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4 years ago

Which factors has the largest influence on the formation and direction of the asiatic monsoons?

Physics

1 answer:

cricket20 [7]4 years ago

8 0

Answer:

The factors which have the largest influence on the formation and direction of the asiatic monsoons are:

Moisture-laden or <em>Humid </em> winds. The presence of abundant water bodies around the subcontinent such as the Arabian Sea, Bay of Bengal, and Indian Ocean. These help moisture accumulate in the winds during the hot season.
Droplet formation
Land form barriers (that is Highlands) such as Himalayas, the western Ghats of India, and the Annamese Highlands of Vietnam

Explanation:

The term monsoon comes from the Arabic word <em>mausim</em>, meaning

season.

As a meteorological term, monsoon refers to the directional shifting of winds from one season to the next.

It may also be referred to as a seasonal prevailing wind in the region of South and South-East Asia, blowing from the south-west between May and September and bringing rain (the wet monsoon ), or from the north-east between October and April (the dry monsoon).

Usually, the monsoon occurs when a humid wind blowing from the ocean toward the land in the summer shifts to a dry, cooler wind blowing seaward off the land in the winter, and it involves a full 180° direction change in the wind.

Orographic barriers (that is highlands in the path of the winds) force the winds to rise. Precipitation then occurs on the windward side of the highlands because of adiabatic cooling and condensation of the moist rising air.

Cheers!

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Here is a force diagram of an object in water. The weight of the object is 15N and the buoyancy force is 17N. Will the object fl

lawyer [7]

Answer:

Object will float.

Explanation:

Total force on the body = Weight of body + Buoyancy force on body.

Weight of body = 15 N downwards = 15 N

Buoyancy force on body = 17 N upwards = -17 N

Total force on body = 15 - 17 = -2 N = 2 N upwards

So, the body will float.

Object will float.

8 0

3 years ago

Discuss whether any work is being done by each of the following agents and, if so, whether the work is positive or negative: (a)

Aleks04 [339]

Answer:

a) As the chicken is still, the displacement is zero, which implies that the work is zero.

b) as the person is still there is no displacement therefore the work is zero

c) Lagraua applies a vertical force and the displacement is vertical, therefore the Work is positive

d) the force of gravity is directed downwards and the displacement is upwards, therefore the angle between it is 180º and the 180º fly is -1. Consequently the lock is negative

e) when the person meticulously feels the upward force and the displacement is downward, therefore the work is negative

Explanation:

Work is defined by the expression

W = F. r

bold letters indicate vectors, we can write this expression as a module

W= F r cos θ

where is at the angle between force and displacement.

Let's apply this expression to the different cases

a) As the chicken is still, the displacement is zero, which implies that the work is zero.

b) as the person is still there is no displacement therefore the work is zero

c) Lagraua applies a vertical force and the displacement is vertical, therefore the Work is positive

d) the force of gravity is directed downwards and the displacement is upwards, therefore the angle between it is 180º and the 180º fly is -1. Consequently the lock is negative

e) when the person meticulously feels the upward force and the displacement is downward, therefore the work is negative

7 0

3 years ago

The fragment of an asteroid or any interplanetary material is known as a

Elodia [21]

The fragment of an asteroid or any interplanetary material is known as A. METEROID

8 0

4 years ago

The temperature and time t given in hours from 0 to 24 after midnight in downtown mathville is given by t=10-5 sin(pi/12 t) degr

Dvinal [7]

Answer:

$T_A_v_g=9.918192559$^{\circ}C$

Explanation:

The problem tell us that the temperature as function of time in downtown mathville is given by:

$T(t)=10-5*sin(\frac{\pi}{12t})$

The average temperature over a given interval can be calculated as:

$T_a_v_g=\frac{T_o+T_f}{2}$

Where:

$T_o=Initial\hspace{3}temperature\\T_f=Final\hspace{3}temperature$

So, the initial temperature in this case, would be the temperature at noon, and the final temperature would be the temperature at midnight:

Therefore:

$T_o=T(12)=10-5*sin(\frac{\pi}{12*12}) =9.890925575$^{\circ}C$

$T_f=T(24)=10-5*sin(\frac{\pi}{12*24}) =9.945459543$^{\circ}C$

Hence, the average temperature between noon and midnight is:

$T_A_v_g=\frac{9.890925575+9.945459543}{2}=9.918192559$^{\circ}C$

3 0

3 years ago

NASA is designing a Mars-lander that will enter the Martian atmosphere at high speed. To land safely it must slow to a constant

Viktor [21]

Answer:

a) maximum mass of the Mars lander to ensure it can land safely is 200 kg

b) area of the parachute required is 480 m² which is larger than 400 m²

c) area of the parachute should be 12.68 m²

Explanation:

Given the data in the question;

V = 20 m/s

A = 200 m²

drag co-efficient CD = 1.855

g = 3.71 m/s²

density of the atmospheric pressure β = 0.01 kg/m³

a. Calculate the maximum mass of the Mars lander to ensure it can land safely?

Drag force FD = 1/2 × CD × β × A × V²

we substitute

FD = 1/2 × 1.855 × 0.01 kg/m × 200 m² × ( 20 m/s )²

FD = 742 N

we know that;

FD = Fg

Fg = gravity force

Fg = mg

FD = mg

m = FD/g

we substitute

m = 742 N / 3.71 m/s²

m = 200 kg

Therefore, the maximum mass of the Mars lander to ensure it can land safely is 200 kg

b. The mission designers consider a larger lander with a mass of 480 kg. Show that the parachute required would be larger than 400 m²;

Given that;

M = 480 kg

Show that the parachute required would be larger than 400 m²

we know that;

FD = Fg = Mg = 480 kg × 3.71 m/s²

FD = 1780.8 N

Now, FD = 1/2 × CD × β × A × V², we solve for A

A = FD / 0.5 × CD × β × V²

we substitute

A = 1780.8 / 0.5 × 1.855 × 0.1 × (20)²

A = 1780.8 / 3.71

A = 480 m²

Therefore, area of the parachute required 480 m² which is larger than 400 m²

c. To test the lander before launching it to Mars, it is tested on Earth where g = 9.8 m/s^2 and the atmospheric density is 1.0 kg m-3. How big should the parachute be for the terminal speed to be 20 m/s, if the mass of the lander is 480 kg?

Given that;

g = 9.8 m/s²,

β" = 1 kg/m³

v" = 20 m/s

M" = 480 kg

we know that;

FD = Fg = M"g

FD = 480 kg × 9.8 m/s² = 4704 N

from the expression; FD = 1/2 × CD × β × A × V²

A = FD / 0.5 × CD × β" × V"²

we substitute

A = 4704 / 0.5 × 1.855 × 1 × (20)²

A = 4704 / 371

A = 12.68 m²

Therefore area of the parachute should be 12.68 m²

3 0

3 years ago