Technician A says the term hot wire refers to the section of the circuit after the load or electric device. Technician says the ground wire refers to the section of the circuit after the load or electric device. Who is right?

Answer:

what are you in third grade??

Explanation:

the answer is  the air pressure inside of the bottle reaches the same air pressure as outside of the bottle. K?? got it love???I hope soo

Answer:

the air pressure inside of the bottle reaches the same air pressure as outside of the bottle.

Explanation:

Answer:

actually, it was John Doe and Jane Doe, as they were created as testers on the exact same say when Roblox was created.

Explanation:

The first  player was builderman. Option C

The first player was one whose user handle was builderman.

However, the account was terminated. It was replaced with

‘Builderman’ is the account that belonged to the CEO and Co-Founder of , David Baszucki. He manages all of the admins.

When you make a new account, he's automatically your friend.

Note that ‘John Doe’ and ‘Jane Doe’ were two test accounts that were created by David.

The accounts were later rumored to hack others’

Learn about username at: https://brainly.com/question/28344005

#SPJ6

Answer:

  a. vehicle color

Explanation:

One might expect that the cost of owning a vehicle would not be a function of its color. There is no reason to believe that a blue car gets better gas mileage than a green car of the same make, model, and equipment. So, the appropriate choice is ...

  a. vehicle color

__

However, vehicle color may play a role in ownership costs if more money is spent on washing a white car than would be spent on a black or beige car. Similarly, a light-colored car may require less use of an air-conditioner in the summer sun than does a dark-colored car, ultimately affecting fuel cost. It isn't always obvious what the features of a vehicle are that contribute to ownership cost.

Answer:

V= 90h cm³ where h is the height of the rectangular prism.

Explanation:

The formula for volume of a rectangular prism is ;

V=l*w*h  where;

V=volume in cm³

l= length of prism=10cm

w =width of the prism = 9 cm

Assume the height of the prism as h cm then the volume will be;

V= 10* 9*h

V= 90h cm³

when the value of height of the prism is given, substitute that value with h to get the actual volume of the prism.

Answer:

(a) X = 17.48 m

(b) Y = 8.903 m

(c) X = 28.33 m

(d) Y = 8.68 m

(e) X = 44.68 m

(f) Y = 0 m

Explanation:

Given;

magnitude of initial velocity, v = 21.1 m/s

angle of projection, θ = 39.9°

the horizontal component of the velocity, [tex]v_x = vcos \theta[/tex]

[tex]v_x = 21.1(cos 39.9^0) = 16.187 \ m/s[/tex]

the vertical component of the velocity, [tex]v_y = vsin \theta[/tex]

[tex]v_y = 21.1(sin 39.9^0) = 13.535 \ m/s[/tex]

(a)  the horizontal components of its displacement;

[tex]X = v_xt + \frac{1}{2} gt^2[/tex]

gravitational influence on horizontal direction is negligible, g = 0

[tex]X = v_xt \\\\X = (16.187)(1.08)\\\\X = 17.48 \ m[/tex]

(b) the vertical components of its displacement;

[tex]Y = v_yt + \frac{1}{2} gt^2\\\\Y = (13.535*1.08) + \frac{1}{2} (-9.8)(1.08)^2\\\\Y = 14.618 -5.715\\\\Y = 8.903 \ m\\[/tex]

(c) the horizontal components of its displacement;

[tex]X = v_xt\\\\ X = (16.187)(1.75)\\\\X = 28.33 \ m[/tex]

(d) the vertical components of its displacement;

[tex]Y = v_yt + \frac{1}{2} gt^2\\\\Y = (13.535 *1.75) + \frac{1}{2} (-9.8)(1.75)^2\\\\Y = 8.68 \ m[/tex]

(e)  the horizontal components of its displacement;

[tex]X = v_xt \\\\X = (16.187*5.09)\\\\X = 82.39 \ m\\[/tex]

(f) the vertical components of its displacement;

[tex]Y = v_yt + \frac{1}{2}gt^2\\\\ Y = (13.535*5.09)+ \frac{1}{2}(-9.8)(5.09)^2\\\\Y = -58.1 \ m[/tex]

this displacement is not possible because it is beyond zero vertical level; it shows that the stone will hit the ground before 5.09 s.

