Weather exam 2 Flashcards

Question

1. Radiation fog

Answer 1

a. Forms during clear calm nights. b. Earth’s surface cools the overlying air to Td. c. Additional cooling causes condensation to occur.

Answer 2

a. Forms both day and night. c. Earth’s surface cools the overlying air to Td. d. Additional cooling causes condensation to occur. b. Air with Td greater than surface T advected by wind over surface. e. Often occurs during snowmelt

Answer 3

a. Forms over warm bodies of water. c. Water warms the overlying air. d. Water evaporates into the air. b. Cold air advected by the wind over the water. e. Warm moist air rises via convection es increases e, Td and RH increase f. Rising air quickly cooled by surrounding cold air to Td. g. Additional cooling causes condensation to occur.

Answer 4

Lift initially subsaturated (T > Td) air Cool to Td via expansional cooling Cloud forms when T < Td

Answer 5

1. Warm Clouds: T > 32ᵒF a. All liquid, no ice b. Collision and Coalescence i. Larger droplets collide and merge/coalesce with smaller droplets ii. Efficiency increases as larger droplet size increases

Answer 6

a. Ice crystals, “supercooled” droplets and vapor coexist in clouds when -40ᵒF ≤ T ≤ 32ᵒF i. deposition nuclei Hexagonal Rare b. Cold clouds composed almost entirely supercooled droplets

Answer 7

a. esi i. sublimation = deposition (slow) b. esl i. evaporation = condensation (fast) 3. Within a cold cloud, there exists a single c. For a cloud where T < 32ᵒF, esl > esi b. e, where esl > esi a. T, where -40ᵒF ≤ T ≤ 32ᵒF i. RHl = (e/esl) = 100% saturation ii. Rhi = (e/esi) > 100% supersaturation

Answer 8

1. If e = esi a. RHl < 100% b. RHi = 100% equilibrium c. Droplets evaporate 2. If e = esl a. RHl = 100% equilibrium b. RHi > 100% c. Crystals grow (via deposition) ❄

Answer 9

1. Given a cloud where a. -40ᵒF ≤ T ≤ 32ᵒF b. e = esl i. RHl = 100% ii. RHi > 100% iii. Vapor deposits on crystal Crystals GROW e DECREASES

Answer 10

2. Given a cloud where a. -40ᵒF ≤ T ≤ 32ᵒF b. esi < e < esl i. RHl < 100% ii. RHi > 100% e INCREASES iv. Droplets EVAPORATE iii. Crystal growth decreases

Answer 11

3. Given a cloud where a. -40ᵒF ≤ T ≤ 32ᵒF b. e = esl (same as 1) i. RHl = 100% ii. RHi > 100% iii. Vapor deposits on crystal Crystal growth INCREASES e DECREASES 4. Repeat Steps 1-3 until ALL droplets EVAPORATE

Answer 12

5. Given a cloud where only ice crystals exist a. -40ᵒF ≤ T ≤ 32ᵒF b. e = esi Bergeron Process (5) i. RHi = 100% deposition = sublimation crystal growth STOPS

Answer 13

...the process by which ice crystals grow at the expense of supercooled droplets in a cold cloud

Answer 14

Snow, freezing rain, sleet

Answer 15

When air temperature remains below freezing throughout the atmosphere

Answer 16

Partly frozen drops refreeze

Answer 17

Rain drops become "Supercooled" in cold air and freeze on contact

Answer 18

Clockwise and out

Answer 19

Counter-Clockwise and in

Answer 20

. Pressure Gradient Force (PGF) b. Coriolis force (f) c. Friction (Fr)

Answer 21

a. Arrow has both direction and magnitude b. Vectors have both left- and right-hand sides

Answer 22

a. only force that can act on air at rest b. directed from high to low pressure c. points in direction of rate of greatest pressure decrease d. perpendicular to the isobars e. magnitude increases as the pressure gradient increases i. PGF increases as distance between isobars decreases

Answer 23

a. “Apparent” deflection of large-scale motions b. caused by latitudinal dependence of Earth’s radial velocity (RV) (2r cos/24hrs; r = 3959 miles (6371 km);  = latitude) c. RV = 1036 mph at equator, 777.8 mph at BG (41.37N), 0 mph at North Pole

Answer 24

a. Always acts 90ᵒ to the right (left) of the wind in Northern (Southern) Hemisphere b. Increases with wind speed (equals 0 when the wind is calm) c. For a given V, f = 0 at the equator and increases with latitude

Answer 25

a. PGF initiates the wind (V) b. f acts 90ᵒ to the right of V c. V accelerates and turns to the right to strike a balance between PGF and f d. Acceleration and turning continue until balance is achieved

Answer 26

a. Wind that results from a balance between PGF and f i. Always parallel to isobars ii. Low pressure to the left iii. High pressure to the right iv. Explains 80-85% real wind

Answer 27

a. Causes the following sequence (in order) i. Vg decreases (becomes V) iii. PGF becomes dominant force ii. f decreases iv. V turns toward PGF

Answer 28

a. Surface high and low locations determined by upper level circulation pattern.

Answer 29

Column pressure is greater over warm air than cold air.

Answer 30

Warm column height is greater then cold column height.

