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2 years ago

How deep does electrical conduit need to be buried?

1 answer:

7 0

In general, bury metal conduits at least 6 inches below the soil surface. You may also run them at a depth of 4 inches under a 4-inch concrete slab. Under your driveway, the conduits must be below a depth of 18 inches, and under a public road or alleyway, they must be buried below 24 inches.

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For unrestrained cube made from linear, isotropic, homogeneous material the temperature increase causes strain in_____ direction

LenKa [72]

Answer: The answer is four; four

Explanation: This is because of the mixture of material used and the number of directions it causes strain I directly proportional to the number of times it causes stress.

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2 years ago

svp [43]

Answer:

серйозних порушень точності,

∵∴⊕∴∵

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2 years ago

An airliner is flying at 34,000 ft cruise altitude on a standard day. Calculate the pressure difference between the cabin and th

nadya68 [22]

Answer:

$\Delta P=61,952.8\ lb/ft^2$

Explanation:

Given

Airline flying at 34,000 ft.

Cabin pressurized to an altitude 8,000 ft.

We know that at standard condition ,density of air

$\rho =0.074\ lb/ft^3$

We know that pressure difference

ΔP=ρ g ΔZ

Here ΔZ=34,000-8,000 ft

ΔZ=26,000 ft

$g= 32.2\ ft/s^2$

ΔP=0.074 x 32.2 x 26,000

$\Delta P=61,952.8\ lb/ft^2$

So pressure difference will be $\Delta P=61,952.8\ lb/ft^2$ .

7 0

3 years ago

Describe the Optical Fiber Network( i need 2 paragraphs, but give me about 5 sentences or like 3)

frez [133]

Answer: Fiber Optic Network Fiber-optic networks have been used for decades to transmit large volumes of traffic across the country. The economics of fiber networks have only recently allowed for connecting the fiber directly to the home, creating a fiber-to-the-home (FTTH) network.

Explanation:

6 0

3 years ago

Read 2 more answers

Race car is accelerating and has a velocity of 10 m/s @ t=0. It completes a lap on a circular track of 400 m in 14 seconds. Calc

wariber [46]

Answer:

component of acceleration are a = 3.37 m/s² and ar = 22.74 m/s²

magnitude of acceleration is 22.98 m/s²

Explanation:

given data

velocity = 10 m/s

initial time to = 0

distance s = 400 m

time t = 14 s

to find out

components and magnitude of acceleration after the car has travelled 200 m

solution

first we find the radius of circular track that is

we know distance S = 2πR

400 = 2πR

R = 63.66 m

and tangential acceleration is

S = ut + 0.5 ×at²

here u is initial speed and t is time and S is distance

400 = 10 × 14 + 0.5 ×a (14)²

a = 3.37 m/s²

and here tangential acceleration is constant

so velocity at distance 200 m

v² - u² = 2 a S

v² = 10² + 2 ( 3.37) 200

v = 38.05 m/s

so radial acceleration at distance 200 m

ar = $\frac{v^2}{R}$

ar = $\frac{38.05^2}{63.66}$

ar = 22.74 m/s²

so magnitude of total acceleration is

A = $\sqrt{a^2 + ar^2}$

A = $\sqrt{3.37^2 + 22.74^2}$

A = 22.98 m/s²

so magnitude of acceleration is 22.98 m/s²

8 0

2 years ago