When the stone hits the ground Y = 0 m

At zero vertical displacement (Y = 0 m), the time at that position will be calculated as;

[tex]Y = v_yt+\frac{1}{2}gt^2\\\\0 = (13.535t) + \frac{1}{2}(-9.8)t^2\\\\0= 13.535t - 4.9t^2\\\\0 = t(13.535-4.9t)\\\\t = 0 \ \ or \\\\13.535-4.9t = 0\\\\4.9t = 13.535\\\\t = 2.76 \ s[/tex]

The horizontal displacement at this time is given by;

X = vₓt

X = (16.187 x 2.76)

X = 44.68 m

Answer:

The pump delivers 32.737 kilowatts to the water.

Explanation:

We can describe the system by applying the Principle of Energy Conservation and the Work-Energy Theorem, the pump system, which works at steady state and changes due to temperature are neglected, is represented by the following model:

[tex]\dot W_{in} + \dot m \cdot g \cdot (z_{1}-z_{2}) + \frac{1}{2}\cdot \dot m \cdot (v_{1}^{2}-v_{2}^{2})+\dot m \cdot [(u_{1}+P_{1}\cdot \nu_{1})-(u_{2}+P_{2}\cdot \nu_{2})]-\dot E_{losses} = 0[/tex] (Eq. 2)

Where:

[tex]\dot m[/tex] - Mass flow, measured in kilograms per second.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

[tex]z_{1}[/tex], [tex]z_{2}[/tex] - Initial and final heights, measured in meters.

[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final flow speeds at pump nozzles, measured in meters per second.

[tex]u_{1}[/tex], [tex]u_{2}[/tex] - Initial and final internal energies, measured in joules per kilogram.

[tex]P_{1}[/tex], [tex]P_{2}[/tex] - Initial and final pressures, measured in pascals.

[tex]\nu_{1}[/tex], [tex]\nu_{2}[/tex] - Initial and final specific volumes, measured in cubic meters per kilogram.

Then, we get this expression:

[tex]\dot W_{in} + \dot m \cdot g \cdot (z_{1}-z_{2}) +\frac{1}{2}\cdot \dot m \cdot (v_{1}^{2}-v_{2}^{2}) +\dot m\cdot \nu \cdot (P_{1}-P_{2})-\dot E_{losses} = 0[/tex]  (Ec. 3)

We note that specific volume is the reciprocal of density:

[tex]\nu = \frac{1}{\rho}[/tex] (Ec. 4)

Where [tex]\rho[/tex] is the density of water, measured in kilograms per cubic meter.

The initial pressure of water ([tex]P_{1}[/tex]), measured in pascals, can be found by Hydrostatics:

[tex]P_{1} = P_{atm} + \rho\cdot g \cdot \Delta z[/tex] (Ec. 5)

Where:

[tex]P_{atm}[/tex] - Atmospheric pressure, measured in pascals.

[tex]\Delta z[/tex] - Depth of the entrance of the inlet pipe with respect to the limit of the water reservoir.

If we know that [tex]p_{atm} = 101325\,Pa[/tex], [tex]\rho = 1000\,\frac{kg}{m^{3}}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]\Delta z = 6\,m[/tex], then:

[tex]P_{1} = 101325\,Pa+\left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}})\cdot (6\,m)[/tex]

[tex]P_{1} = 160167\,Pa[/tex]

And the specific volume of water ([tex]\nu[/tex]), measured in cubic meters per kilogram, is: ([tex]\rho = 1000\,\frac{kg}{m^{3}}[/tex])

[tex]\nu = \frac{1}{1000\,\frac{kg}{m^{3}} }[/tex]

[tex]\nu = 1\times 10^{-3}\,\frac{m^{3}}{kg}[/tex]

The power losses due to friction is found by this expression:

[tex]\dot E_{losses} = \dot m \cdot g\cdot h_{losses}[/tex]

Where [tex]h_{losses}[/tex] is the total friction head loss, measured in meters.

The mass flow is obtained by this:

[tex]\dot m = \rho \cdot \dot V[/tex] (Ec. 6)

Where [tex]\dot V[/tex] is the volumetric flow, measured in cubic meters per second.

