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2 years ago

1. Fuel lines are constructed from

1 answer:

8 0

Fuel lines are normally made of double wall steel tubing. For fire safety, a fuel line must be able to withstand the constant and severe vibration produced by the engine and road surface. Lines are placed away from exhaust pipes , mufflers, and manifolds, so that excessive heat will not cause vapor lock.

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Select the correct answer. Jude is a mechanical engineer. He works in the automobile industry. He is creating a prototype of an

Nikolay [14]

Answer:

Bid bureau of investigation

6 0

3 years ago

List 5 characteristics of metal?

RSB [31]

Answer:

- Son buenos conductores de electricidad.

- Excelentes conductores de calor.

- Tienen una alta tenacidad.

- Se encuentran en estado solido, exepto por el mercurio.

- Y se destaca la baja electronegatividad.

3 0

3 years ago

Using Pascal’s Law and a hydraulic jack, you want to lift a 4,000 lbm rock. The large cylinder has a diameter of 6 inches.

jolli1 [7]

Answer:

a diameter of D₂ = 0.183 inches would be required

Explanation:

appyling pascal's law

P applied to the hydraulic jack = P required to lift the rock

F₁*A₁ = F₂*A₂

since A₁= π*D₁²/4 , A₂= π*D₂²/4

F₁*π*D₁²/4 = F₂* π*D₂²/4

F₁*D₁²=F₂*D₂²

D₂ = D₁ *√(F₁/F₂)

replacing values

D₂ = D₁ *√(F₁/F₂) = 6 in * √(120 lbf/(4000 lbm * 32.174 (lbf/lbm)) = 0.183 inches

6 0

4 years ago

An aircraft is flying at 300 mph true airspeed has a 50 mph tailwind. What is its ground speed?

Free_Kalibri [48]

Answer:

304.13 mph

Explanation:

Data provided in the question :

The Speed of the flying aircraft = 300 mph

Tailwind of the true airspeed = 50 mph

Now,

The ground speed will be calculated as:

ground speed = $\sqrt{300^2+50^2}$

The ground speed = $\sqrt{92500}$

The ground speed = 304.13 mph

Hence, the ground speed is 304.13 mph

8 0

4 years ago

Problem 4 You are designing a circuit to drive LED1, using the following circuit. The datasheet for the LED specifies that VF =

HACTEHA [7]

Answer:

a) I_LED= 1/6 A b) Vf= 2.5V

Explanation:

Consider circuit in the attachment.

a) We will simplify current source in paraller with resistor to a voltage source in series with a resistor(see attachment 2)

Solving the circuit in attachment 2 using mesh analysis

-9+2I1+4(I1-I2)-4+2I1=0

8I1 - 4I2= 13 ............... eq 1

4+4(I2-I1)+ I2 + 2=0

4I1- 5I2 = 6 ............ eq 2

I1= 41/24 ; I2 = 1/6; I2= I_LED

b) Solving the circuit in attachment 2 again, this time I2=0

8I1 - 4I2= 13

8I1- 4(0)=13

I1= 13/8

Vf= 4(I1- I2) -4

I2=I_LED=0

Vf= 2.5 V

4 0

4 years ago