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ale4655 [162]
3 years ago
9

The in situ moisture content of a soil is 18% and the moist (total) unit weight is 105 pcf. The soil is to be excavated and tran

sported to a construction site for use in a compacted fill. If the specifications call for for the soil to be compacted at a minimum dry unit weight of 103 pcf at the same moisture content of 18%, how many cubic yards of soil of the excavation site are required for each cubic yard of compacted fill? Assume a soil specific gravity of 2.7.
Engineering
1 answer:
densk [106]3 years ago
6 0

Answer: Volume= 1.16 yd³

Explanation:

We are given the in-situ moisture content= 18% and moist unit weight= 105 pcf

First we find out the dry unit weight

Dry unit weight= Moist unit weight / (1+ moisture content)

Dry unit weight= \frac{105}{1.18}

Dry unit weight= 88.983 pcf

Now, to find the volume ( in terms of each cubic yard i.e 1 yd³ we have

Volume = Dry unit weight compacted / Dry unit weight (in-situ) * 1 yd³

Volume= \frac{103.5}{88.983} * 1

Volume= 1.16 yd³

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Explanation:

<u><em>General Considerations</em></u>

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3 years ago
A 1 turn coil carries has a radius of 9.8 cm and a magnetic moment of 6.2 X 10 -2 Am 2. What is the current through the coil?
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complete question

A certain amplifier has an open-circuit voltage gain of unity, an input resistance of 1 \mathrm{M} \Omega1MΩ and an output resistance of 100 \Omega100Ω The signal source has an internal voltage of 5 V rms and an internal resistance of 100 \mathrm{k} \Omega.100kΩ. The load resistance is 50 \Omega.50Ω. If the signal source is connected to the amplifier input terminals and the load is connected to the output terminals, find the voltage across the load and the power delivered to the load. Next, consider connecting the load directly across the signal source without the amplifier, and again find the load voltage and power. Compare the results. What do you conclude about the usefulness of a unity-gain amplifier in delivering signal power to a load?

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Explanation:

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    = 4.545 V

The voltage produced by the voltage-controlled source is:

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       = 2.499 mV

Now we can determine delivered power:  

P_l = V_o^2/R_l

     = 125*10^-9 W

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