Answer:
the volume of water that will be required to bring these soils to the optimum moisture content is 1859 kL
Explanation:
Given that;
volume of cut = 25,100 m³
Volume of dry soil fill = 23,300 m³
Weight of the soil will be;
⇒ 93% × 18.3 kN/m³ × 23,300 m³
= 0.93 × 426390 kN 3
= 396,542.7 kN
Optimum moisture content = 12.9 %
Required amount of moisture = (12.9 - 8.3)% = 4.6 %
So,
Weight of water required = 4.6% × 396,542.7 = 18241 kN
Volume of water required = 18241 / 9.81 = 1859 m³
Volume of water required = 1859 kL
Therefore, the volume of water that will be required to bring these soils to the optimum moisture content is 1859 kL
Answer:
Trojan horse
Explanation:
A trojan horse attack is a type of malware that misleads users, as it appears unsuspicious at first, but actually presents a threat to the user. A common example is that of an email that contains a malicious attachment. Another common example is that of a fake advertisement. The name comes from the Greek story of the Trojan horse that led to the fall of the city of Troy.
Answer:
over a rive or fast moving water or canyon
Explanation: you would use a suspension bridge in an area where you can't put supports down.
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?
Answer:
3.03 V 0.184 W
2.499 mV 125*10^-9 W
Explanation:
First, apply voltage-divider principle to the input circuit: 1
*5
= 4.545 V
The voltage produced by the voltage-controlled source is:
A_voc*V_i = 4.545 V
We can find voltage across the load, again by using voltage-divider principle:
V_o = A_voc*V_i*(R_o/R_l+R_o)
= 4.545*(100/100+50)
= 3.03 V
Now we can determine delivered power:
P_L = V_o^2/R_L
= 0.184 W
Apply voltage-divider principle to the circuit:
V_o = (R_o/R_o+R_s)*V_s
= 50/50+100*10^3*5
= 2.499 mV
Now we can determine delivered power:
P_l = V_o^2/R_l
= 125*10^-9 W
Delivered power to the load is significantly higher in case when we used amplifier, so a unity gain amplifier can be useful in situation when we want to deliver more power to the load. It is the same case with the voltage, no matter that we used amplifier with voltage open-circuit gain of unity.
Answer:
False
Explanation:
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