Match the use of the magnetic field to its respective description.​

Answer:

"Tempering Process" seems to be the appropriate choice.

Explanation:

Answer:

b

Explanation:

Answer:

b

Explanation:

Answer:

the third anwser is right on edgeinuity

Explanation:

Answer:

Technology can pull students’ focus away from their tasks.

Explanation:

Answer:

THE thing is 435

Explanation:

Some of the trade-offs that must be made while transitioning from an individually tailored solution to such an enterprise-level solution are about as continues to follow.

As a result, the aforementioned response seems to be correct.

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Answer:

[tex]\mathbf{y_1 =0.1750}[/tex]

[tex]\mathbf{y_2 = 0.8088}[/tex]

[tex]\mathbf{y_3 = 0.0161}[/tex]

Explanation:

Given that:

The temperature of the ideal solution formed by the compounds = 353K

= (353 - 273)° C

= 80° C

The data below shows the Antoine constant obtained for 1-Butanol, Benzene, and Phenol.

Compound               A                   B                C

1 - Butanol            15.3144          3212.43      182.739

Benzene              13.7819          2726.81      217.572

Phenol                 14.4397          3507.80     175.400

By the application of the Antoine constant, we can find the vapor pressure of each corresponding component at the given temperature of 80° C.

The Antoine equation is expressed as:

[tex]In (P^*) = A - \dfrac{B}{T+C}[/tex]

[tex]P_1 ^* = exp \bigg [15.314 - \dfrac{3212.43}{80+182.739} \bigg ][/tex]

[tex]P_1 ^* = exp \bigg [15.314 - \dfrac{3212.43}{262.739} \bigg ][/tex]

[tex]P_1 ^* = 21.9267 \ kPa[/tex]

[tex]P_2^* = exp \bigg [13.7819 - \dfrac{2726.81}{80+217.572} \bigg ][/tex]

[tex]P_2^* = exp \bigg [13.7819 - \dfrac{2726.81}{297.572} \bigg ][/tex]

[tex]P_2^* = 101.3287 \ kPa[/tex]

[tex]P_3^* = exp \bigg [14.4387 - \dfrac{3507.80}{80+175.400} \bigg ][/tex]

[tex]P_3^* = exp \bigg [14.4387 - \dfrac{3507.80}{255.4} \bigg ][/tex]

[tex]P_3^* =2.0221 \ kPa[/tex]

However, since the ideal solution has equimolar composition.

Then:

component (1) = 1 mole ,

component (2) = 1 mole,

component (3) = 1 mole,

The total mole = 1 + 1 + 1 = 3,

Thus, mole fraction = mole of component/total mole,

mole fraction of component (1) [tex]x_1[/tex] = 1/3,

mole fraction of component (2) [tex]x_2[/tex] = 1/3,

mole fraction of component (3) [tex]x_3[/tex] = 1/3,

i.e.

[tex]x_1 =x_2=x_3 = \dfrac{1}{3}[/tex]

Using Raoult's law, The total pressure is computed as:

[tex]P = x_1P_1^* + x_2 P_2^* + x_3P_3^*[/tex]

[tex]P = \dfrac{1}{3}(21.9267) + \dfrac{1}{3} (101.3287) + \dfrac{1}{3} (2.0221)[/tex]

P = 41.7592 kPa

and;

[tex]P_1 *x_1 = y_1P[/tex]

[tex]\implies y_1 = \dfrac{P_1\times x_1}{P}[/tex]

Thus:

[tex]y_1 = \dfrac{21.9267 \times \dfrac{1}{3}}{41.7592}[/tex]

[tex]\mathbf{y_1 =0.1750}[/tex]

[tex]y_2 = \dfrac{P_2\times x_2}{P}[/tex]

[tex]y_2 = \dfrac{101.3287 \times \dfrac{1}{3}}{41.7592}[/tex]

[tex]\mathbf{y_2 = 0.8088}[/tex]

[tex]y_3 = \dfrac{P_3\times x_3}{P}[/tex]

[tex]y_3 = \dfrac{2.0221\times \dfrac{1}{3}}{41.7592}[/tex]

[tex]\mathbf{y_3 = 0.0161}[/tex]

