Answer:
E = 7333.33 mm
Explanation:
The annual evapotranspiration (E) amount can be calculated using the water budget equation:
(1)
<u>Where</u>:
<em>P: is the precipitation = 2500 mm, </em>
<em>Q(in): is the water flow into the river of the farmland = 5 m³/s, </em>
<em>ΔS: is the change in water storage = 2.5x10⁶ m³, </em>
<em>Q(out): is the water flow out of the river of the farmland = 4 m³/s.</em>
<em>Δt: is the time interval = 1 year = 3.15x10⁷ s </em>
<em>A: is the surface area of the farmland = 6.0x10⁶ m² </em>
Solving equation (1) for ET we have:
Therefore, the annual evapotranspiration amount is 7333.33 mm.
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Answer:
Expression for output voltage = V = Vo sin wt
Explanation:
Amplitude of output voltage = Vo = 19 V
Frequency = f = 240 Hz
Now we know that
Angular frequency = w = 2 π (1 / T) = 2 π 240 = 480 π Hz
Expression for output voltage = V = Vo sin wt
Expression for output voltage = V = 19 (sin 480 π) t
Answer:
A calliper is a device used to measure the dimensions of an object.
Many types of calipers permit reading out a measurement on a ruled scale, a dial, or a digital display. Some calipers can be as simple as a compass with inward or outward-facing points, but no scale.
I'm not sure what a "Vapier" Calliper is, but I assume it's about the same thing.
Answer:
The total system active power P = P_1 + P_2 + P_3 = 34.91 KW
Explanation:
Load 1: Active power P_1 = 20 HP = 14.91kW;
Reactive power 
Load 2: Active power P_2 = 20 kW;
Reactive power Q2 = 0 since the load is purely resistive.
Load 3: Active power 
= 0 due to purely capacitiveload
Reactive power 
= -20 Var
a) since all three loads are connected in parallel therefore
The total system active power P = P_1 + P_2 + P_3 = 34.91 KW
Total system reactive power Q = Q_1 + Q_2 + Q_3 = 11.18 + 0 -20 = -8.82 kVar
Since Q = 0, the power factor is unity.
Supply current per phase is given by
