1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
krok68 [10]
3 years ago
4

Consider a 0.15-mm-diameter air bubble in a liquid. Determine the pressure difference between the inside and outside of the air

bubble if the surface tension at the air-liquid interface is a. 0.1 N/m. b. 0.12 N/m.
Engineering
1 answer:
Rzqust [24]3 years ago
8 0

Answer:

a)\Delta P= 2666.66 Pa

b)\Delta P= 3200 Pa

Explanation:

Given that

Diameter ,d= 0.15 mm

We know that pressure difference is given as

\Delta P=\dfrac{4\sigma }{d}

Now by putting the values

When surface tension 0.1 N/m  :

The surface tension ,\sigma=0.1\ N/m

\Delta P=\dfrac{4\times 0.1 }{0.15}\times 10^3\ Pa

\Delta P= 2666.66 Pa

When surface tension 0.12 N/m  :

The surface tension ,\sigma=0.12\ N/m

\Delta P=\dfrac{4\times 0.12 }{0.15}\times 10^3\ Pa

\Delta P= 3200 Pa

a)\Delta P= 2666.66 Pa

b)\Delta P= 3200 Pa

You might be interested in
Air exits a compressor operating at steady-state, steady-flow conditions at 150 oC, 825 kPa, with a velocity of 10 m/s through a
ioda

Answer:

a) Qe = 0.01963 m^3 / s , mass flow rate m^ = 0.1334 kg/s

b) Inlet cross sectional area = Ai = 0.11217 m^2 , Qi = 0.11217 m^3 / s    

Explanation:

Given:-

- The compressor exit conditions are given as follows:

                  Pressure ( Pe ) = 825 KPa

                  Temperature ( Te ) = 150°C

                  Velocity ( Ve ) = 10 m/s

                  Diameter ( de ) = 5.0 cm

Solution:-

- Define inlet parameters:

                  Pressure = Pi = 100 KPa

                  Temperature = Ti = 20.0

                  Velocity = Vi = 1.0 m/s

                  Area = Ai

- From definition the volumetric flow rate at outlet ( Qe ) is determined by the following equation:

                   Qe = Ae*Ve

Where,

           Ae: The exit cross sectional area

                   Ae = π*de^2 / 4

Therefore,

                  Qe = Ve*π*de^2 / 4

                  Qe = 10*π*0.05^2 / 4

                  Qe = 0.01963 m^3 / s

 

- To determine the mass flow rate ( m^ ) through the compressor we need to determine the density of air at exit using exit conditions.

- We will assume air to be an ideal gas. Thus using the ideal gas state equation we have:

                   Pe / ρe = R*Te  

Where,

           Te: The absolute temperature at exit

           ρe: The density of air at exit

           R: the specific gas constant for air = 0.287 KJ /kg.K

             

                ρe = Pe / (R*Te)

                ρe = 825 / (0.287*( 273 + 150 ) )

                ρe = 6.79566 kg/m^3

- The mass flow rate ( m^ ) is given:

               m^ = ρe*Qe

                     = ( 6.79566 )*( 0.01963 )

                     = 0.1334 kg/s

- We will use the "continuity equation " for steady state flow inside the compressor i.e mass flow rate remains constant:

              m^ = ρe*Ae*Ve = ρi*Ai*Vi

- Density of air at inlet using inlet conditions. Again, using the ideal gas state equation:

               Pi / ρi = R*Ti  

Where,

           Ti: The absolute temperature at inlet

           ρi: The density of air at inlet

           R: the specific gas constant for air = 0.287 KJ /kg.K

             

                ρi = Pi / (R*Ti)

                ρi = 100 / (0.287*( 273 + 20 ) )

                ρi = 1.18918 kg/m^3

Using continuity expression:

               Ai = m^ / ρi*Vi

               Ai = 0.1334 / 1.18918*1

               Ai = 0.11217 m^2          

- From definition the volumetric flow rate at inlet ( Qi ) is determined by the following equation:

                   Qi = Ai*Vi

Where,

           Ai: The inlet cross sectional area

                  Qi = 0.11217*1

                  Qi = 0.11217 m^3 / s    

- The equations that will help us with required plots are:

Inlet cross section area ( Ai )

