How does a rudder help maneuver an airplane?

A 600-ha farmland receives annual rainfall of 2500 mm. There is a river flowing through the farmland with an inflow rate of 5 m3

shepuryov [24]

Answer:

E = 7333.33 mm

Explanation:

The annual evapotranspiration (E) amount can be calculated using the water budget equation:

$P*A + Q_{in}*\Delta t = E*A + \Delta S + Q_{out}*\Delta t$ (1)

Where:

P: is the precipitation = 2500 mm,

Q(in): is the water flow into the river of the farmland = 5 m³/s,

ΔS: is the change in water storage = 2.5x10⁶ m³,

Q(out): is the water flow out of the river of the farmland = 4 m³/s.

Δt: is the time interval = 1 year = 3.15x10⁷ s

A: is the surface area of the farmland = 6.0x10⁶ m²

Solving equation (1) for ET we have:

$E = \frac{P*A + Q_{in}*\Delta t - \Delta S - Q_{out}*\Delta t}{A}$

$E = \frac{2.5 m \cdot 6.0 \cdot 10^{6} m^{2} + 5 m^{3}/s \cdot 3.15 \cdot 10^{7} s - 2.5 \cdot 10^{6} m^{3} - 4 m^{3}/s \cdot 3.15 \cdot 10^{7} s}{6.0\cdot 10^{6} m^{2}}$

$E = 7333.33 mm$

Therefore, the annual evapotranspiration amount is 7333.33 mm.

I hope it helps you!

3 0

4 years ago

Technician A says that much of the lighter GMAW welding work in a body shop can be done with a 115-volt welder. Technician B say

Alina [70]

Answer:

c. Both Technicians A and B

8 0

3 years ago

Suppose an underground storage tank has been leaking for many years, contaminating a groundwater and causing a contaminant conce

Ghella [55]

Answer: the cancer risk for an adult male drinking the water for 10 years is 2.88 × 10⁻⁶

Explanation:

firstly, we find the time t required to travel for the contaminant to the well;

Given that, contamination flowing rate = 0.5 ft/day

Distance of well from the site = 1 mile = 5280 ft

so t = 5280 / 0.5 = 10560 days

k is given as 1.94 x 10⁻⁴ 1/day

next we find the Pollutant concentration Ct in the well

Ct = C₀ × e^-( 1.94 x 10⁻⁴ × 10560)

Ct = 0.3 x e^-(kt)

Ct= 0.0386 mg/L

next we determine the chronic daily intake, CDI

CDI = (C x CR x EF x ED) / (BW x AT)

where C is average concentration of the contaminant(0.0368mg/L), CR is contact rate (2L/day), EF is exposure frequency (350days/Year), ED is exposure duration (10 years), BW is average body weight (70kg).

now we substitute

CDI = (0.0368 x 2 x 350 x 10) / ((70x 365) x 70)

= 257.7 / 1788500

= 0.000144 mg/Kg.day

CDI = 1.44 x 10⁻⁴ mg/kg.day

Finally we calculate the cancer risk, R

Slope factor SF is given as 0.02 Kg.day/mg

Risk, R = I x SF

= 1.44 x 10⁻⁴ mg/kg.day x 0.02Kg.day/mg

R = 2.88 × 10⁻⁶

therefore the cancer risk for an adult male drinking the water for 10 years is 2.88 × 10⁻⁶

7 0

3 years ago

Determine the resolution of a manometer required to measure the velocity of air at 50 m/s using a pitot-static tube and a manome

oksano4ka [1.4K]

Answer:

a) Δh = 2 cm, b) Δh = 0.4 cm

Explanation:

Let's start by using Bernoulli's equation for the Pitot tube, we define two points 1 for the small entry point and point 2 for the larger diameter entry point.

P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

Point 1 is called the stagnation point where the fluid velocity is reduced to zero (v₁ = 0), in general pitot tubes are used in such a way that the height of point 2 of is the same of point 1

y₁ = y₂

subtitute

P₁ = P₂ + ½ ρ v₂²

P₁ -P₂ = ½ ρ v²

where ρ is the density of fluid

now we measure the pressure on the included beforehand as a pair of communicating tubes filled with mercury, we set our reference system at the point of the mercury bottom surface

ΔP =ρ_{Hg} g h - ρ g h

ΔP = (ρ_{Hg} - ρ) g h

as the static pressure we can equalize the equations

ΔP = P₁ - P₂

(ρ_{Hg} - ρ) g h = ½ ρ v²

v = $\sqrt{\frac{2 (\rho_{Hg} - \rho) g}{\rho } } \ \sqrt{h}$

in this expression the densities are constant

v = A √h

A = $\sqrt{\frac{2(\rho_{Hg} - \rho ) g}{\rho } }$

They indicate the density of mercury rhohg = 13600 kg / m³, the density of dry air at 20ºC is rho air = 1.29 kg/m³

we look for the constant

A = $\sqrt{\frac{2( 13600 - 1.29) \ 9.8}{1.29} }$

A = 454.55

we substitute

v = 454.55 √h

to calculate the uncertainty or error of the velocity

h = $\frac{1}{454.55^2} \ v^2$

Δh = $\frac{dh}{dv}$ Δv

$\frac{\Delta h}{h } = 2 \ \frac{\Delta v}{v}$

Suppose we have a height reading of h = 20 cm = 0.20 m

a) uncertainty 2.5 m / s ( 0.05)

$\frac{\delta v}{v} = 0.05$

$\frac{\Delta h}{h}$ = 2 0.05

Δh = 0.1 h

Δh = 0.1 20 cm

Δh = 2 cm

b) uncertainty 0.5 m / s ( Δv/v= 0.01)

$\frac{\Delta h}{h}$ = 2 0.01

Δh = 0.02 h

Δh = 0.02 20

Δh = 0.1 20 cm

Δh = 0.4 cm = 4 mm

5 0

3 years ago

Stefan Lano needs displays that will show the musical instrument inventory in his chain of music stores that caters to musicians

AlexFokin [52]

Answer:

attached below is an example of the form

Explanation:

This type of form-fill can be described/developed based on the requirements or information needed by the organization in order to perfectly fulfill an order and also to retain customers by making the form very easy and interactive attached below is an example of such form fill interface

4 0

3 years ago