Answer:
Explanation:
<u><em>General Considerations</em></u>
The design of the yard will affect the natural surface and subsurface drainage pattern of a watershed or individual hillslope. Yard drainage design has as its basic objective the reduction or elimination of energy generated by flowing water. The destructive power of flowing water increases exponentially as its velocity increases. Therefore, water must not be allowed to develop sufficient volume or velocity so as to cause excessive wear along ditches, below culverts, or along exposed running surfaces, cuts, or fills.
A yard drainage system must satisfy two main criteria if it is to be effective throughout its design life:
1. It must allow for a minimum of disturbance of the natural drainage pattern.
2.It must drain surface and subsurface water away from the roadway and dissipate it in a way that prevents excessive collection of water in unstable areas and subsequent downstream erosion
The diagram below ilustrate diffrent sturcture of yard to be consider before planing to utiliza rainwater
Answer:
The current through the coil is 2.05 A
Explanation:
Given;
number of turns of the coil, N = 1
radius of the coil, r = 9.8 cm = 0.098 m
magnetic moment of the coil, P = 6.2 x 10⁻² A m²
The magnetic moment is given by;
P = IA
Where;
I is the current through the coil
A is area of the coil = πr² = π(0.098)² = 0.03018 m²
The current through the coil is given by;
I = P / A
I = (6.2 x 10⁻² ) / (0.03018)
I = 2.05 A
Therefore, the current through the coil is 2.05 A
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.