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Base course aggregate has a target density of 121.8 lb/ft3 in place It will be laid down and compacted in a rectangular street r

irakobra [83]

Answer:

total weight of the aggregate = 3594878.28 lbs

Explanation:

given data

density = 121.8 lb/ft³

area = 1000 ft x 60 ft x 6 in = 1000 ft x 60 ft x 0.5 ft

moisture = 3.5 %

compaction = 95%

solution

we get here first volume of the space that is filled with the aggregate that is

volume = 1000 ft x 60 ft x 0.5 ft = 30,000 cu ft

now we get fill space with aggregate that compact to 95% of dry density.

so we fill space with aggregate of density that is = 95% of 121.8

= 115.71 lb/ cu ft

so now dry weight of aggregate is

dry weight of aggregate = 30,000 × 115.71 = 3471300 lb

when we assume that moisture percentage is by weight

then weight of the moisture in aggregate will be

weight of the moisture in aggregate = 3.56 % of 3471300 lb

weight of the moisture in aggregate = 123578.28 lbs

and

we get total weight of the aggregate to fill space that is

total weight of the aggregate = 3471300 lb +123578.28 lb

total weight of the aggregate = 3594878.28 lbs

4 0

3 years ago

- Find the file that has the string "You Got Me" in your home directory. - When you have the correct command to do that, add the

Scilla [17]

Answer:

Explanation: see attachment below

6 0

4 years ago

4. Which of these is typically NOT a function of the flywheel?

Sunny_sXe [5.5K]

Answer:D

Explanation:

6 0

4 years ago

You have an ac voltage of 10 sin(60t). Design a full wave bridge rectifier which will give you an output voltage that varies a m

Masteriza [31]

As there are 10 V, for Vp1, that is the peak-voltage of the source:

$Vp1=10*\sqrt{2}=14.14 V$

Then, transformer's theory says that the relation of transformations is:

V1/V2=a

Where V1 is the voltage in the primary and V2 in the secondary.

V1=14.14 V

V2=8.55 V

a=1.65

Then, with the 8.5 V, we find the real peak-voltage, taking in account that in the diodes we have a drop of 0.7 V each, so:

8.5 -1.4=7.1 V

And this will be called VpL

Now we proceed to calculate the mean voltage:

$V_{mean}=VpL-(\frac{Vr}{2} )$

Where Vr is the ripple voltage, we asume that is 1 V

So, Vmean = 6.6 V

Then we have

Vmean/R= I mean

We have that R=1000 Ohm

Imedia=6.6 V/1000 Ohm

Imedia=6.6 mAmps

Finally, we can calculate the capacitor:

C=Q/Vr

C=Imean/(Vr*2f)

Where f is 60Hz

C=6.6mA/(1V*120)

C=5.5 uFarads

Therefore:

C=5.5 uFarads that works at 12 V

6 0

3 years ago

Calculate the osmotic pressure of seawater containing 3.5 wt % NaCl at 25 °C . If reverse osmosis is applied to treat seawater,

AlladinOne [14]

Answer:

Highest osmotic pressure that membrane may experience is

' =58.638 atm

Explanation:

Suppose sea-water taken is M= 1 kg

Density of water = 1000 kg/m3

Therefore Volume of water= Mass,M/Density of water

V= 1 kg/(1000 kg/m3)

V= 10-3 m3= 1 Litre

Since mass of Nacl is 3.5 wt%,Therefore in 1 kg of water

Mass present of NaCl= m= 0.035*1000 g

m= 35 g

Since molecular weight of NaCl= 58.44 g/mol =M.W.

Thus its Number of moles of Nacl= m/M.W

nNaCl= 35g/58.44 gmol-1

= 0.5989 mol

ans since volume of solution is 1 L thus concentration of NaCl is ,C= number of moles/Volume of solution in Litres

C= 0.5989mol/ 1L

=0.5989 M

Since 1 mol NaCL disssociates to form 2 moles of ions of Na+ andCl- Thus van't hoff factor i=2

And osmotic pressure = iCRT ------------------------------(1)( Where R= 0.0821 L.atm/mol.K and T= 25oC= 298.15 K)

Putting in equation 1 ,we get = 2*(0.5989 mol/L)*(0.0821 L.atm/mol.K)*298.15 K

=29.319 atm

Now as the water gets filtered out of the membrane,the water's volume decreases and concentration C of NacL increases, thus osmotic pressure also increases.Thus, at 50% water been already filtered out, the osmotic pressure at the membrane will be maximum

Thus Volume of water left after 50% is filtered out as fresh water= 0.5 L (assuming no salt passes through semi permeable membrane)

Thus New concentration of NaCl C'= 2*C

C'=2*0.5989 M

=1.1978 M

and Since Osmotic pressure is directly proportional to concentration, Thus As concentration C doubles to C', Osmotic Pressure ' also doubles from ,

Thus,Highest osmotic pressure that membrane may experience is, '=2*

=2*29.319 atm

' =58.638 atm

3 0

3 years ago