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3 years ago

"Write a statement that outputs variable numItems. End with a newline. Program will be tested with different input values."

Engineering

1 answer:

kirill [66]3 years ago

6 0

Answer:

The solution code is written in Java.

System.out.println(numItems);

Explanation:

Java println() method can be used to display any string on the console terminal. We can use println() method to output the value held by variable numItems. The numItems is passed as the input parameter to println() and this will output the value of numItems to console terminal and at the same time the output with be ended with a newline automatically.

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Describe the three primary duties for engineer

Wewaii [24]

Answer:Prepare plans with detailed drawings that include project specifications and cost estimates.

Design and execute engineering experiments to create workable solutions.

Develop engineering calculations, diagrams and technical reports.

Explanation:

4 0

3 years ago

6. After removing the vacuum hose from the fuel-pressure regulator, you see gasoline has dripped out of the hose. Technician A s

lubasha [3.4K]

Answer:

Technician B is correct.

Explanation:

7 0

3 years ago

A four-cylinder, four-stroke internal combustion engine operates at 2800 RPM. The processes within each cylinder are modeled as

marysya [2.9K]

Answer:

1) 287760.4 Hp

2) 18410899.5 kPa

Explanation:

The parameters given are;

p₁ = 14.7 lbf/in² = 101325.9 Pa

v₁ = 0.0196 ft³ = 0.00055501 m³

T₁ = 80°F = 299.8167 K

k = 1.4

Assumptions;

1) Air standard conditions are appropriate

2) There are negligible potential and kinetic energy changes

3) The air behaves as an ideal gas and has constant specific heat capacities of temperature and pressure

1) Process 1 to 2

Isentropic compression

T₂/T₁ = (v₁/v₂)^(1.4 - 1) = 10^0.4

p₂/p₁ = (v₁/v₂)^(1.4)

p₂ = p₁×10^0.4 = 101325.9*10^0.4 = 254519.153 Pa

T₂ = 299.8167*10^0.4 = 753.106 K

p₃ = 1080 lbf/in² = 7,446,338 Pa

Stage 2 to 3 is a constant volume process

p₃/T₃ = p₂/T₂

7,446,338/T₃ = 254519.153/753.106

T₃ = 7,446,338/(254519.153/753.106) = 22033.24 K

T₃/T₄ = (v₁/v₂)^(1.4 - 1) = 10^0.4

T₄ = 22033.24/(10^0.4) = 8771.59 K

The heat supplied, Q₁ = cv(T₃ - T₂) = 0.718*(22033.24 -753.106) = 15279.14 kJ

The heat rejected = cv(T₄ - T₁) = 0.718*(8771.59 - 299.8167) = 6082.73 kJ

W(net) = The heat supplied - The heat rejected = (15279.14 - 6082.73) = 9196.41 kJ

The power = W(net) × RPM/2*1/60 = 9196.41 * 2800/2*1/60 = 214582.9 kW

The power by the engine = 214582.9 kW = 287760.4 Hp

2) The mean effective pressure, MEP = W(net)/(v₁ - v₂)

v₁ = 0.00055501 m³

v₁/v₂ = 10

v₂ = v₁/10 = 0.00055501/10 = 0.000055501

MEP = 9196.41/(0.00055501 - 0.000055501) = 18410899.5 kPa

4 0

3 years ago

A rectangular concrete (n=0.013) channel 20 feet wide, on a 2.5% slope, is discharging 400 ft3 /sec into a stilling basin. The s

Sonbull [250]

Answer:

Length of stilling basin = 32.9 feet

Height of end sill = 6.58 feet

Explanation:

Discharge = Q = 400 ft^3 /sec

Slope = 2.5 ft

Width = 20 feet

n = 0.013

we will assume the depth of flow as "d"

Q = 1/n (R)^2/3 (slope)^1/2 A ( here R is the hydraulic Radius)

by substituting the given data in above formula we get:

400 = 1/0.013 * (R)^2/3 * sqrt (2.5/100) * 20d

R = A/P

here, A is the flow area and P is the wetted perimeter

400*0.013 = (20d/(20+20d))^2/3 * sqrt(2.5/100) * 20d

d = 1.42 feet

Depth of stilling channel before the jump will be = d1 = 8 feet

Depth of stilling channel after the jump will be = d2 = 1.42 feet

Length of stilling basin = 5(d2 - d1)

= 5( 8 - 1.42)

Length of stilling basin = 32.9 feet

Now calculating the height of end sill:

Jump height = (8 - 1.42)

Height of end sill = 6.58 feet

8 0

3 years ago

What is the meaning of 52k?

amm1812

Answer:

52K is a form of money or amount. It basically means 52,000.

Explanation:

K stands for thousand.

4 0

3 years ago