Answer 31

Height varies on constant pressure (500 mb) surface Isoheights (solid blue) 3. Isotherms (dashed red) 4. Station models (T, Td, wind)

Answer 32

Rate of pressure decrease with height is larger in cold air than in warm air. The height of an upper level pressure surface is higher in the warm column/lower in the cold

Answer 33

The pressure on an upper level horizontal surface (levels 1 and 2) is larger in the warm column than the cold column. At any upper level, the PGF points from warm to cold.

Answer 34

a. The N-S PGF increases with increasing height. b. Wind speed increases with increasing height. i. Not just a result of decreasing friction!

Answer 35

Upper level heights decrease with increasing latitude (colder to the north). Upper level PGF rules mirror those at the surface. a. PGF directed from high to low heights toward greatest height decrease.

Answer 36

PGF perpendicular to the isoheight lines. c. PGF magnitude increases as distance between isoheights decreases.

Answer 37

a. PGF initiates wind (V). b. Coriolis force (f) acts 90ᵒ to the right of V. c. Wind (V) accelerates and turns to the right. d. Acceleration and turning continue until balance is achieved.

Answer 38

Wind is geostrophic (Vg). ii. Upper level wind westerly component ⇾Equator to pole temperature decrease!!! ⇾Low heights/cold to left of Vg ⇾High heights/warm to right of Vg

Answer 39

a. Ridges: i. Axis of high heights and warm column temperatures. b. Troughs: i. Axes of low heights and cold column temperatures.

Answer 40

a. Meridional wave pattern i. High amplitude waves ⇾Cold south in troughs ⇾Warm north in ridges

Answer 41

a. Zonal wave pattern i. Low amplitude waves ⇾Cold air north ⇾Warm air south

Answer 42

Heights slope down toward the north as column becomes progressively colder.

Answer 43

a. Clockwise circulation around highs/ ridges b. Counterclockwise circulation around lows/troughs

Answer 44

a. Centrifugal or Centripetal force??? i. Motion is tangent to the ride. ii. Ride pushes person IN along the circle. iii. Centrifugal force (brown) is apparent/fake. iv. Centripetal force (white) is real! ⇾Centripetal force ALWAYS directed inward toward middle of circle.

Answer 45

a. Centripetal force (Ce) vector points toward the center of the circle.

Answer 46

Ce acts in the same direction as the Coriolis force (f). i. f becomes large relative to the PGF. ⇾V increases and becomes faster than geostrophic, or “supergeo- strophic”(SUP).

Answer 47

Ce acts in the same direction as the PGF. i. f becomes small relative to the PGF. ⇾V decreases and becomes slower than geostrophic, or “subgeostrophic” (SUB).

Answer 48

a. Results from a balance between PGF, Coriolis force (f) and centripetal force (Ce). i. Supergeostrophic (SUP) in ridges. ii. Subgeostrophic (SUB) in troughs.

Answer 49

a. Supergeostrophic (SUP) in ridges. b. Subgeostrophic (SUB) in troughs.G c. Geostrophic (G) in areas of straight flow between ridges and troughs. d. So, Vgr accelerates and decelerates as air moves through the wave pattern.

Answer 50

Downstream of ridge Where the winds are geostrophic (G) I. Mass converges (CON) Ii. Column weight increases Iii. Surface High pressure Iv. Sinking air

Answer 51

Sinking mid-tropospheric air I. compressional warming Ii. relative humidity decrease Iii. relatively cloud-free near high center

Answer 52

a. Downstream of trough b. Where the winds are geostrophic (G) i. Mass diverges (DIV) ii. Column weight decreases iii. Surface LOW pressure iv. Rising Air

Answer 53

a. Rising mid-tropospheric air ii. relative humidity increase iii. clouds and precipitation near low center

Answer 54

Shape indicates “stability” and intensity of vertical motion.

Answer 55

determined by the response of an object displaced from its initial position

Answer 56

Object displaced from initial position returns to that position

Answer 57

Object displaced from initial position accelerates away from that position

Answer 58

Object displaced from initial position remains where displaced

Answer 59

small hypothetical volume

Answer 60

Defined by 3 interactions with surrounding environment 1. flexible boundaries 2. no MASS exchange with surrounding environment 3. no ENERGY exchange with surrounding environment ...ADIABATIC Expansional Cooling as Parcels Rise ENVIRONMENT (TE = PE + KE = constant)

Answer 61

Expansional Cooling as Parcels Rise

Answer 62

Dry adiabatic lapse rate (d) = 10C/km (Expansional cooling ➣Moist adiabatic lapse rate (m) = 6C/km (Expansional Cooling - Latent Heat Release) ➣Environmental lapse rate (e) = ?C/km CP

Answer 63

Vertical temperature profile of environment (i.e., 𝚪e) Rate of temperature change in the environment is NOT constant

Answer 64

if the temperature of the parcel air is colder than that of the environment then parcel more dense than environment parcel sinks to surface STABLE

Answer 65

If the temperature of the parcel air is warmer than that of the environment then UNSTABLE

Answer 66

if the temperature of the parcel air is equal to that of the environment then parcel and environment densities are identical parcel remains at 1 km NEUTRAL

Weather exam 2 Flashcards

(90 cards)