0}[/tex]

Thus,

[tex]0.109=\frac{C_{0}-0.022 }{6.7-0.022}[/tex]

Hence

[tex]C_{0}=0.75[/tex]%wt

Answer:

4.2019 mts , 4.598 mts

Explanation:

center of flotation = 2 m fwd.

MCTC = 120 tonnes

Determine the position of  200 t of cargo to be loaded

first step : determine the change trim = trimming moment / MCTC

                                                              = 134 * 1 / 120 = 0.11 mts

∴ бtf = 0.11 * ( 2/100 ) = 0.0022 mts ( given that the ship trims forward )

бtf = - 0.0022 mts

also ; бTa = 0.0019 mts (  + due to increase in draught )

determine the Final position

Ta = 4.2 ( draught at AP )  + бTa

     = 4.2 +  0.0019

     = 4.2019 mts

T[tex]_{f}[/tex] = 4.6 ( draught at Fp )  + бtf

                                = 4.6 + ( - 0.0022 )  = 4.598 mts

Correct question is;

A commercial refrigerator with refrigerant-134a as the working fluid is used to keep the refrigerated space at −30°C by rejecting its waste heat to cooling water that enters the condenser at 18°C at a rate of 0.25 kg/s and leaves at 26°C. The refrigerant enters the condenser at 1.2 MPa and 65°C and leaves at 42°C. The inlet state of the compressor is 60 kPa and −34°C and the compressor is estimated to gain a net heat of 450 W from the surroundings. Determine (a) the quality of the refrigerant at the evaporator inlet, (b) the mass flow rate of the refrigerant.

Answer:

A) Quality = 0.48

B) Mass flow rate; m' = 0.0455 kg/s

Explanation:

A) From the refrigerant R-144 table I attached,

At P=60kpa and interpolating at - 34°C,we obtain enthalpy;

h1 = 230.03 Kj/kg

Also at P= 1.2MPa which is 1200kpa and interpolating at 65°C,we obtain enthalpy ;

h2 = 295.16 Kj/Kg

Also at P= 1.2MPa which is 1200kpa and interpolating at 42°C,we obtain enthalpy ;

h3 = 111.23 Kj/Kg

h4 is equal to h3 and thus h4 = 111.23 Kj/kg

We want to find the refrigerant quality at the evaporation inlet which is state 4 and P= 60 Kpa.

Thus, from the table attached, we see that hf = 3.84 at that pressure and hg = 227.8

Now, to find the quality of the refrigerant, we'll use the formula,

x4 = (h4 - hg) /(hf - hg)

Where x4 is the quality of the refrigerant. Thus;

x4 = (111.23 - 3.84)/(227.8 - 3.84) = 0.48

B) The mass flow rate of the refrigerant can be determined by applying a 1st law energy balance across the condenser. Thus, the water properties can be obtained by using a saturated liquid at the given temperatures;

So using the first table in the image i attached; interpolating at 18°C; hw1 = hf = 75.54 kJ/kg

Also interpolating at 26°C; hw2 = hf = 109.01 kJ/kg

Now;

(m')(h2 − h3)= (m_w)(hw2 − hw1)

m' is mass flow rate

Making m' the subject, we get;

m' = [(m_w)(hw2 − hw1)]/(h2 − h3)

m' = [(0.25 kg/s)(109.01 − 75.54) kJ/kg] /(295.13 − 111.37) kJ/kg

m' = 8.3675/183.76

m' = 0.0455 kg/s

Answer:

I think D is the correct answer.

Explanation:

hope its correct

Answer:

Hydraulic conductivity =  878.53 gpd / ft^2

Explanation:

Calculate the hydraulic conductivity of the confined aquifer

Thickness of aquifer ( b )  = 100 ft

Pump rate ( Q ) = 1500 gpm

r₂ =  700 ft

r₁  = 70 ft

h1 = 1 ft , h2 = 10 ft  

To determine the hydraulic conductivity we will apply the relation below

Q = 2πKb (( h2 - h1) / In(r₂ / r₁ ) )

1500 = 2 * π * k * 100  (( -1 -(- 10 ) / In ( 700 / 70 ) )

∴ K ( hydraulic conductivity ) = [ 1500 / (( 9 / 2.30 )  ] / 2π*100

                                                = 383.33 / 628.32

K = 0.610087 gpm /ft^2

   = 878.53 gpd / ft^2

Answer:

