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Alborosie
2 years ago
10

Find the remaining trigonometric functions of 0 if

Engineering
1 answer:
garik1379 [7]2 years ago
8 0

Answer:

cosΘ=−√558

tanΘ=−3√5555

cscΘ=83

secΘ=−8√5555

cotΘ=−√553

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At a retirement party, a coworker described terry as dedicated
denis23 [38]

Answer:

At a retirement party, a coworker described Terry as dedicated, hardworking, and dependable. He also said that Terry was a great leader, knew the computer system, and kept the company's finances in order

8 0
3 years ago
A four-cylinder, four-stroke internal combustion engine operates at 2800 RPM. The processes within each cylinder are modeled as
Ulleksa [173]

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
(Blank) welding involves manual welding with equipment anomalously controls one or more of the windy conditions while (blank) we
shepuryov [24]

Answer:

D.

Explanation:

In automated welding, defined as “welding with equipment that requires only occasional or no observation of the weld, and no manual adjustment of the equipment controls,” the welder's involvement is limited to activating the machine to initiate the welding cycle and observing the weld on an intermittent basis.

8 0
3 years ago
The boy in the wagon begins throwing bricks out of the wagon to simulate rocket propulsion. The wagon begins at rest, and the bo
Digiron [165]

Q:What velocity does the boy attain if he throws the bricks one at a time?

Answer:Linear velocity since it moves back and firth and does not rotate like angular velocity.

5 0
3 years ago
A hydrauliic jack is rated at 5000 pound capacity. The area of the large piston on the jack is 4.45 Square inches. Calculate the
sergeinik [125]

Answer:

1123.6 pounds/ square inch.

Explanation:

Fluid pressure is the ratio of force or weight applied by the fluid per unit area.

i.e Fluid pressure = \frac{weight}{area}

The maximum load of the jack is obtained at its maximum capacity = 5000 pounds

Area of the large piston on the jack = 4.45 square inches

Thus,

Fluid pressure = \frac{5000}{4.45}

                        = 1123.5955

Fluid pressure = 1123.6 pounds/ square inch

Thu, the fluid pressure in the jack at maximum load is 1123.6 pounds/ square inch.

7 0
2 years ago
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