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3 years ago

10

Hello I need some help with this please. Pick a problem in your school or community that you think could be solved with technolo

gy. Describle the problem and design a technological solution.

2 answers:

bezimeni [28]3 years ago

7 0

Answer:

An AI operated automatic garbage collection system

Explanation:

There is always an issue in my neighbourhood with the garbagemen coming on time so having an automatic system will help in the overall efficiency in the task

Anna007 [38]3 years ago

6 0

An online agenda/calendar

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Pie charts should have no more than eight segments. True or False?

Pepsi [2]

Answer:

Explanation:

Pie charts generally should have no more than eight segments.

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3 years ago

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8) Microsoft Windows is

Alisiya [41]

The Answer Is C. A collection of computer programs or applications along with its related data
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2 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

The A-36 solid steel shaft is 3.3 m long and has a diameter of 50 mm. It is required to transmit 35 kW of power from the engine

spin [16.1K]

Answer:

Explanation:

Answer is in the following attachment

8 0

3 years ago

Air at 1600 K, 30 bar enters a turbine operating at steady state and expands adiabatically to the exit, where the pressure is 2.

djyliett [7]

Solution :

The isentropic efficiency of the turbine is given as :

$\eta = \frac{\text{actual work done}}{\text{isentropic work done}}$

$=\frac{m(h_1-h_2)}{m(h_1-h_{2s})}$

$=\frac{h_1-h_2}{h_1-h_{2s}}$

The entropy relation for the isentropic process is given by :

$0=s^\circ_2-s^\circ_1-R \ln \left(\frac{P_2}{P_1}\right)$

$\ln \left(\frac{P_2}{P_1}\right)=\frac{s^\circ_2-s^\circ_1}{R}$

$\frac{P_2}{P_1}=exp\left(\frac{s^\circ_2-s^\circ_1}{R}\right)$

$\left(\frac{P_2}{P_1}\right)_{s=constant}=\frac{P_{r2}}{P_{r1}}$

Now obtaining the properties from the ideal gas properties of air table :

At $T_1 = 1600 \ K,$

$P_{r1}=791.2$

$h_1=1757.57 \ kJ/kg$

Calculating the relative pressure at state 2s :

$\frac{P_{r2}}{P_{r1}}=\frac{P_2}{P_1}$

$\frac{P_{r2}}{791.2}=\frac{2.4}{30}$

$P_{r2}=63.296$

Obtaining the properties from Ideal gas properties of air table :

At $P_{r2}=63.296$ , $T_{2s}\approx 860 \ K$

Considering the isentropic relation to calculate the actual temperature at the turbine exit, we get:

$\eta=\frac{h_1-h_2}{h_1-h_{2s}}$

$\eta=\frac{c_p(T_1-T_2)}{c_p(T_1-T_{2s})}$

$\eta=\frac{T_1-T_2}{T_1-T_{2s}}$

$0.9=\frac{1600-T_2}{1600-860}$

$T_2= 938 \ K$

So, at $T_2= 938 \ K$ , $h_2=975.66 \ kJ/kg$

Now calculating the work developed per kg of air is :

$w=h_1-h_2$

= 1757.57 - 975.66

= 781 kJ/kg

Therefore, the temperature at the exit is 938 K and work developed is 781 kJ/kg.

4 0

3 years ago