Answer:

a) the average length per chip is 3 mm

b) the metal removal rate MR is 90 mm³/sec

c) number of chips formed per unit time for the portion of the operation when the wheel is engaged in the work is 4,000,000 chips/min

Explanation:

Given that;

wheel diameter = 150 mm

infeed = 0.06 mm

wheel speed = 1600 m/min = 16000000 mm/s

work speed = 0.30 m/s = 300 mm/s

cross feed = 5 mm

active grits per area = 50 grits/cm²

a)

Average length per chip

Average chip length is given by

Lc = √fd

f is infeed and d is diameter of the wheel

so we substitute

Lc = √( 0.06 × 150

Lc = √ 9

Lc = 3 mm

Therefore the average length per chip is 3 mm

b)

metal removal rate MR is expressed as;

MR = fvc

v is work speed, c is cross feed and f is infeed

so we substitute

MR = 0.06 × 300 × 5

MR = 90 mm³/sec

Therefore the metal removal rate MR is 90 mm³/sec

c)

number of chips formed per unit time is expressed as

No = Vw × c × G

Vw is wheel speed and G is active grits per area

so we substitute

No = 1600000 × 5 × 50/10²

= 4,000,000 chips/min

Therefore number of chips formed per unit time for the portion of the operation when the wheel is engaged in the work is 4,000,000 chips/min

Answer:

a) the metal removal rate is 14.4 in³/min

b) the cutting time is 0.98 min

Explanation:

Given the data from the question

first we find the rpm for the spindle of the drilling tool, using the equation

Ns = 12V/πD

V is the cutting speed(120 fpm) and D is the diameter of the hole( 2 in)

so we substitute

Ns = 12 × 120 / π2

Ns = 1440 / 6.2831

Ns = 229.18 rmp

Now we find the metal removal rate using the equation

MRR = (πD²/4) Fr × Ns

Fr is the feed rate( 0.02 ipr ),

so we substitute

MRR = ((π × 2²)/4) × 0.02 × 229.18

MRR = 14.3998 ≈ 14.4 in³/min

Therefore the metal removal rate is 14.4 in³/min

Next we find the allowance for approach of the tip of the drill

A = D/2

A = 2/2

= 1 in

now find the time required to drill the hole

Tm = (L + A) / (Fr × Ns)

Lis the the depth of the hole( 3.5 in)

so we substitute our values

Tm = (3.5 + 1) / (0.02 × 229.18  )

Tm = 4.5 / 4.5836

Tm = 0.98 min

Therefore the cutting time is 0.98 min

Answer:

% increase = 26.32%

Explanation:

From conservation of mass, we can say that;

Mass flow rate at inlet = mass flow rate at exit.

Thus;

m'1 = m'2

Formula for mass flow rate is;

m' = ρV'

Where V' is volumetric flow rate = Av

Thus;

m' = ρAv

Where;

ρ is density

A is area

v is velocity

Therefore from m'1 = m'2, we can say that;

ρ1•A1•v1 = ρ2•A2•v2

Since the duct has a constant diameter, then A1 = A2

Thus, we now have;

ρ1•v1 = ρ2•v2

Making v2 the subject, we have;

v2 = ρ1•v1/ρ2

Now, since we want to find the percent increase in the velocity of the air as it flows through the dryer,we would use;

% increase = ((v2 - v1)/v1) × 100%

We have v2 = ρ1•v1/ρ2

Thus;

% increase = ((ρ1•v1/ρ2) - v1)/v1) × 100%

Factorizing v1 out, we have;

% increase = ((ρ1/ρ2) - 1)/1) × 100%

We are given;

ρ1 = 1.2 kg/m³

ρ2 = 0.95 kg/m³

Thus;

% increase = ((1.2/0.95) - 1)/1) × 100%

% increase = 26.32%

Answer:

to keep the cables away from any belts so that when starting the car back up, they do not get tangled or caught up into the engine

Explanation:

When jump-starting a car, always remember jumper cables typically have two clamps, one with the label “positive” in red and “negative” in black.

Jump-starting a car means starting a car when it is off for a long time, and it should be started by giving heat to a battery. The cables of the car are attached to the battery box and the current is given.

The steps to jump-starting a car are:

Thus, when jump-starting a car, always remember the cables are in the right position.