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Answer:

i) 154°F

ii) 0.0114 mm

iii) 0.075 btu/s

iv) 0.080 in^3/s

Explanation:

i)Determine the average film Temperature

( from Viscosity-temperature chart in US customary units for SAE10 )

at Temp =  154°F

absolute viscosity = 4.25 rev

and ΔT = 2 ( 154 - operating temp ) = 28°F

where : operating temp = 140°F as given in question

also from the chart applying Raimondi and Boyd boundary conditions

ΔT = 29°F  hence we can pick 154°F as the average film temperature

ii) Calculate the minimum film thickness

Cmin = Bore diameter - Journal shaft diameter / 2

         = 3.003 - 3 / 2 = 0.0015 in

Given that : h₀ / Cmin = 0.76

there h₀ = 0.0015 * 0.76 = 0.0114 mm

iii)Determine the heat loss rate

Also known as power loss ratio ( H ) = ( 2π*w*f*r*N ) / ( 778 *12 )

( 2π * 0.0132 * 675 * 1.5 * 10 ) / ( 9336 )

Heat loss rate = 0.075 btu/s

iv)Calculate lubricant side-flow rate for minimum clearance assembly

Side flow rate = 0.315 * Total volume flow rate

                       = 0.315 * ( 3.8 * 1.5 * 0.0015 * 10 * 3 )

                      = 0.080 in^3/s.

Answer:

(c) electronic

Explanation:

for pf, this is correct....Here is the sentence from the pf lesson

" The primary energy source for the controller within buildings is typically electric, pneumatic, or

electronic

."

Answer:

Both Tech A and Tech B are correct.

Explanation:

In order to ensure that some intake manifolds have coolant running through them to assist them to warm up and to also ensure that an electric heater grid is used in some intake manifolds to warm up the intake air, a preheat grid is positioned between the intake manifold and throttle body.

This therefore implies that both Tech A and Tech B are correct. That is, it is true that coolant circulates through some intake manifolds to ensure that they warm up. And it is also true that some intake manifolds use an electric heater grid to warm up the intake air. This however occurs when a preheat grid is positioned between the intake manifold and throttle body.

Answer:

I think D is the correct answer

Explanation:

hope it is

Answer:

70,900

Explanation:

Given :

True stress (psi) _____ True strain (psi)

48400 ______________ 0.11

60400 ______________ 0.19

Using ratio simplification :

Let :

s = True stress ; t = true strain

s1 = 48400

s2 = 60400

t1 = 0.11

t2 = 0.19

True stress, s0 ; needed to produce a True plastic strain, tp = 0.26

(s0 - s1) / (s2 - s1) = (tp - t1) / (t2 - t1)

(s0 - 48400)/(60400 - 48400) = (0.26 - 0.11)/(0.19 - 0.11)

(s0 - 48400)/12000 = 0.15/0.08

Cross multiply :

0.08(s0 - 48400) = 0.15 * 12000

0.08s0 - 3872 = 1800

0.08s0 = 1800 + 3872

0.08s0 = 5672

s0 = 5762 / 0.08

s0 = 70,900

The true stress required to produce a true plastic strain of 0.26 is 70,900

Answer:

M/A = 0.425 g/cm²

Explanation:

Given the following data;

Mass = 192 grams

Dimensions = 7 * 10 inches.

To find the mass per unit area;

First of all, we would determine the area of the lead sheet;

Area = 7 * 10

Area = 70 in²

Conversion:

1 square inch = 6.452 square centimeters

70  inches = 70 * 6.452 = 451.64 square centimeters

Next, we find the mass per unit area;

M/A = 192/451.64

M/A = 0.425 g/cm²

Answer:

A)

  D = 158.42 kmol/h

  B =  191.578 kmol/h

B) Rmin = 1.3095

Explanation:

a) Determine the distillate and bottoms flow rates ( D and B )

F = D + B ----- ( 1 )

Given data :

F = 350 kmol/j

Xf = 0.45 mole

yD ( distillate comp ) = 0.97

yB ( bottom comp ) = 0.02

back to equation 1

350(0.45) = 0.97 D + 0.02 B  ----- ( 2 )

where; B = F - D

Equation 2 becomes

350( 0.45 ) = 0.97 D + 0.02 ( 350 - D )  ------ 3

solving equation 3

D = 158.42 kmol/h

resolving equation 2

B = 191.578 kmol/h

B) Determine the minimum reflux ratio Rmin

The minimum reflux ratio occurs when the enriching line meets the q line in the VLE curve

first we calculate the value of the enriching line

Y =( Rm / R + 1 m ) x   +  ( 0.97 / Rm + 1 )

q - line ;  y =  ( 9 / 9-1 ) x -  xf/9-1

therefore ; x = 0.45

Finally Rmin

=  (( 0.97 / (Rm + 1 ))  = 0.42

0.42 ( Rm + 1 ) = 0.97

∴ Rmin = 1.3095

C. Semiconductors.

They are made up of what is called a porous silicon material that is very light and extremely heat resistant.

