All categories

3 years ago

Adding moisture is the ONLY way to obtain saturated air from unsaturated air. a)-True b) False

Engineering

1 answer:

Aleksandr [31]3 years ago

7 0

Answer:

The statement is false.

Explanation:

A unsaturated mass of air can become saturated in 3 ways:

1) Adding moisture to the initially unsaturated air till it becomes saturated.

2) Mixing of 2 different air masses initially unsaturated but the mixture becomes saturated.

3) Reducing the temperature of the surrounding so that the capacity to hold moisture of the air mass decreases until it becomes saturated.

You might be interested in

A DOHC V-6 has how many camshafts?

viktelen [127]

Answer:

two camshafts

Explanation:

4 0

3 years ago

Small droplets of carbon tetrachloride at 68 °F are formed with a spray nozzle. If the average diameter of the droplets is 200 u

Licemer1 [7]

Answer:

the difference in pressure between the inside and outside of the droplets is 538 Pa

Explanation:

given data

temperature = 68 °F

average diameter = 200 µm

to find out

what is the difference in pressure between the inside and outside of the droplets

solution

we know here surface tension of carbon tetra chloride at 68 °F is get from table 1.6 physical properties of liquid that is

σ = 2.69 × $10^{-2}$ N/m

so average radius = $\frac{diameter}{2}$ = 100 µm = 100 × $10^{-6}$ m

now here we know relation between pressure difference and surface tension

so we can derive difference pressure as

2π×σ×r = Δp×π×r² .....................1

here r is radius and Δp pressure difference and σ surface tension

Δp = $\frac{2 \sigma }{r}$

put here value

Δp = $\frac{2*2.69*10^{-2}}{100*10^{-6}}$

Δp = 538

so the difference in pressure between the inside and outside of the droplets is 538 Pa

7 0

3 years ago

Write the following decorators and apply them to a single function (applying multiple decorators to a single function): 1. The f

natita [175]

Answer:

Complete question is:

write the following decorators and apply them to a single function (applying multiple decorators to a single function):

1. The first decorator is called strong and has an inner function called wrapper. The purpose of this decorator is to add the html tags of and to the argument of the decorator. The return value of the wrapper should look like: return “” + func() + “”

2. The decorator will return the wrapper per usual.

3. The second decorator is called emphasis and has an inner function called wrapper. The purpose of this decorator is to add the html tags of and to the argument of the decorator similar to step 1. The return value of the wrapper should look like: return “” + func() + “.

4. Use the greetings() function in problem 1 as the decorated function that simply prints “Hello”.

5. Apply both decorators (by @ operator to greetings()).

6. Invoke the greetings() function and capture the result.

Code :

def strong_decorator(func):

def func_wrapper(name):

return "{0}".format(func(name))

return func_wrapper

def em_decorator(func):

def func_wrapper(name):

return "{0}".format(func(name))

return func_wrapper

@strong_decorator

@em_decorator

def Greetings(name):

return "{0}".format(name)

print(Greetings("Hello"))

Explanation:

5 0

3 years ago

A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 5.6 mm; the spe

ankoles [38]

Answer:

F = 8849 N

Explanation:

Given:

Load at a given point = F = 4250 N

Support span = L = 44 mm

Radius = R = 5.6 mm

length thickness of tested material = 12 mm

First compute the flexural strength for circular cross section using the formula below:

σ $_{fs} = F_{f} L / \pi R^{3}$

σ = FL / π R³

Putting the given values in the above formula:

σ = 4250 ( 44 x 10⁻³ ) / π ( 5.6 x 10⁻³ ) ³

= 4250 ( 44 x 10⁻³ ) / 3.141593 ( 5.6 x 10⁻³ ) ³

= 4250 (44 x 1 /1000 )) / 3.141593 ( 5.6 x 10⁻³ ) ³

= 4250 ( 11 / 250 ) / 3.141593 ( 5.6 x 10⁻³ ) ³

= 187 / 3.141593 ( 5.6 x 1 / 1000 ) ³

= 187 / 3.141593 (0.0056)³

= 338943767.745358

= 338.943768 x 10⁶

σ = 338 x 10⁶ N/m²

Now we compute the load i.e. F from the following formula:

$F_{f}$ = 2 σ $_{fs}$ d³/3 L

F = 2σd³/3L

= 2(338 x 10⁶)(12 x 10⁻³)³ / 3(44 x 10⁻³)

= 2 ( 338 x 1000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

= 2 ( 338000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

= 676000000 ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

= 676000000 ( 12 x 1/1000 )³ / 3 ( 44 x 10⁻³)

= 676000000 ( 3 / 250 )³ / 3 ( 44 x 10⁻³)

= 676000000 ( 27 / 15625000 ) / 3 ( 44 x 10⁻³)

= 146016 / 125 / 3 ( 44 x 1 / 1000 )

= ( 146016 / 125 ) / (3 ( 11 / 250 ))

= 97344 / 11

F = 8849 N

4 0

3 years ago

When the psychologist simply records the relationship between two variables...

Wewaii [24]

When a psychologist simply records the relationship between two variables without manipulating them, it is called a correlational study.

The observed relationship does not by itself reveal which variable causes the other. This is the directionally problem. Also, the relationship may be due to a third variable controlling both of the observed variables.

8 0

3 years ago