If we know that [tex]\rho = 1000\,\frac{kg}{m^{3}}[/tex] and [tex]\dot V = 0.061\,\frac{m^{3}}{s}[/tex], then:

[tex]\dot m = \left(1000\,\frac{kg}{m^{3}}\right)\cdot \left(0.061\,\frac{m^{3}}{s} \right)[/tex]

[tex]\dot m = 61\,\frac{kg}{s}[/tex]

Then, the power loss due to friction is: ([tex]h_{losses} = 5\,m[/tex])

[tex]\dot E_{losses} = \left(61\,\frac{kg}{s}\right)\cdot \left(9.807\,\frac{m}{s^{2}} \right) \cdot (5\,m)[/tex]

[tex]\dot E_{losses} = 2991.135\,W[/tex]

Now, we calculate the inlet and outlet speed by this formula:

[tex]v = \frac{\dot V}{\frac{\pi}{4}\cdot D^{2} }[/tex] (Ec. 7)

Inlet nozzle ([tex]\dot V = 0.061\,\frac{m^{3}}{s}[/tex], [tex]D = 0.12\,m[/tex])

[tex]v_{1} = \frac{0.061\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.12\,m)^{2} }[/tex]

[tex]v_{1} \approx 5.394\,\frac{m}{s}[/tex]

Oulet nozzle ([tex]\dot V = 0.061\,\frac{m^{3}}{s}[/tex], [tex]D = 0.05\,m[/tex])

[tex]v_{2} = \frac{0.061\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.05\,m)^{2} }[/tex]

[tex]v_{2} \approx 31.067\,\frac{m}{s}[/tex]

([tex]\dot m = 61\,\frac{kg}{s}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]z_{2} = 2\,m[/tex], [tex]z_{1} = -6\,m[/tex], [tex]v_{2} \approx 31.067\,\frac{m}{s}[/tex], [tex]v_{1} \approx 5.394\,\frac{m}{s}[/tex], [tex]P_{2} = 101325\,Pa[/tex], [tex]P_{1} = 160167\,Pa[/tex], [tex]\dot E_{losses} = 2991.135\,W[/tex])

[tex]\dot W_{in} = \left(61\,\frac{kg}{s}\right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot [2\,m-(-6\,m)]+\frac{1}{2}\cdot \left(61\,\frac{kg}{s}\right) \cdot \left[\left(31.067\,\frac{m}{s} \right)^{2}-\left(5.394\,\frac{m}{s} \right)^{2}\right] +\left(61\,\frac{kg}{s}\right)\cdot \left(1\times 10^{-3}\,\frac{m^{3}}{kg} \right)\cdot (101325\,Pa-160167\,Pa)+2991.135\,W[/tex]

[tex]\dot W_{in} = 32737.518\,W[/tex]

The pump delivers 32.737 kilowatts to the water.

Answer:

A computo is a box and it works really well if its good

Answer:false

Explanation:

Bc

It is given that volume is constant i.e 3000 L.

So, [tex]\dfrac{T_1}{P_1}=\dfrac{T_2}{P_2}[/tex]

Now,

[tex]T_1=28 +273 = 301\ K\\\\T_2=58 + 273 = 331\ K\\\\P_1 = 21.1\ MPa[/tex]

Putting all values in above equation, we get :

[tex]\dfrac{301}{21.1}=\dfrac{331}{P_2}\\\\P_2=\dfrac{331\times 21.1}{301}\ MPa\\\\P_2=23.20\ MPa[/tex]

Therefore, the new pressure will be 23.20 MPa.

Hence, this is the required solution.

Answer:

The torque at the end of the beam is 2.5 Nm

Explanation:

Given;

length of beam, r = 0.5 m

applied force, F = 5 N

The torque at the end of the beam is given by;

τ = F x r

where;

τ  is the torque

F is applied force

r is length of the beam

τ = 5 x 0.5

τ = 2.5 Nm

Therefore, the torque at the end of the beam is 2.5 Nm

Answer:

the first one is the correct answer

Answer:

the first one would be correct

Explanation:

Complete question is;

Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a 100 mm thick plate (ρ = 7830 kg/m3, Cp = 550 J/kg K, k = 48 W/m K). The plate initially is at 200 °C and is to be heated to a minimum temperature of 550 °C. Heating is effected in a gas-fired furnace where the products of combustion at T∞ = 800 °C maintain a convection heat transfer coefficient of h = 250 W/m.K on both surfaces of the plate. How long should the plate be left in the furnace?

Answer:

860 seconds

Explanation:

We are given;

Initial Temperature; Ti = 200 °C

Minimum Temperature; T_i = 550 °C

T∞ = 800 °C

convection coefficient; h = 250 W/m².K

ρ = 7830 kg/m³

Cp = 550 J/kg K

k = 48 W/m K

Plate thickness = 100mm

Thus,L = 100/2 = 50mm = 0.05 m

Let's find the biot number from the formula;

Bi = hL/K

Bi = (250 × 0.05)/48

Bi = 0.2604

Now, lowest temperature in the slab is given as;

θ_o = (T_o - T∞)/(T_i - T∞)

θ_o = (550 - 800)/(200 - 800)

θ_o = 0.4167

Now, from online tables calculation, we can find the root of the biot number.