Answer:

composites

Explanation:

Answer: Hello your question is poorly written attached below is the complete question

answer :

: max value to avoid failure = 59 MPa

; max value to avoid failure = 34.064 Mpa

Explanation:

Attached below is the detailed solution of the given problem

For Tresca criterion : max value to avoid failure = 59 MPa

For Von-Nissen criterion ; max value to avoid failure = 34.064 Mpa

Answer:

Technician A is correct

Explanation:

Technician A is correct because Toe misalignment in vehicles can be caused by either wear and tear of steering linkages or by a bent trailing arm or a bent tie rod and nit by a bent inverted U- channel hence Technician A is correct

Answer:

1.2 Newtons upwards

Explanation:

because the friction is opposite of the magnitude.

The container that traps the most heat is a shoebox covered in aluminum foil.

An aluminum foil is used for food storage, and to wrap items, such as meats, to minimize moisture loss while cooking.

Aluminum foil is also an excellent insulator. This is due to the fact that it reduces heat emission by reflecting it back.

Therefore, in a shoebox container, the aluminum foil helps to trap most heat better than a plastic wrap or wax paper.

Learn more about aluminum foil here:

https://brainly.com/question/277354

#SPJ2

Answer:

12.332 KW

The positive sign indicates work done by the system ( Turbine )

Explanation:

Stagnation pressure( P1 ) = 900 kPa

Stagnation temperature ( T1 ) = 658K

Expanded stagnation pressure ( P2 ) = 100 kPa

Expansion process is  Isentropic, also assume steady state condition

mass flow rate ( m ) = 0.04 kg/s

Calculate the Turbine power

Assuming a steady state condition

( p1 / p2 )^(r-1/r)  = ( T1 / T2 )

= (900 / 100)^(1.4-1/1.4) = ( 658 / T2 )

=  ( 9 )^0.285 = 658 / T2

∴ T2 = 351.22 K

Finally Turbine Power / power developed can be calculated as

Wt = mCp ( T1 - T2 )

    = 0.04 * 1.005 ( 658 - 351.22 )

    = 12.332 KW

The positive sign indicates work done by the system ( Turbine )

Answer:

The receiving end Voltage ( VR )  will reduce significantly

Explanation:

Determine what happens to the receiving end voltage

Given that ; Load on transmission line is increased and Load has a lagging power factor

Increase in load = Increase in reactive/real power drawn from the transmission line and this will cause an increase in the current drawn as well.

Vs = VR + j XL I

Vs = sending end voltage

VR = receiving end voltage

jXL I = voltage drop across reactance

The rotating and shifting of Vs to a new power angle ( lagging power factor )  without a change in magnitude will cause the value of the VR receiving end voltage to drop significantly

Answer:

For the Top Side

- Strain ε  = 0.00021739

- Elongation is 0.00260868 cm

For The Right side

- Strain ε  = 0.00021739

-Elongation is 0.00347826 cm

Explanation:

Given the data in the question;

Length of the squared titanium plate = 12 cm by 12 cm = 0.12 m by 0.12 m

Thickness = 5 mm = 0.005 m

Force to the Top F[tex]_t[/tex] = 15 kN = 15000 Newton

Force to the right F[tex]_r[/tex] = 20 kN = 20000 Newton

elastic modulus, E = 115 GPa = 115 × 10⁹ pascal

Now, For the Top Side;

- Strain = σ/E = F[tex]_t[/tex]  / AE

we substitute

= 15000 / ( 0.12 × 0.005 × (115 × 10⁹) )

= 15000 / 69000000

Strain ε  = 0.00021739

- Elongation

Δl = ε × l

we substitute

Δl = 0.00021739 ×  12 cm

Δl = 0.00260868 cm

Hence, Elongation is 0.00260868 cm

For The Right side

- Strain = σ/E = F[tex]_r[/tex]  / AE

we substitute

Strain = 20000 / ( 0.12 × 0.005 × (115 × 10⁹) )

= 20000 / 69000000

Strain ε = 0.000289855

- Elongation

Δl = ε × l

we substitute

Δl = 0.000289855×  12 cm

Δl = 0.00347826 cm

Hence, Elongation is 0.00347826 cm