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2 years ago

A. Why is the stratosphere considered a "stable" layer in the atmosphere?

Physics

1 answer:

vovangra [49]2 years ago

7 0

Answer:

A. Stratosphere is said to be stable layer of the atmosphere when cool air sinks and warm air rises.Due to the fact that cool air has tendency to sink ,the air is not going fluctuating up and down in the stratosphere. This means that the air remains stationary or particles remains there for a very long duration.

B. If the lifted index is negative then the parcel temperature is warmer than the actual temperature. In addition, the parcel that is less warm than the surrounding will be less dense and will rise.

C. The water vapor come from different kinds of fronts; gust fronts from existing storms as their downdraft hits the surface, spreads and lifts air in front, upper air disturbances and surface heating by solar radiation making an unbalanced vertical profile .

D. the threshold used by storm chasers to assess if the dew point temperature is high enough to produce large thunderstorms is moisture ,the surface dew point needs to be 55 degrees fahrenheit or greater for a surface based thunderstorm to occur.

E. Wind shear is the change in wind direction or speed with height in an atmosphere.

Explanation:

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You are riding on a roller coaster that starts from rest at a height of 25.0 m and moves along a frictionless track. however, af

djyliett [7]

I attached the missing picture.
We can figure this one out using the law of conservation of energy.
At point A the car would have potential energy and kinetic energy.
$A: mgh_1+\frac{mv_1^2}{2}$
Then, while the car is traveling down the track it loses some of its initial energy due to friction:
$W_f=F_f\cdot L$
So, we know that the car is approaching the point B with the following amount of energy:
$mgh_1+\frac{mv_1^2}{2}- F_fL$
The law of conservation of energy tells us that this energy must the same as the energy at point B.
The energy at point B is the sum of car's kinetic and potential energy:
$B: mgh_2+\frac{mv_2}{2}$
As said before this energy must be the same as the energy of a car approaching the loop:
$mgh_2+\frac{mv_2}{2}=mgh_1+\frac{mv_1^2}{2}- F_fL$
Now we solve the equation for $v_1$ :
$v_1^2=2g(h_2-h_1)+v_2^2+\frac{2F_fL}{m}\\
v_1^2=39.23\\
v_1=\sqrt{39.23}=6.26\frac{m}{s}$

4 0

2 years ago

Read 2 more answers

What is the mass of a large dog that weighs 441 newtons

PtichkaEL [24]

Weight, w = mg. g ≈ 9.8 m/s². m = mass in kg. w is weight in N

441 N = m* 9.8

9.8m = 441

m = 441/9.8

m = 45 kg.

Mass of the dog is = 45 kg

5 0

2 years ago

Suppose you are talking by interplanetary telephone to your friend, who lives on the Moon. He tells you that he has just won a n

Savatey [412]

Answer:

The friend on moon will be richer.

Explanation:

We must calculate the mass of gold won by each person, to tell who is richer. For that purpose we will use the following formula:

W = mg

m = W/g

where,

m = mass of gold

W = weight of gold

g = acceleration due to gravity on that planet

FOR FRIEND ON MOON:

W = 1 N

g = 1.625 m/s²

Therefore,

m = (1 N)/(1.625 m/s²)

m(moon) = 0.6 kg

FOR ME ON EARTH:

W = 1 N

g = 9.8 m/s²

Therefore,

m = (1 N)/(9.8 m/s²)

m(earth) = 0.1 kg

Since, the mass of gold on moon is greater than the mass of moon on earth.

Therefore, the friend on moon will be richer.

7 0

2 years ago

Describe the energy transformations that take place when a skier starts skiing down a hill but after a time is brought to rest b

Andrews [41]

The skier will transform their gravitational energy into mostly kinetic energy (with a minor amount transformed into heat from the friction of the skis across the snow and air friction). Once the skier hits the snowdrift, their kinetic energy is transferred into the snow which moves when they strike it due to the kinetic energy that is now in the snow. Along with again a minor amount of heat energy transferred as they move through the snowdrift.

6 0

2 years ago

g The international space station has an orbital period of 93 minutes at an altitude (above Earth's surface) of 410 km. A geosyn

krok68 [10]

Answer:

r = 4.21 10⁷ m

Explanation:

Kepler's third law It is an application of Newton's second law where the forces of the gravitational force, obtaining

T² = ( $\frac{4\pi }{G M_s}$ ) r³ (1)

in this case the period of the season is

T₁ = 93 min (60 s / 1 min) = 5580 s

r₁ = 410 + 6370 = 6780 km

r₁ = 6.780 10⁶ m

for the satellite

T₂ = 24 h (3600 s / 1h) = 86 400 s

if we substitute in equation 1

T² = K r³

K = T₁²/r₁³

K = $\frac{ 5580^2}{ (6.780 10^6)^2}$

K = 9.99 10⁻¹⁴ s² / m³

we can replace the satellite values

r³ = T² / K

r³ = 86400² / 9.99 10⁻¹⁴

r = ∛(7.4724 10²²)

r = 4.21 10⁷ m

this distance is from the center of the earth

7 0

2 years ago