Thus, root of the biot number Bi = 0.2604 is;

ζ1 = 0.488 rad

Also, C1 is gotten as 1.0396

Now,formula for thermal diffusivity is;

α = k/ρc

α = 48/(7830 × 550)

α = 1.115 × 10^(-5) m²/s

Also, from online tables, f(ζ1) = 0.401

Thus, we can find the time the plate should the plate be left in the furnace from;

-(ζ1)²(αt/L²) = In 0.401

-(ζ1)²(αt/L²) = -0.9138

t = (-0.9138 × 0.05²)/-(0.488² × 1.115 × 10^(-5))

t ≈ 860 s

In this exercise we want to calculate the time, in seconds, of the time left on the plate in the furnace, like this:

860 seconds

Organizing the information given in the statement we find that:

Using the formula given below, we will have how to calculate the number of biot, like this:

[tex]B = hL/K\\B = (250 * 0.05)/48\\B = 0.2604[/tex]

calculating the angle that corresponds to the temperature difference as:

[tex]\theta_o = (T_o - T)/(T_i - T)\\\theta_o = (550 - 800)/(200 - 800)\\\theta_o = 0.4167[/tex]

Using the formula below, we will have:

[tex]\alpha = k/\rho c\\\alpha = 48/(7830 * 550)\\\alpha = 1.115 * 10^{-5}[/tex]

Combining all the information from the previous calculations, we have that the time will be calculated as:

[tex]-(\phi)^2(\alpha t/L^2) = In 0.401\\-(\phi )^2(\apha t/^2) = -0.9138\\t = (-0.9138 * 0.05^2)/-(0.488^2 * 1.115 * 10^{-5})\\t = 860 s[/tex]

See more about time at brainly.com/question/2570752

Answer:

wat?

Explanation:

The question should be labeled in the psychology section, but what I assume the question means is some sort of paradoxical reverse psychology method of braincell loss. Even photosynthesis doesn't understand what it means. The question probably is what is the question, not that there's no help.

Hope this helps (a little)!

Answer:

A. 72.53 mol/min

B. 76.2%

Explanation:

Please check attachment to see the diagram of the components I drew.

I first converted flow rate to a molar flow rate. I got 11.5mol/ min after converting. Check attachment

A.) When we create a balance on all these components

n1 + n2 + n3

n1 = n2 + 11.5

Creating a balance on hexane

n1y1C6H14 = n2y2C6H14 + n3y3C6H14

n1(0.18) = n2(0.05) + 11.5mol/min(1.00)

When we put the first equation into the second one:

(n2+11.5mol/min)0.18 + n2(0.05) + 11.5mol/min(1.00)

= 0.18n2 + 2.07min/mol = 0.5n2 + 11.5mol/min

Collect like terms

0.18n2 - 0.05n2 = 11.5 - 2.07min/mol

= 0.13n2 = 9.43

n2 = 9.43/0.13

n2 = 72.53 mol/min

72.53 is the flow rate of the gas stream leaving the burner.

B.) n1 = n2 + 11.5mol/min

n1 = 72.53mol/min + 11.5mol/min

n1 = 84.93mol/min

Amount of hexane entering condenser

n1c6H14 = y1C6H14 = 0.18(84.03mol/min)

= 15.1mol/min

The percentage condensed

= (11.5mol/min)/(15.1mol/min) * 100

= 76.2% of hexane is recovered as liquid

Answer:

  f(x) = x -3

Explanation:

Solve for y, then replace y by f(x).

  x - y = 3

  x - 3 = y . . . . . add y-3 to both sides

  f(x) = x -3

Answer:

(x) = x – 3

Explanation:

edge

Answer:

ben black diyorum ve Iyi srdsler dilerim

#elmaliturta23

Answer:

$1,783,805

Explanation:

Okay, if we are going to answer thus question efficiently, we need to take time to understand each and every sentence in the question. Hence, from the question we have the following data or information and they are;

1. The initial deposit in which Juan deposited on his 26th birthday = $7500.

2. After the statement in [1] above, Juan decided to be depositing $1000 more than the one he deposited on his 26th birthday.