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Answer:

Explanation:

Given that:

v(t) = 339.4 sin(377t + 90°) V

i(t) = 5.657 sin (377t + 60°) A

v = 339.4 ∠ 90°      [tex]v_m \angle\phi_1[/tex]

i = 5.657∠60°         [tex]I_m \angle \phi_2[/tex]

The phase difference [tex]\phi = 90 -60[/tex] = 30

The average power [tex]P_{avg}[/tex] can be expressed as:

[tex]P_{avg} = \dfrac{v_m}{\sqrt{2}}\dfrac{I_m}{\sqrt{2}} \times cos (30)[/tex]

[tex]P_{avg} = \dfrac{339.4}{\sqrt{2}}*\dfrac{5.657}{\sqrt{2}} \times cos (30)[/tex]

[tex]\mathbf{P_{avg} = 831.38 \ watts}[/tex]

The reactive power Q is as follow;

[tex]Q = \dfrac{v_m}{\sqrt{2}} * \dfrac{I_m}{\sqrt{2}} \ sin \phi\\[/tex]

[tex]Q = \dfrac{339.4}{\sqrt{2}}*\dfrac{5.657}{\sqrt{2}} \times sin (30)[/tex]

Q = 479.99 VAR

The complex power S =  P + jQ

The complex power S = 831.38 W + j479.99 VAR

Answer:

Both of them are wrong

Explanation:

The two technicians have given the wrong information about the wires.

This is because firstly, a higher rating of AWG means it is smaller in diameter. Thus, the diameter of a 18 AWG wire is smaller than that of a 12 AWG wire and that makes the assertion of the technician wrong.

Also, the higher the resistance, the smaller the cross sectional area meaning the smaller the diameter. A wire with bigger cross sectional area will have a smaller resistance

So this practically makes the second technician wrong too

Answer:

Concentric circles

Explanation:

Concentric circles are two or more circles which have the same center point. The region between two concentric circles is called an annulus.

Answer:

Both Technician A and Technician B are correct

Explanation:

Air tools and electric tools are both power tools as they are used to make work easier. Air tools generally use an air compressor that powers the motor of the tool making it possible to use it while electric tools as the name implies are powered by an electric source which in this case is batteries. An example of an air tool is the nail gun which can be used by furniture makers to drive nails and they are often louder than electric tools because of vibrations caused by the compressor making it necessary to use ear protection when using the tool for ear safety.

Technician B  is also correct because it is always advisable to use impact sockets while using impact guns due to the ability of the impact sockets to withstand the force caused by operating impact guns and make work neater when nuts and bolts are being loosened or tightened.

Answer:

what are the options available?

Answer:

  0.9104

Explanation:

Suitable technology can tell you the probability.

P(-1.5≤Z≤2) ≈ 0.9104

__

A phone app gives the probability as 0.9104426667829628.

Answer:

Dam they be thinking we racist

Explanation:

Answer:

XDDDD

Explanation:

XD I LOVE THIS MEME IM SAVING THIS TO MY GALLERY XD

Answer:

MATLAB Code is written below with comments in bold, starting with % sign.

MATLAB Code:

function L = Lkm(mpg)

 L = mpg*1.60934/3.78541;  %Conversion from miles per gallon to km per   liter

 L = L^(-1);  %Conversion to liter per km

 L = L*100;   %Conversion to liter per 100 km

end

Explanation:

A function named Lkm is defined with an output variable "L" and input argument "mpg". So, in argument section, we give function the value in miles per gallon, which is stored in mpg. Then it converts it into km per liter by following formula:

L = (mpg)(1.60934 km/1 mi)(1 gallon/3.78541 liter)

Then this value is inverted to convert it into liter per km, in the next line. Then to find out liter per 100 km, the value is multiplied by 100 and stored in variable "L"

Test Run:

>> Lkm(100)

ans =

   2.3522

Answer:

43.2%

Explanation:

Given that,

Heat absorbed by a carnot heat engine, [tex]Q_1=235\ kW[/tex]

Heat rejected to the atmosphere, [tex]Q_2=164\ kW[/tex]

We need ti find the thermal efficiency of the heat engine. It is equal to the ratio of output work to the energy supplied. Its mathematical form is given by :

[tex]\eta=1-\dfrac{Q_1}{Q_2}\\\\\eta=1-\dfrac{235}{164}\\\\\eta=-0.432[/tex]

or

[tex]\eta=-43.2\%[/tex]

The egative value of efficiency shows work is done by the engine.

