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3 years ago

Where would outdoor Air quality monitors need to be placed to properly record data?

1 answer:

3 0

it is ideal to place indoor sensors near the typical breathing zone height (3 – 6 ft). They should be placed away from air pollution sources, like a toaster, and air pollution sinks, like air cleaners, to get a more representative measure of indoor air quality.

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Create a variable pounds to store weight in pounds. Convert this to kilograms and assign the result to a variable kilos. The con

vodka [1.7K]

Answer:

>>pounds=13.2

>>kilos=pounds/2.2

Explanation:

Using Matlab to write the program, consider at any time when the weight in pounds is 13.2 lb, this variable of weight is created in MATLAB by typing >>pounds=13.2. To convert it from lb to Kg, we simply divide it by 2.2 hence the second command to created is kilos. For this, the output of the program will be 6 Kg.

5 0

3 years ago

A hot air balloon is used as an air-vehicle to carry passengers. It is assumed that this balloon is sealed and has a spherical s

monitta

Answer:

a. $\dfrac{D_{1}}{ D_{2}} = \left (\dfrac{ \left{D_1} }{ {D_2}} \right )^{-3\times n}$ which is constant therefore, n = constant

b. The temperature at the end of the process is 109.6°C

c. The work done by the balloon boundaries = 10.81 MJ

The work done on the surrounding atmospheric air = 10.6 MJ

Explanation:

p₁ = 100 kPa

T₁ = 27°C

D₁ = 10 m

v₂ = 1.2 × v₁

p ∝ α·D

α = Constant

$v_1 = \dfrac{4}{3} \times \pi \times r^3$

$\therefore v_1 = \dfrac{4}{3} \times \pi \times \left (\dfrac{10}{2} \right )^3 = 523.6 \ m^3$

v₂ = 1.2 × v₁ = 1.2 × 523.6 = 628.32 m³

Therefore, D₂ = 10.63 m

We check the following relation for a polytropic process;

$\dfrac{p_{1}}{p_{2}} = \left (\dfrac{V_{2}}{V_{1}} \right )^{n} = \left (\dfrac{T_{1}}{T_{2}} \right )^{\dfrac{n}{n-1}}$

We have;

$\dfrac{\alpha \times D_{1}}{\alpha \times D_{2}} = \left (\dfrac{ \dfrac{4}{3} \times \pi \times \left (\dfrac{D_2}{2} \right )^3}{\dfrac{4}{3} \times \pi \times \left (\dfrac{D_1}{2} \right )^3} \right )^{n} = \left (\dfrac{ \left{D_2} ^3}{ {D_1}^3} \right )^{n}$

$\dfrac{D_{1}}{ D_{2}} = \left (\dfrac{ \left{D_2} }{ {D_1}} \right )^{3\times n} = \left (\dfrac{ \left{D_1} }{ {D_2}} \right )^{-3\times n}$

$\dfrac{ D_{1}}{ D_{2}} = \left ( 1.2 \right )^{n} = \left (\dfrac{ \left{D_2} ^3}{ {D_1}^3} \right )^{n}$

$log \left (\dfrac{D_{1}}{ D_{2}}\right ) = -3\times n \times log\left (\dfrac{ \left{D_1} }{ {D_2}} \right )$

n = -1/3

Therefore, the relation, pVⁿ = Constant

b. The temperature T₂ is found as follows;

$\left (\dfrac{628.32 }{523.6} \right )^{-\dfrac{1}{3} } = \left (\dfrac{300.15}{T_{2}} \right )^{\dfrac{-\dfrac{1}{3}}{-\dfrac{1}{3}-1}} = \left (\dfrac{300.15}{T_{2}} \right )^{\dfrac{1}{4}}$

T₂ = 300.15/0.784 = 382.75 K = 109.6°C

c. $W_{pdv} = \dfrac{p_1 \times v_1 -p_2 \times v_2 }{n-1}$

$p_2 = \dfrac{p_{1}}{ \left (\dfrac{V_{2}}{V_{1}} \right )^{n} } = \dfrac{100\times 10^3}{ \left (1.2) \right ^{-\dfrac{1}{3} } }$

p₂ = 100000/0.941 = 106.265 kPa

$W_{pdv} = \dfrac{100 \times 10^3 \times 523.6 -106.265 \times 10^3 \times 628.32 }{-\dfrac{1}{3} -1} = 10806697.1433 \ J$

The work done by the balloon boundaries = 10.81 MJ

Work done against atmospheric pressure, Pₐ, is given by the relation;

Pₐ × (V₂ - V₁) = 1.01×10⁵×(628.32 - 523.6) = 10576695.3 J

The work done on the surrounding atmospheric air = 10.6 MJ

4 0

3 years ago

A spherical balloon is filled with a gas. The outer diameter of the balloon is 20 in. and the thick-ness is 0.012 in. Calculate

coldgirl [10]

Answer:

1.4 psi

Explanation:

Before diving into the solution to the question above, let's pick out the parameters needed in solving this problem from the question.

=> The measurement for the outer diameter of the balloon = 20 inches, the measurement for the thickness = 0.012 in, the allowable tensile stress = 1ksi and the allowable shear stress in the balloon = 0.3 ksi.

The first thing to determine is the inner diameter = 20 - 2 × 0.012 in = 19.976 in.

Therefore, the tensile stress:

1000 = k × [19.976/2]÷ 2 × 0.012 = 2.4 psi.

Also, the sheer stress which is also the maximum permissible pressure in the balloon can be calculated below as:

0.3 × 1000 = k × [19.976/2]/ 4 × 0.012 = 1.4 psi.

4 0

3 years ago

You are an employee of Crewdson, Inc. Under the new policy, if you arrive to work late one morning, you will be required to

EastWind [94]

Answer:

C.leave work late that evening

Explanation:

As I'm the Employee of Crrewdson, Inc.

since I came late for work in the morning, because of the late train

As per the company's new policy, everyone should have to complete there work hours duration.

so I can't explain that situation to my Manager.

Therefore I had to work late that evening for completion of my work hours duration for that day.

5 0

3 years ago

Use the diagram to determine total resistance. (Round the FINAL answer to one decimal place.) Note: Use the rated voltage and wa

Valentin [98]

Answer:

22.6622 Ω ≈ 22.7 Ω
5.29516 A ≈ 5.3 A

Explanation:

For an electric motor, 1 hp = 746 W, so 1/2 hp = 373 W. Then the total wattage of the load is ...

Pl = 373 +100 +100 +75 = 648 W

The equivalent resistance is ...

R = V^2/P = (120 V)^2/(648 W) = 22 2/9 Ω

Adding the wiring resistance, we get a circuit resistance of ...

22.2222 +0.22 +0.22 = 22.6622 Ω . . . . total circuit resistance

Then the circuit current is ...

120 V/(22.6622 Ω) ≈ 5.29516 A

Circuit resistance is about 22.7 Ω; circuit current is about 5.3 A.

_____

Exact values are 5099/225 ohms and 27000/5099 amps. Of course, these assume the motor and lamps are purely resistive, which they are not.

3 0

3 years ago