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
blagie [28]
3 years ago
10

How would you describe the stability of the atmosphere if you noted a dry adiabatic rate of 10ºC/1000 meters, a wet adiabatic ra

te of 6.5ºC/1000 meters, and an environmental lapse rate of 7.8ºC/1000 meters?
Engineering
2 answers:
Nady [450]3 years ago
8 0

Answer:

conditionally unstable

Explanation:

Given environmental lapse rate (ELR) = 7.8°C/1000 m

Wet adiabatic lapse rate (WALR) = 6.5°C/1000 m  

Dry adiabatic lapse rate (DALR) = 10°C/1000 m  

The stability of the atmosphere can be described in three ways which are;  

a.      Absolutely stable atmosphere: Here the lifted air whether saturated or unsaturated is colder and hence heavier than the surrounding air and it would return to its initial potion when released. In this case the lifted air will descent back to its original position. It occurs when WALR is greater than ELR.  

b.     Absolutely unstable: here both saturated (and moist) and unsaturated (dry) air parcels are warmer than the surrounding and hence lighter. This means both moist and dry air parcels will rise continuously when released. This occurs when DALR is less than ELR.  

c.      Conditionally unstable atmosphere: This occurs where moist air is unstable while dry air is stable. That is moist or saturated air is warmer than the surrounding and hence it will continuously rise when released while dry air is colder than the surrounding and hence it will descent back to its original position when released.  It occurs when DALR is greater than ELR and ELR is greater than WALR (DALR>ELR>WALR).  

The situation given in the question is "c". 10°C/1000 m > 7.8°C/1000 m > 6.5°C/1000 m

muminat3 years ago
3 0

Answer:

conditional instability (Γd >  Γe > Γw)

Explanation:

Given;

dry adiabatic rate, Γd = 10ºC/1000 meters

wet adiabatic rate, Γw= 6.5ºC/1000 meters

environmental lapse rate, Γe = 7.8ºC/1000 meters

Stability of the atmosphere can be described as Absolute stability, Absolute instability or conditional instability.

Conditions for Absolute stability:

Γd > Γw > Γe

Conditions for Absolute instability:

Γe > Γd > Γw

Conditions for conditional instability:

Γd >  Γe > Γw

Thus, conditional instability satisfies the given values of the atmospheric condition: Γd (10) > Γe (7.8) >  Γw (6.5)

You might be interested in
2. The following segment of carotid artery has an inlet velocity of 50 cm/s (diameter of 15 mm). The outlet has a diameter of 11
ahrayia [7]

This question is incomplete, the missing diagram is uploaded along this answer below.

Answer:

the forces required to keep the artery in place is 1.65 N

Explanation:

Given the data in the question;

Inlet velocity V₁ = 50 cm/s = 0.5 m/s

diameter d₁ = 15 mm = 0.015 m

radius r₁ = 0.0075 m

diameter d₂ = 11 mm = 0.011 m

radius r₂ = 0.0055 m

A₁ = πr² = 3.14( 0.0075 )² =  1.76625 × 10⁻⁴ m²

A₂ = πr² = 3.14( 0.0055 )² =  9.4985 × 10⁻⁵ m²

pressure at inlet P₁ = 110 mm of Hg = 14665.5 pascal

pressure at outlet P₂ = 95 mm of Hg = 12665.6 pascal

Inlet volumetric flowrate = A₁V₁ = 1.76625 × 10⁻⁴ × 0.5 = 8.83125 × 10⁻⁵ m³/s

given that; blood density is 1050 kg/m³

mass going in m' = 8.83125 × 10⁻⁵ m³/s × 1050 kg/m³ = 0.092728 kg/s

Now, using continuity equation

A₁V₁ = A₂V₂

V₂ = A₁V₁ / A₂ = (d₁/d₂)² × V₁

we substitute

V₂ =  (0.015 / 0.011 )² × 0.5

V₂ = 0.92975 m/s

from the diagram, force balance in x-direction;

0 - P₂A₂ × cos(60°) + Rₓ = m'( V₂cos(60°) - 0 )    

so we substitute in our values

0 - (12665.6 × 9.4985 × 10⁻⁵)  × cos(60°) + Rₓ = 0.092728( 0.92975 cos(60°) - 0 )    

