Which of the following is an essential component of reinforced concrete?

A horizontal curve of a two-lane undivided highway (12-foot lanes) has a radius of 678 feet to the center line of the roadway. A

OLEGan [10]

Answer:

maximum speed for safe vehicle operation = 55mph

Explanation:

Given data :

radius ( R ) = 678 ft

old building located ( m )= 30 ft

super elevation = 0.06

<u>Determine the maximum speed for safe vehicle operation </u>

firstly calculate the stopping sight distance

m = R ( 1 - cos $\frac{28.655*S}{R}$ ) ---- ( 1 )

R = 678

m ( horizontal sightline ) = 30 ft

back to equation 1

30 = 678 ( 1 - cos (28.655 *s / 678 ) )

( 1 - cos (28.655 *s / 678 ) ) = 30 / 678 = 0.044

cos $\frac{28.65 *s }{678}$ = 1.044

hence ; 28.65 * s = 678 * 0.2956

s = 6.99 ≈ 7 ft

next we will calculate the design speed ( u ) using the formula below

S = 1.47 ut + $\frac{u^2}{30(\frac{a}{3.2} )-G1}$ ---- ( 2 )

t = reaction time, a = vehicle acceleration, G1 = grade percentage

assuming ; t = 2.5 sec , a = 11.2 ft/sec^2, G1 = 0

back to equation 2

6.99 = 1.47 * u * 2.5 + $\frac{u^2}{30[(11.2/32.2)-0 ]}$

3.675 u + 0.0958 u^2 - 6.99 = 0

u ( 3.675 + 0.0958 u ) = 6.99

5 0

3 years ago

A sheet of steel 3-mm thick has nitrogen atomospheres on both sides at 900 C and is permitted to achieve a steady-state di usion

kati45 [8]

Answer:

$X_B = 1.8 \times 10^{-3} m = 1.8 mm$

Explanation:

Given data:

Diffusion constant for nitrogen is $= 1.85\times 10^{-10} m^2/s$

Diffusion flux $= 1.0\times 10^{-7} kg/m^2-s$

concentration of nitrogen at high presuure = 2 kg/m^3

location on which nitrogen concentration is 0.5 kg/m^3 ......?

from fick's first law

$J = D \frac{C_A C_B}{X_A X_B}$

Take C_A as point on which nitrogen concentration is 2 kg/m^3

$x_B = X_A + D\frac{C_A -C_B}{J}$

Assume X_A is zero at the surface

$X_B = 0 + ( 12\times 10^{-11} ) \frac{2-0.5}{1\times 10^{-7}}$

$X_B = 1.8 \times 10^{-3} m = 1.8 mm$

4 0

3 years ago

6. Find the heat flow in 24 hours through a refrigerator door 30.0" x 58.0" insulated with cellulose fiber 2.0" thick. The tempe

Ilia_Sergeevich [38]

Answer:

The heat flow in 24 hours through the refrigerator door is approximately 1,608.57 BTU

Explanation:

The given parameters are;

The duration of the heat transfer, t = 24 hours = 86,400 seconds

The area of the refrigerator door, A = 30.0" × 58.0" = 1,740 in.² = 1.122578 m²

The material of the insulator in the door = Cellulose fiber

The thickness of the insulator in the door, d = 2.0" = 0.0508 m

The temperature inside the fridge = 38° F = 276.4833 K

The temperature of the room = 78°F = 298.7056 K

The thermal conductivity of cellulose fiber = 0.040 W/(m·K)

By Fourier's law, the heat flow through a by conduction material is given by the following formula;

$\dfrac{Q}{t} = \dfrac{k \cdot A \cdot (T_2 - T_1) }{d}$

$Q = \dfrac{k \cdot A \cdot (T_2 - T_1) }{d} \times t$

Therefore, we have;

$Q = \dfrac{0.04 \times 1.122578 \times (298.7056 - 276.4833 ) }{0.0508} \times 86,400 =1,697,131.73522$

The heat flow in 24 hours through the refrigerator door, Q = 1,697,131.73522 J = 1,608.5705140685 BTU

7 0

3 years ago

________ is not within other loop.Immersive Reader

djyliett [7]

Answer:

I think node

Explanation:

thank you

have a great day

8 0

3 years ago

Take my points already been warned once

ANTONII [103]

Answer:

bet

Explanation:

give me em

5 0

3 years ago

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