                Ai = m^ / ρi*Vi  

                Ai = 0.1334 / 1.18918*Vi

                Ai ( V ) = 0.11217 / Vi   .... Eq 1

Inlet flow rate ( Qi ):

                Qi = 0.11217 m^3 / s ... constant  Eq 2

               

6 0
3 years ago
Given that the skin depth of graphite at 100 (MHz) is 0.16 (mm), determine (a) the conductivity of graphite, and (b) the distanc
Andrei [34K]

Answer:

the answer is below

Explanation:

a) The conductivity of graphite (σ) is calculated using the formula:

\delta=\frac{1}{\sqrt{\pi f \mu \sigma} }\\\\\sigma =\frac{1}{\pi f \mu \delta^2}

where f = frequency = 100 MHz, δ = skin depth = 0.16 mm = 0.00016 m, μ = 0.0000012

Substituting:

\sigma =\frac{1}{\pi *10^6* 0.0000012*0.00016^2}=0.99*10^4\ S/m

b) f = 1 GHz = 10⁹ Hz.

\alpha=\sqrt{\pi f \mu \sigma} = \sqrt{0.0000012*10^9*\pi*0.99*10^5}=1.98*10^4\ Np/m\\\\20log_{10}  e^{-\alpha z}=-30\ dB\\\\(-\alpha z)log_{10}  e=-1.5 \\\\z=\frac{-1.5}{log_{10}  e*-\alpha} =1.75*10^{-4}\ m=0.175\ mm

4 0
3 years ago
Two resistors, with resistances R1 and R2, are connected in series. R1 is normally distributed with mean 65 and standard deviati
Sedbober [7]

Answer:

n this question, we are asked to find the probability that  

R1 is normally distributed with mean 65  and standard deviation 10

R2 is normally distributed with mean 75  and standard deviation 5

Both resistor are connected in series.

We need to find P(R2>R1)

the we can re write as,

P(R2>R1) = P(R2-R1>R1-R1)

P(R2>R1) = P(R2-R1>0)

P(R2>R1) = P(R>0)

Where;

R = R2 - R1

Since both and are independent random variable and normally distributed, we can do the linear combinations of mean and standard deviations.

u = u2-u1

u = 75 - 65 = 10ohm

sd = √sd1² + sd2²

sd = √10²+5²

sd = √100+25 = 11.18ohm

Now we will calculate the z-score, to find  P( R>0 )

Z = ( X -u)/sd

the z score of 0 is

z = 0 - 10/11.18

z= - 0.89

4 0
4 years ago
"Transportation is the way of expanding business activies" justify this statement with long answer​
Leokris [45]

Answer:

Transportation methods ensure deliveries to and from your facility flow smoothly and arrive at their designated destinations on time. Because of the importance of transportation to your business's success, it's vital to include this factor in your supply chain management strategy.

So,transportation is the way of expanding business activities.

4 0
3 years ago
A __________ is a single lane used by drivers to enter and exit a freeway. a) Climbing lane b) Weave lane c) Thru lane d) Offset
Rufina [12.5K]

A Weave lane is a single lane used by drivers to enter and exit a freeway.

<h3>What is this lane about?</h3>

Weave lanes are known to be lanes that acts as an entrance and exits for a lot of cars in highways.

Hence, one can say that A Weave lane is a single lane used by drivers to enter and exit a freeway.

Learn more about Weave lane from

brainly.com/question/10828527

#SPJ1

6 0
2 years ago
Other questions:
  • When a traffic officer directs you proceed through the intersection
    6·1 answer
  • What is the reading of this 0-1" micrometer? Enter your
    9·1 answer
  • Two synchronous motors having subtransient reactances of 0.8 and 0.25 p.u., respectively, on a base of 500 V 2 MVA are connected
    13·1 answer
  • For an automotive spark-ignition engine, the combustion duration (from time of ignition through completion) is approximately one
    11·1 answer
  • A(n) ____________ is a side view that shows height.
    12·1 answer
  • The first electricity producing wind turbine was invented in what year?*
    6·1 answer
  • In software engineering how do you apply design for change?
    13·1 answer
  • 6. Question
    13·1 answer
  • What is the answer to life the universe and everything <br> (worth 95 points!)
    13·2 answers
  • In one study the critical stress intensity factor for human bone was calculated to be 4.05 MN/m3/2. If the value of Y in Eq. (2.
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!