0 - 0.6014925 + Rₓ =  0.043106929 - 0

Rₓ = 0.043106929 + 0.6014925

Rₓ = 0.6446 N

Also, we do the same force balance in y-direction;

P₁A₁ - P₂A₂ × sin(60°) + R_y = m'( V₂sin(60°) - 0.5 )  

we substitute

⇒ (14665.5 × 1.76625 × 10⁻⁴) - (12665.6 × 9.4985 × 10⁻⁵) × sin(60°) + R_y = 0.092728( 0.92975sin(60°) - 0.5 )

⇒ 1.5484 + R_y = 0.092728( 0.305187 )

⇒ 1.5484 + R_y = 0.028299    

R_y = 0.028299 - 1.5484

R_y = -1.52 N

Hence reaction force required will be;

R = √( Rₓ² + R_y² )

we substitute

R = √( (0.6446)² + (-1.52)² )

R = √( 0.41550916 + 2.3104 )

R = √( 2.72590916 )

R = 1.65 N

Therefore, the forces required to keep the artery in place is 1.65 N

 

7 0
2 years ago
How could the location of tests affect the performance of a catapult ?
Luba_88 [7]

Answer:

It could affect how far the projectile travels

Explanation:

Facing Uphill: Moves less far

Downhill: Moves further

3 0
3 years ago
Read 2 more answers
The phrase "positive to positive, negative to ground" is correct when jump starting a car.
Artist 52 [7]

The correct answer is A; True.

Further Explanation:

This is a correct phrase that is important to learn when owning any type of vehicle. When a car battery is dead, it can usually be jump started by using another cars battery or a portable battery charger. It is extremely important to put the positive battery cable on the positive battery post. Then the negative cable will be placed on the negative car battery post and the negative ground wire can be anywhere on the car except on the battery.

The car needs to be connected properly for a few minutes before trying to start the car. This helps the car battery to get enough "juice" to start. If the battery cables are placed wrong this can cause sparks to come out of the cables/battery and cause bodily harm.

Learn more about car batteries at brainly.com/question/7734062

#LearnwithBrainly

7 0
3 years ago
PLS HELP ME
Oksana_A [137]

Answer:

The Euler buckling load of a 160-cm-long column will be 1.33 times the Euler buckling load of an equivalent 120-cm-long column.

Explanation:

160 - 120 = 40

120 = 100

40 = X

40 x 100 / 120 = X

4000 / 120 = X

33.333 = X

120 = 100

160 = X

160 x 100 /120 = X

16000 / 120 = X

133.333 = X

4 0
3 years ago
Which of the eight diagnostic steps for locating an engine performance problem is performed first?
Kay [80]

Answer:

D. Perform a thorough visual inspection.

4 0
2 years ago
Other questions:
  • Write SQL queries to answer the following questions: What are the names of the course(s) that student Altvater took during the s
    13·1 answer
  • At the instant shown, slider block B is moving with a constant acceleration, and its speed is 150 mm/s. Knowing that after slide
    13·1 answer
  • In case of damaged prestressed concrete I girders which are used for restoring strength?
    9·1 answer
  • A light pressure vessel is made of 2024-T3 aluminum alloy tubing with suitable end closures. This cylinder has a 90mm OD, a 1.65
    8·1 answer
  • Anne-Marie Cole runs the sales division for a local auto insurance firm. One of her key duties is to calculate her company's mar
    11·2 answers
  • Item110pointseBook HintPrintReferences Check my work Check My Work button is now disabled5Item 1Item 1 10 pointsAn ideal Diesel
    10·1 answer
  • You are analyzing an open-return wind tunnel that intakes air at 20 m/s and 320K. When the flow exits the wind tunnel it is movi
    14·2 answers
  • A local surf report provides the height of the wave from the trough to the crest of the wave. How does this relate to the wave’s
    11·1 answer
  • Help me, iv been having problems with ads going in my phones storage files, what can i do to stop this?
    14·2 answers
  • What are the horizontal structures beneath a slab that help transfer the load from the slab to the columns?
    14·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!