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neonofarm [45]
4 years ago
15

An ideal air mass source region must meet two criteria. First, it must be an extensive and physically uniform area. The second c

riterion is that the area is characterized by a general __________ of atmospheric circulation so that air will stay over the region long enough to come to some measure of equilibrium with the surface. A. agitation B. stagnation C. saturation D. increase
Physics
1 answer:
Usimov [2.4K]4 years ago
7 0

Answer:

The correct option is : B. stagnation

Explanation:

Air mass is a volume of air, which is described on the basis of water vapor content and temperature. It has the ability to adapt to the characteristics of surface below it and can travel thousands of miles from its source region.

An ideal source region of an air mass must fulfill the following two criteria:

1. It should be a large and uniform area.

2. the area should have stagnant circulation of atmospheric air .

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A bus contains a 1440 kg flywheel (a disk that has a 0.63 m radius) and has a total mass of 10200 kg. Calculate the angular velo
CaHeK987 [17]

Answer:\omega =93.51 rad/s

Explanation:

Given

mass of Flywheel m_1=1440 kg

mass of bus m_b=10200 kg

radius of Flywheel r=0.63 m

final speed of bus v=21 m/s

Conserving Energy i.e.

0.9(Rotational Energy of Flywheel)= change in Kinetic Energy of bus

Let \omegabe the angular velocity of Flywheel

0.9\cdot \frac{I\omega ^2}{2}=\frac{m_bv^2}{2}

I=moment\ of\ Inertia =mr^2=1440\cdot 0.63^2=571.536 kg-m^2

0.9\cdot \frac{571.536\cdot \omega ^2}{2}=\frac{10200\cdot 21^2}{2}

\omega ^2=21^2\times \frac{10200}{0.9\times 571.536}

\omega =21\times 4.45=93.51 rad/s

8 0
3 years ago
The product of charge through, and potential across, an electrical device is:
melomori [17]
The option is Work.

The product of charge and potential is equal to the energy. Adn, as we know work is related to energy as the capacity to do work.  

Alos, because, Potential is given as, V = E/q
or E = Vq
Thus, t<span>he product of charge through, and potential across, an electrical device is:work
</span>  
5 0
3 years ago
Read 2 more answers
if vector u has lenght 70 and direction 40 degrees, and vector v has length 85 and direction 335 degrees what is the length and
Anastaziya [24]

Answer:

Magnitude of resultant = 131.15

Direction of resultant = 3.97°

Explanation:

||u|| = 70

θ = 40°

\vec{u}_x=||u||cos\theta \\\Rightarrow \vec{u}_x=70cos40=53.62

\vec{u}_y=||u||sin\theta \\\Rightarrow \vec{u}_y=70sin40=44.99

||v|| = 85

θ = 335°

\vec{v}_x=||v||cos\theta \\\Rightarrow \vec{v}_x=85cos335=77.03

\vec{v}_y=||v||sin\theta \\\Rightarrow \vec{v}_y=85sin335=-35.92

Resultant

R=\sqrt{R_x^2+R_y^2}\\\Rightarrow R=\sqrt{(\vec{u}_x+\vec{v}_x)^2+(\vec{u}_y+\vec{v}_y)^2}\\\Rightarrow R =\sqrt{(70cos40+85cos335)^2+(70sin40+85sin335)^2}\\\Rightarrow R =131.15

\theta=tan^{-1}\frac{R_y}{R_x}\\\Rightarrow \theta=tan^{-1}\frac{70sin40+85sin335}{70cos40+85cos335}\\\Rightarrow \theta=tan^{-1}0.069=3.97^{\circ}

Magnitude of resultant = 131.15

Direction of resultant = 3.97°

4 0
3 years ago
Which fundamental force has a small range and is always an attractive force?
erica [24]

Answer:

gravitational force is a fundamental force and also , it does have a small range and is always an attractive force.

Explanation:

5 0
3 years ago
Calculate the electric potential at point A, the middle of the rectangle, and at point B, the middle of the right-hand side of t
dsp73

Answer:

With the help of formula.

Explanation:

We can calculate the electric potential of any point through the formula of electric potential which is given below.

Electric potential =  Coulomb constant x charge/ distance of separation.

Symbolically it can be written as,  V = k q/ r where

V = electric potential  

k = Coulomb constant

q = charge

r = distance of separation

If we have all these data, we can simply put the data in the formula and we will get the value of electric potential.

5 0
3 years ago
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