Answer:

i) specific thrust = 11.065

ii) F = attached below

iii) Isp vs [tex]\pi[/tex]c  = attached below

iv)  [tex]\pi c1 = 16.38* 10^8\\\pi c2 = 24.47 * 10^8[/tex]

Explanation:

Given data :

To = 220k

Po = 20 kPa

Mo = 0.8

Hf = 42800 KJ/kg

Answer:

required feedback resistance ( R2 ) = 100 k Ω

Explanation:

Given data :

Voltage gain = 100

input resistance ( R1 ) = 1 k ohms

calculate feedback resistance required

voltage gain of differential amplifier

[tex]\frac{Vout}{V2 - V1 } = \frac{R2}{R1}[/tex]

= Voltage gain =  R2/R1

= 100 = R2/1

hence required feedback resistance ( R2 ) = 100 k Ω

Answer:

Explanation:

is there a book or smg

Answer:

Hw = 1.01 meters

Explanation:

Given data:

flow rate = 10 m^3

depth of flow in channel = 2 m

Determine the height of a suppressed rectangular weir ( Hw ) using the following expressions

expression for the elevation of of water surface above crest of weir

H = 2 - [tex]H_{w}[/tex]  ------ ( 1 )

expression for the height of the weir ( Hw )

Hw = 2 - [tex]( \frac{Q}{C_{w} b}) ^{\frac{3}{2} }[/tex]   ---------- ( 2 )

expression for the weir coefficient ( Cw )

Cw = [tex]\frac{2}{3} C_{d} \sqrt{2g}[/tex]   -------------- ( 3 )

expression for the coefficient of discharge ( Cd )

Cd = 0.611 + 0.075 [tex]\frac{H}{Hw}[/tex]   ---------- ( 4 )

Finally to determine the value of Hw we apply the trial and error method

in the trial and error method the value of LHS = RHS for the number chosen to be true

Answer:

The radius of the specimen is assumed to be 9.724 mm

Explanation:

Given that:

For a cylindrical specimen of a brass alloy;

The length = 104 mm, Elongation = 5.20 mm and the tensile load = 101000 N

Let's first determine the radius of the cylindrical brass alloy from the knowledge of the cross-sectional area of a cylinder.

[tex]A_0 = \pi r ^2[/tex]

[tex]r = \sqrt{\dfrac{A_o}{\pi}}[/tex]

[tex]r = \sqrt{\dfrac{\bigg (\dfrac{F}{\sigma} \bigg )}{\pi}}[/tex]

[tex]r = \sqrt{\dfrac{F}{ \sigma \pi}}[/tex]

To estimate the tensile stress:

We need to first determine the strain relating to elongation at 5.20 mm

[tex]Strain \ \ \varepsilon= \dfrac{\Delta l}{l_o}[/tex]

[tex]Strain \ \ \varepsilon= \dfrac{5.20}{104}[/tex]

Strain ε = 0.05

Using the stress-strain plot; let assume that under the circumstances; [tex]\sigma[/tex] = 340 MPa for stress corresponding to 0.05 strain

Thus;

The cylindrical brass alloy radius [tex]r = \sqrt{\dfrac{F}{ \sigma \pi}}[/tex]

[tex]r =\sqrt{ \dfrac{101000}{(340\times 10^{6})\pi}[/tex]

r = 0.009724 m

r = 9.724 mm

Answer:

Wain for a safe gap

Explanation:

Answer:

A Normal tape is very weak under  tensile force and when a small fracture is caused, it will affect the whole tape and the tape will fail easily. while a fiber-reinforced tape will not fail easily under same stress and this is the major advantage of a fiber-reinforced tape has over a normal tape

Explanation:

The fracture behavior would be different for a fiber-reinforced tape in the following way :

* It's behavior during tensile stress and its fracture behavior.

A Normal tape is very weak under  tensile force and when a small fracture is caused, it will affect the whole tape and the tape will fail easily. while a fiber-reinforced tape will not fail easily under same stress and this is the major advantage of a fiber-reinforced tape has over a normal tape

Answer:

ok

Explanation:

Answer:

[tex]62.14\ \text{miles}[/tex]

[tex]6213727.37\ \text{miles}[/tex]

Explanation:

The distance of the chain would be the product of the dislocation density and the volume of the metal.