Therefore, we are given that there is 5% in the interest rate. Thus, after his thirty-fifth deposit he will have;

$7500[90.3203] + [5.516] × $1000 × 200.581] = $1,783,805

Answer:

Shallow foundations, often called footings, are usually embedded about a metre or so into soil. ... Another common type of shallow foundation is the slab-on-grade foundation where the weight of the structure is transferred to the soil through a concrete slab placed at the surface.

Explanation:

Because I said so.

Answer:

The steady rate of heat transfer through the glass window is 707.317 watts.

Explanation:

A figure describing the problem is included below as attachment. From First Law of Thermodynamics we get that steady rate of heat transfer through the glass window is the sum of thermal conductive and convective heat rates, all measured in watts:

[tex]\dot Q_{total} = \dot Q_{cond} + \dot Q_{conv, in} + \dot Q_{conv, out}[/tex] (Eq. 1)

Given that window is represented as a flat element, we can expand (Eq. 1) as follows:

[tex]\dot Q_{total} = \frac{T_{i}-T_{o}}{R}[/tex] (Eq. 2)

Where:

[tex]T_{i}[/tex], [tex]T_{o}[/tex] - Indoor and outdoor temperatures, measured in Celsius.

[tex]R[/tex] - Overall thermal resistance, measured in Celsius per watt.

Now, we know that glass window is configurated in series and overall thermal resistance is:

[tex]R = R_{cond} + R_{conv, in}+R_{conv, out}[/tex] (Eq. 3)

Where:

[tex]R_{cond}[/tex] - Conductive thermal resistance, measured in Celsius per watt.

[tex]R_{conv, in}[/tex], [tex]R_{conv, out}[/tex] - Indoor and outdoor convective thermal resistances, measured in Celsius per watt.

And we expand the expression as follows:

[tex]R = \frac{l}{k\cdot w\cdot d} + \frac{1}{h_{i}\cdot w\cdot d} + \frac{1}{h_{i}\cdot w\cdot d}[/tex]

[tex]R = \frac{1}{w\cdot d}\cdot \left(\frac{l}{k}+\frac{1}{h_{i}}+\frac{1}{h_{o}} \right)[/tex] (Eq. 4)

Where:

[tex]w[/tex] - Width of the glass window, measured in meters.

[tex]d[/tex] - Length of the glass window, measured in meters.

[tex]l[/tex] - Thickness of the glass window, measured in meters.

[tex]k[/tex] - Thermal conductivity, measured in watts per meter-Celsius.

[tex]h_{i}[/tex], [tex]h_{o}[/tex] - Indoor and outdoor convection coefficients, measured in watts per square meter-Celsius.

If we know that [tex]w = 2.4\,m[/tex], [tex]d = 1.5\,m[/tex], [tex]l = 0.006\,m[/tex], [tex]k = 0.78\,\frac{W}{m\cdot ^{\circ}C}[/tex], [tex]h_{i} = 10\,\frac{W}{m^{2}\cdot ^{\circ}C}[/tex] and [tex]h_{o} = 25\,\frac{W}{m^{2}\cdot ^{\circ}C}[/tex], the overall thermal resistance is:

[tex]R = \left[\frac{1}{(2.4\,m)\cdot (1.5\,m)}\right] \cdot \left(\frac{0.006\,m}{0.78\,\frac{W}{m\cdot ^{\circ}C} }+\frac{1}{10\,\frac{W}{m^{2}\cdot ^{\circ}C} }+\frac{1}{25\,\frac{W}{m^{2}\cdot ^{\circ}C} } \right)[/tex]

[tex]R = 0.041\,\frac{^{\circ}C}{W}[/tex]

Now, we obtain the steady rate of heat transfer from (Eq. 2): ([tex]R = 0.041\,\frac{^{\circ}C}{W}[/tex], [tex]T_{i} = -5\,^{\circ}C[/tex], [tex]T_{o} = 24\,^{\circ}C[/tex])

[tex]\dot Q_{total} = \frac{24\,^{\circ}C-(-5\,^{\circ}C)}{0.041\,\frac{^{\circ}C}{W} }[/tex]

[tex]\dot Q_{total} = 707.317\,W[/tex]

The steady rate of heat transfer through the glass window is 707.317 watts.

Types; 2 Auto body repair; 3 Auto glass repair; 4 See also; 5 References ... Independent automobile repair shops in the US may also achieve certification through manufacturer sponsored ...