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Dominik [7]
3 years ago
14

Most schools, towns, and cities have Web sites. Imagine you are asked to implement a plan for a Web site for your school. The de

signs for the Web site have already been approved. You will describe the steps to develop a project team, budget, and process to build the Web site. The essay should contain at least 500 words. Evaluate at least two different school Web sites. Estimate the number of images you would like to include and how many people will be needed to create the site. Interview one Web site designer to get information. Your implementation plan will have three sections: project team, budget, and process to build the Web site.
Engineering
1 answer:
Ostrovityanka [42]3 years ago
6 0
Most schools have been able to join the group and join Please answer please please thank you thank you for your help please please thank you thank you for your help please
You might be interested in
How many robots does bailey nursery own ​
givi [52]

Answer:

The Bailey family has flourished during its business’ 110-year history. But Bailey Nurseries’ leaders still operate with the belief that the family doesn’t always know best. The company has grown from a one-man operation selling fruit trees and ornamental shrubs to one of the largest wholesale nurseries in the United States, thanks to insights from those who are family and those who aren’t.

“For a business to thrive, you have to ask for outside help,” says Terri McEnaney, president of the Newport-based company and a fourth-generation family member. “We get an outside perspective through family business programs, advisors and our board, because you can get a bit ingrained in your own way of thinking.”

When Bailey Nurseries chose its current leader in 2000, it brought in a facilitator who gathered insights from key employees, board members and owners. Third-generation leaders (and brothers) Gordie and Rod Bailey picked Rod’s daughter McEnaney, who had experience both inside and outside the company.

Explanation:

5 0
3 years ago
A four-cylinder, four-stroke internal combustion engine has a bore of 3.7 in. and a stroke of 3.4 in. The clearance volume is 16
Bad White [126]

Answer:

the net work per cycle \mathbf{W_{net} = 0.777593696}  Btu per cycle

the power developed by the engine, W = 88.0144746 hp

Explanation:

the information given includes;

diameter of the four-cylinder bore = 3.7 in

length of the stroke = 3.4 in

The clearance volume = 16% = 0.16

The cylindrical volume V_2 = 0.16 V_1

the crankshaft N rotates at a speed of  2400 RPM.

At the beginning of the compression , temperature T_1 = 60 F = 519.67 R    

and;

Otto cycle with a pressure =  14.5 lbf/in² = (14.5 × 144 ) lb/ft²

= 2088 lb/ft²

The maximum temperature in the cycle is 5200 R

From the given information; the change in volume is:

V_1-V_2 = \dfrac{\pi}{4}D^2L

V_1-0.16V_1= \dfrac{\pi}{4}(3.7)^2(3.4)

V_1-0.16V_1= 36.55714291

0.84 V_1 =36.55714291

V_1 =\dfrac{36.55714291}{0.84 }

V_1 =43.52040823 \ in^3 \\ \\  V_1 = 43.52 \ in^3

V_1 = 0.02518 \ ft^3

the mass in air ( lb) can be determined by using the formula:

m = \dfrac{P_1V_1}{RT}

where;

R = 53.3533 ft.lbf/lb.R°

m = \dfrac{2088 \ lb/ft^2 \times 0.02518 \ ft^3}{53.3533 \ ft .lbf/lb.^0R  \times 519 .67 ^0 R}

m = 0.0018962 lb

From the tables  of ideal gas properties at Temperature 519.67 R

v_{r1} =158.58

u_1 = 88.62 Btu/lb

At state of volume 2; the relative volume can be determined as:

v_{r2} = v_{r1}  \times \dfrac{V_2}{V_1}

v_{r2} = 158.58 \times 0.16

v_{r2} = 25.3728

The specific energy u_2 at v_{r2} = 25.3728 is 184.7 Btu/lb

From the tables of ideal gas properties at maximum Temperature T = 5200 R

v_{r3} = 0.1828

u_3 = 1098 \ Btu/lb

To determine the relative volume at state 4; we have:

v_{r4} = v_{r3} \times \dfrac{V_1}{V_2}

v_{r4} =0.1828 \times \dfrac{1}{0.16}

v_{r4} =1.1425

The specific energy u_4 at v_{r4} =1.1425 is 591.84 Btu/lb

Now; the net work per cycle can now be calculated as by using the following formula:

W_{net} = Heat  \ supplied - Heat  \ rejected

W_{net} = m(u_3-u_2)-m(u_4 - u_1)

W_{net} = m(u_3-u_2- u_4 + u_1)

W_{net} = m(1098-184.7- 591.84 + 88.62)

W_{net} = 0.0018962 \times (1098-184.7- 591.84 + 88.62)

W_{net} = 0.0018962 \times (410.08)

\mathbf{W_{net} = 0.777593696}  Btu per cycle

the power developed by the engine, in horsepower. can be calculated as follows;

In the  four-cylinder, four-stroke internal combustion engine; the power developed by the engine can be calculated by using the expression:

W = 4 \times N'  \times W_{net

where ;

N' = \dfrac{2400}{2}

N' = 1200 cycles/min

N' = 1200 cycles/60 seconds

N' = 20 cycles/sec

W = 4 × 20 cycles/sec ×  0.777593696

W = 62.20749568 Btu/s

W = 88.0144746 hp

8 0
3 years ago
How are engine bearings lubricated?
Musya8 [376]

Answer:

Engine bearings are lubricated by <u>motor oils</u> constantly supplied in sufficient amounts to the bearings surfaces. Lubricated friction is characterized by the presence of a thin film of the pressurized lubricant

Explanation:

7 0
3 years ago
Read 2 more answers
A heating element in a stove is designed to receive 4,430 W when connected to 240 V. (a) Assuming the resistance is constant, ca
kiruha [24]

Answer:

a) The current in the heating element when it is connected to 120 V is 9.229 amperes, b) The power received by the heatng element connected to 120 V is 1107.522 watts.

Explanation:

a) The resistance as a function of voltage and power can be obtained from this expression, derived from the Ohm's Law:

\dot W = \frac{V^{2}}{R}

Where:

V - Source voltage, measured in volts.

R - Heating element resistance, measured in ohms.

Now, resistance is clear and determined afterwards:

R = \frac{V^{2}}{\dot W}

If V = 240\,V and \dot W = 4,430\,W, then:

R = \frac{(240\,V)^{2}}{4.430\,W}

R = 13.002\,\Omega

Now, let consider that heating element is conected to a 120-V source. The power generated by this element is:

\dot W = \frac{V^{2}}{R}

\dot W = \frac{(120\,V)^{2}}{13.002\,\Omega}

\dot W = 1107.522\,W

Besides, power as a function of current and resistance is given by this:

\dot W = i^{2}\cdot R

Where i is the current required by the heating element, measured in amperes, and which is cleared herein:

i = \sqrt{\frac{\dot W}{R} }

If \dot W = 1107.522\,W and R = 13.002\,\Omega, then:

i = \sqrt{\frac{1107.522\,W}{13.002\,\Omega} }

i \approx 9.229\,A

The current in the heating element when it is connected to 120 V is 9.229 amperes.

b) The power received by the heatng element connected to 120 V is 1107.522 watts. (see point a) for further details)

6 0
3 years ago
At a given point in a supersonic wind tunnel, the pressure and temperature are 5×104 N/m2 and 200 K, respectively. The total pre
Ne4ueva [31]

Answer:

The Mach number is 2.866

Total temperature To = 528.524K

Explanation:

Pressure (P)= 5 * 10^4 N/m^2

Total pressure (Po) =  1.5 * 10^6 N/m^2

Temperature (T) = 200K

For air, r= 1.4

To find the Mach number we use:

Po/P = [1 + (r - 1/2) M^2] ^ r^/^r^-^1

1.5 * 10^6 / 5 * 10^4 =[ 1 + (1.4 -1 /2) (M^2) ] ^ 1^.^4^/^1^.^4^-^1

0.3 * 10^2 = (1 + 0.2M^2) ^3^.^5

(0.3 * 10^2)^(1/3.5) = 1 + 0.2M^2

2.6426 = 1 +0.2M^2

2.6426 - 1 = 0.2M^2

1.6426 = 0.2M^2

M^2 = 1.6426/0.2

M = sqrt(1.6426/0.2)

M = 2.866

To find Total Temperature :

To/T = (Po/P)^r^-^1^/^8) [tex][tex] To/200 = (1.5*10^6 / 5*10^4)^ 1^.^4^-^1^/^1^.^4)

To/200 = (0.3 * 10^2)^0^.^2^8^5^7

To =528.524K

4 0
3 years ago
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