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Shalnov [3]
3 years ago
7

Suppose a thick nimbostratus cloud contains ice crystals and cloud droplets all about the same size. Which precipitation process

will be most important in producing rain from this cloud? Why?
Physics
1 answer:
Ronch [10]3 years ago
7 0

Answer:The most important process would be the ice crystal process

Explanation:

Ice crystal process also called Bergeron process requires numerous small water drops that are supercooled, which is a common feature in clouds between about 0° and -20°C or below, along with a small number of ice crystals. Therefore because the collision-coalescence process requires that cloud droplets be of varying size so that drops will fill at different speeds, the most important process would be the ice crystal process.

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Why does a black hole have a stronger gravitational pull than the star that collapse to form it?​
Studentka2010 [4]

Answer:

We consider Black Holes as an object that possesses extreme gravitational pull, but wait aren’t they have the same mass(or less) as that of their parent star. And we know that gravitational pull ‘F’ is directly proportional to the mass of an object, so if the mass is same(or less) then why do black holes have stronger gravity than the stars they evolved from.

The above consideration that F is directly proportional to the mass is partially correct, one should also mention that F is also inversely proportional to the square of the distance between the considered objects.

F = G*(M*m)/(r^2)

Where:

· F is the force acting on you due to star

· M is the mass of Parent star / Black Hole

· m is the mass of an observer, here it is you

· r is the radial distance between the star and you

We know that black hole formed, has much smaller size than that of its parent star and all that mass is compressed to a much smaller scale. If you consider a Star as having a size of an earth then the black hole formed will have a size of small city.

Let us say that you are standing at an r distance away from a star (r>R1), where R1 is the radius of the star, of course (R1>R2), where R2 is the radius of Black Hole.

The Force by which the star in case 1 attracts you will be equal(or less) to the force by which black hole in case 2. So, there is nothing increase in gravitational pull, it is same(or less) as that of the parent star.

Wait a minute, then why people say that black holes have massive gravitational pull.

The gravitational pull increases as we move closer to the black hole, and when we are at its surface, it is enormous as compare to its star surface, because of the difference in the size.

We know that gravitational pull not only depends upon the mass but also depends upon the radial distance between the concerned objects here, it is you and the black hole.

Here, the size of the black hole is much smaller than that of its parent star, i.e (R1>>>R2), and thus we get F1<<<F2, and that is why we say that the black hole has enormous gravitational pull, such that nothing can escape, not even light.

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Two bumper cars colide into each other and each one bounces bach. wich of Newton's laws is it ?
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Describe a good level of body fat
m_a_m_a [10]

Answer: a good level of body fat can be found using your weight and height as a reference.

Explanation:

6 0
3 years ago
.1 An 8-ft 3 tank contains air at an initial temperature of 808F and initial pressure of 100 lbf/in. 2 The tank develops a small
Alina [70]

Correct temperature is 80°F

Answer:

T_f = 38.83°F

Explanation:

We are given;

Volume; V = 8 ft³

Initial Pressure; P_i = 100 lbf/in² = 100 × 12² lbf/ft²

Initial temperature; T_i = 80°F = 539.67 °R

Time for outlet flow; t_o = 90 s

Mass flow rate at outlet; m'_o = 0.03 lb/s

Final pressure; P_f = 30 lbf/in² = 30 × 12² lbf/ft²

Now, from ideal gas equation,

Pv = RT

Where v is initial specific volume

R is ideal gas constant = 53.33 ft.lbf/°R

Thus;

v = RT/P

v_i = 53.33 × 539.67/(100 × 12²)

v_i = 2 ft³/lb

Formula for initial mass is;

m_i = V/v_i

m_i = 8/2

m_i = 4 lb

Now change in mass is given as;

Δm = m'_o × t_o

Δm = 0.03 × 90

Δm = 2.7 lb

Now,

m_f = m_i - Δm

Thus; m_f = 4 - 2.7

m_f = 1.3 lb

Similarly in above;

v_f = V/m_f

v_f = 8/1.3

v_f = 6.154 ft³/lb

Again;

Pv = RT

Thus;

T_f = P_f•v_f/R

T_f = (30 × 12² × 6.154)/53.33

T_f = 498.5°R

Converting to °F gives;

T_f = 38.83°F

7 0
3 years ago
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