Answer:
<em>Billow clouds provide a visible signal to aviation interests of potentially dangerous turbulent sky since they indicate instability in air currents.</em>
Explanation:
Billow clouds are created in regions that are not stable in a meteorological sense. They are frequently present in places with air flows, and have marked vertical shear and weak thermal separation and inversion (colder air stays on top of warmer air). Billow clouds are formed when two air currents of varying speeds meet in the atmosphere. They create a stunning sight that looks like rolling ocean waves. Billow clouds have a very short life span of minutes but they provide a visible signal to aviation interests of potentially dangerous turbulent sky since they indicate instability in air currents, which although may not affect us on the ground but is a concern to aircraft pilots. The turbulence due to the Billow wave is the only logical explanation for the loss of 500 m in altitude of the plane.
Answer:

Explanation:
We can solve this problem by using Newton's second law of motion, which states that the net force acting on an object is equal to the product between its mass and its acceleration:

where
F is the net force on the object
m is its mass
a is its acceleration
In this problem:
F = 40 N is the force on the object
m = 2 kg is its mass
Therefore, the acceleration of the object is

Arterie
cwamus
capillaries
phloprant
veins
piscas
The apparent weight of a 1.1 g drop of water is 4.24084 N.
<h3>
What is Apparent Weight?</h3>
- According to physics, an object's perceived weight is a characteristic that describes how heavy it is. When the force of gravity acting on an object is not counterbalanced by a force of equal but opposite normality, the apparent weight of the object will differ from the actual weight of the thing.
- By definition, an object's weight is equal to the strength of the gravitational force pulling on it. It follows that even a "weightless" astronaut in low Earth orbit, with an apparent weight of zero, has almost the same weight that he would have if he were standing on the ground; this is because the gravitational pull of low Earth orbit and the ground are nearly equal.
Solution:
N = Speed of rotation = 1250 rpm
D = Diameter = 45 cm
r = Radius = 22.5 cm
M = Mass of drop = 1.1 g
Angular speed of the water = 


Apparent weight is given by


= 4.24084 N
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Question:
The spin cycle of a clothes washer extracts the water in clothing by greatly increasing the water's apparent weight so that it is efficiently squeezed through the clothes and out the holes in the drum. In a top loader's spin cycle, the 45-cm-diameter drum spins at 1250 rpm around a vertical axis. What is the apparent weight of a 1.1 g drop of water?
I think the correct answer from the choices listed above would be the last option. It is the chemicals in the core of the star that cannot be determined from the spectrum of a star. Spectrum shows the different classification of the stars depending on their spectral characteristics. It usually involves the light, the wavelength and the distance.