In which order are the following measuring instruments listed from lowest precision to highest precision:

An aggregate blend is composed of 65% coarse aggregate by weight (Sp. Cr. 2.635), 36% fine aggregate (Sp. Gr. 2.710), and 5% fil

den301095 [7]

Answer:

Explanation:

From the given information:

Addition of all the materials = 65+ 36+ 5 +6 = 112 which is higher than 100 percentage; SO we need to find;

The actual percentage of each material which can be determined as follows:

Percentage of the coarse aggregate will be = 65 × 112/100

= 72.80%

Percentage of the Fine aggregate will be = 36 × 112/100

= 40.32%

Percentage of the filler will be = 5 × 112/100

= 5.6%

Percentage of the asphalt binder will be = 6 × 112/100

= 6.72 %

So; the theoretical specific gravity (Gt) of the mixture can be calculated as follows:

Gt = 100/( 72.80/2.635 + 40.32/2.710 + 5.6/2.748 + 6.72/1.088)

Gt = 100/( 27.628 + 14.878 + 2.039 + 6.177)

Gt = 100/ (50.722)

Gt =1.972

Also;Given that the bulk density = 143.9 lb/ft³

LIke-wsie ; as we know that unit weight of water is =62.43lb/cu.ft

Hence, the bulk specific gravity of the mix (Gm) = 143.9/62.43

=2.305

The percentage of air void = (Gt -Gm )× 100/ Gt

= (1.972 - 2.305) × 100/ 1.972

= -16.89%

The percentage of the asphalt binder is =(6.72/1.088*100)/(72.80+40.32+5.6+6.72)/2.305)

= 617.647/54.42

= 11.35%

Thus; the percentage voids in mineral aggregate = -16.89% + 11.35%

the percentage voids in mineral aggregate = -5.45%

The percent voids filled with asphalt. = 100 × 11.35/-5.45

The percent voids filled with asphalt = - 208.26 %

7 0

3 years ago

A horizontal curve on a two-lane road is designed with a 2,300-ft radius, 12-ft lanes, and a 65-mph design speed. Determine the

Ierofanga [76]

Answer:

distance = 22.57 ft

superelevation rate = 2%

Explanation:

given data

radius = 2,300-ft

lanes width = 12-ft

no of lane = 2

design speed = 65-mph

solution

we get here sufficient sight distance SSD that is express as

SSD = 1.47 ut + $\frac{u^2}{30(\frac{a}{g}\pm G)}$ ..............1

here u is speed and t is reaction time i.e 2.5 second and a is here deceleration rate i.e 11.2 ft/s² and g is gravitational force i.e 32.2 ft/s² and G is gradient i.e 0 here

so put here value and we get

SSD = 1.47 × 65 ×2.5 + $\frac{65^2}{30(\frac{11.2}{32.2}\pm 0)}$

solve it we get

SSD = 644 ft

so here minimum distance clear from the inside edge of the inside lane is

Ms = Rv ( 1 - $cos (\frac{28.65 SSD}{Rv})$ ) .....................2

here Rv is = R - one lane width

Rv = 2300 - 6 = 2294 ft

put value in equation 2 we get

Ms = 2294 ( 1 - $cos (\frac{28.65 \times 664}{2294})$ )

solve it we get

Ms = 22.57 ft

and

superelevation rate for the curve will be here as

R = $\frac{u^2}{15(e+f)}$ ..................3

here f is coefficient of friction that is 0.10

put here value and we get e

2300 = $\frac{65^2}{15(e+0.10)}$

solve it we get

e = 2%

3 0

3 years ago

If an object is near the surface of the earth the variation of its weight with distance from the center of the earth can ofteb b

zalisa [80]

Answer:

h=32.1 km

Explanation:

solution:

using newton law of gravitational attraction and newton second law:

$W=\frac{Gmm_{E} }{r^{2} } \\a=\frac{Gm_{E}}{r^{2}} \\W=ma\\$

$m_{E}= mass of earth$

r= distance between two masses

at sea level

a=g

$r=R_{E}$

$a=\frac{Gm_{E}}{r^{2}}$ .............................(1)

$Gm_{E} =gR_{E}^2$ .........................(2)

by substituting (2) and (1) $a=g\frac{R_{E}^2 }{r^{2} }$ acceleration due to gravity at a distance r from the centre of the earth in terms of g (sea level)

so the weight of the object at a distance r from the centre of the earth (W=ma)

W=mg(Re^2/r^2)..........(3)

h the height above the surface of the earth: r=Re+h

putting the value of r in eq (3)

W=mg(Re/Re+h)^2

W=0.99 mg

solving for height h:

h=Re(1/√0.99)-(1))

h=32.1 km

4 0

3 years ago

Atmospheric air at 25 °C and 8 m/s flows over both surfaces of an isothermal (179C) flat plate that is 2.75m long. Determine the

vekshin1

Answer:

Re=100,000⇒Q=275.25 $\frac{W}{m^2}$

Re=500,000⇒Q=1,757.77 $\frac{W}{m^2}$

Re=1,000,000⇒Q=3060.36 $\frac{W}{m^2}$

Explanation:

Given:

For air $T_∞$ =25°C ,V=8 m/s

For surface $T_s$ =179°C

L=2.75 m ,b=3 m

We know that for flat plate

$Re$ ⇒Laminar flow

$Re>30\times10^5$ ⇒Turbulent flow

Take Re=100,000:

So this is case of laminar flow

$Nu=0.664Re^{\frac{1}{2}}Pr^{\frac{1}{3}}$

From standard air property table at 25°C

Pr= is 0.71 ,K=26.24 $\times 10^{-3}$

So $Nu=0.664\times 100,000^{\frac{1}{2}}\times 0.71^{\frac{1}{3}}$

Nu=187.32 ( $\dfrac{hL}{K_{air}}$ )

187.32= $\dfrac{h\times2.75}{26.24\times 10^{-3}}$

⇒h=1.78 $\frac{W}{m^2-K}$

heat transfer rate =h $(T_∞-T_s)$

=275.25 $\frac{W}{m^2}$

Take Re=500,000:

So this is case of turbulent flow

$Nu=0.037Re^{\frac{4}{5}}Pr^{\frac{1}{3}}$

$Nu=0.037\times 500,000^{\frac{4}{5}}\times 0.71^{\frac{1}{3}}$

Nu=1196.18 ⇒h=11.14 $\frac{W}{m^2-K}$

heat transfer rate =h $(T_∞-T_s)$

=11.14(179-25)

= 1,757.77 $\frac{W}{m^2}$

Take Re=1,000,000:

So this is case of turbulent flow

$Nu=0.037Re^{\frac{4}{5}}Pr^{\frac{1}{3}}$

$Nu=0.037\times 1,000,000^{\frac{4}{5}}\times 0.71^{\frac{1}{3}}$

Nu=2082.6 ⇒h=19.87 $\frac{W}{m^2-K}$

heat transfer rate =h $(T_∞-T_s)$

=19.87(179-25)

= 3060.36 $\frac{W}{m^2}$

7 0

3 years ago

What is the difference between a series circuit and a parallel circuit?

serg [7]

Answer:

In a parallel circuit, the voltage across each of the components is the same, and the total current is the sum of the currents flowing through each component. ... In a series circuit, every device must function for the circuit to be complete. If one bulb burns out in a series circuit, the entire circuit is broken.

Explanation:

7 0

3 years ago

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