Dislocation density = [tex]10^5\ \text{mm}^{-2}[/tex]

Volume of the metal = [tex]1000\ \text{mm}^3[/tex]

[tex]10^5\times 1000=10^8\ \text{mm}\\ =10^5\ \text{m}[/tex]

[tex]1\ \text{mile}=1609.34\ \text{m}[/tex]

[tex]\dfrac{10^5}{1609.34}=62.14\ \text{miles}[/tex]

The chain would extend [tex]62.14\ \text{miles}[/tex]

Dislocation density = [tex]10^{10}\ \text{mm}^{-2}[/tex]

Volume of the metal = [tex]1000\ \text{mm}^3[/tex]

[tex]10^{10}\times 1000=10^{13}\ \text{mm}\\ =10^{10}\ \text{m}[/tex]

[tex]\dfrac{10^{10}}{1609.34}=6213727.37\ \text{miles}[/tex]

The chain would extend [tex]6213727.37\ \text{miles}[/tex]

Answer:

Following are the solution to this question:

Explanation:

let,

[tex]\text{m= mass of the gas}\\\text{M= molar weight of the gas}\\ m_{N_2} = 1 \kg \\T_{N_2}= 298 \ K\\ M_{N_2} = 28 \ \frac{kg}{kmol}\\ P_{N_2}= 300 \ kPa\\m_{O_2} = 3 \ kg \\T_{O_2} = 298 \ K \\M_{O_2}=32 \ \frac{kg}{kmol}\\ P_{O_2} = 500 \ kPa \\T_m= 298 \ K[/tex]

Calculating the mole of nitrogen:  

[tex]N_{N_2} = \frac{m_{N_2}}{M_{N_2}}[/tex]

       [tex]=\frac{1}{28} \\\\= 0.0357 \ kmol[/tex]

Calculating the mole oxygen:

[tex]N_{O_2} = \frac{m_{O_2}}{M_{O_2}}\\\\=\frac{3}{32} \\\\= 0.09375 \ kmol[/tex]

Calculating the total mole:  

[tex]N_m=N_{N_2}+N_{O_2}\\\\= 0.0357+0.09375\\\\= 0.1295 \ kmol[/tex]

Calculating the volume of nitrogen:  

[tex]V_{N_2} =\frac{N_{N_2} \times R_{u}\times T_{N_2}}{P_{N_2}}\\\\= \frac{0.0357 \times 8.314 \times 298}{300}\\\\= 0.295 m^3[/tex]

Calculating the volume of oxygen:

[tex]V_{O_2} =\frac{N_{O_2} \times R_{u}\times T_{O_2}}{P_{O_2}}\\\\= \frac{0.09375 \times 8.314 \times 298}{500}\\\\= 0.465 m^3[/tex]

Calculating the total volume:  

[tex]V_m=V_{N_2}+V_{O_2}\\\\=0.295+0.465 \\\\=0.76 \ m^3[/tex]

Calculating the mixture pressure:

[tex]P_m = \frac{N_{m} \times R_u \times T_m}{V_m} \\\\= \frac{0.1295 \times 8.314 \times 298}{0.76} \\\\= 422.2\ kPa[/tex]

Answer:

Ф = [tex]\frac{Q}{e_{0} } [ \frac{\frac{4\pi }{3 }(R)^3 }{\frac{4}{3}\pi (R)^3 } ][/tex]

Explanation:

Radius of Gaussian surface = R

Charge in the Sphere ( Gaussian surface ) = Q

lets take the radius of the sphere to be equal to radius of the Gaussian surface i.e. R

To determine the net electric flux through the Gaussian surface

we have to apply Gauci law

Ф = 4[tex]\pi r^2 E[/tex]

Ф = [tex]\frac{Q_{enc} |}{e_{0} }[/tex]

    = [tex]\frac{Q}{e_{0} } [ \frac{\frac{4\pi }{3 }(R)^3 }{\frac{4}{3}\pi (R)^3 } ][/tex]

Answer:

1 foot 10 inches

Explanation:

1 foot = 12 inches + 10 inches = 22 inches

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