<h2>
The current rotational period of that star is 10.01 hours.</h2>
Explanation:
Given that,
Initial angular velocity of the star, 
It decelerates, final angular speed, 
Deceleration, 
It is not required to use the rotational kinematics formula. The angular velocity in terms of time period is given by :
T is current rotational period of that star
T = 36036.03 second
or
1 hour = 3600 seconds
So, T = 10.01 hours
So, the current rotational period of that star is 10.01 hours. Hence, this is the required solution. Hence, this is the required solution.
Newton's second law of motion pertains to the behavior of objects for which all existing forces are not balanced. The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object. The acceleration of an object depends directly upon the net force acting upon the object, and inversely upon the mass of the object. As the force acting upon an object is increased, the acceleration of the object is increased. As the mass of an object is increased, the acceleration of the object is decreased.
And the correct answer is A) on the surface of the moon; because Newton's second law provides the explanation for the behavior of objects upon which the forces do not balance. The law states that unbalanced forces cause objects to accelerate with an acceleration that is directly proportional to the net force and inversely proportional to the mass.
So the correct answer is A) on the surface of the moon
Hope I helped.
Answer:
Electromagnetic radiation is an electric and magnetic disturbance traveling through space at the speed of light (2.998 × 108 m/s). It contains neither mass nor charge but travels in packets of radiant energy called photons, or quanta.
<h2>Amoeba / Unicellular</h2><h2>Segmented worm / Earthworm</h2><h2>Unsegment worm / Tapeworm</h2><h2>Snail / Molluscs</h2><h2>Butterfly / A pair of antenna</h2><h2 /><h3><em>Unicellular: </em><u><em>aboema</em></u><em>: a </em><u><em>one-celled</em></u><em>, microscopic organism belonging to any of several families of rhizopods that move and feed using pseudopodia and reproduce by fission</em></h3><h3><em /></h3><h3><em>Segmented worms: segmented worms include the common </em><u><em>earthworm</em></u><em> and leeches.</em></h3><h3><em /></h3><h3><u><em>Unsegented worms:</em></u><em> unsegmented Worms Phylum Platyhelminthes & Nematoda. Worms. Worms are divided into three different phyla: Phylum Platyhelminthes, the flatworms. These include marine flatworms, flukes, and </em><u><em>tapeworms</em></u><em>.</em></h3><h3><em /></h3><h3><u><em>Molluscs</em></u><em>: molluscs examples: – </em><u><em>snails</em></u><em>, slugs, limpets, whelks, conchs, periwinkles, etc. Class Bivalvia – clams, oysters, mussels, scallops, cockles, shipworms, etc. The Class Scaphopoda contains about 400 species of molluscs called tooth or tusk shells, all of which are marine.</em></h3><h3><em /></h3><h3><u><em>Antennas</em></u><em>: </em><u><em>Nearly all insects have a pair of antennae</em></u><em> on their heads. They use their antennae to touch and smell the world around them. ... Insects are the only arthropods that have wings, and the wings are always attached to the thorax, like the legs.</em></h3>
Answer:
The angle for the forward Mach line is 19.47°
The angle for the rearward Mach line is 5.21°
Explanation:
From table A-1 (Modern Compressible Flow: with historical perspective):
(M₁ = 3)
If Po₁ = Po₂
Table A-1:
Table A-5:
v₁ = 49.76°
μ₁ = 19.47°
v₂ = 60.55°
μ₂ = 16°
θ = 60.55 - 49.76 = 10.79°
The angle for the forward Mach line is:
μ₁ = 19.47°
The angle for the rearward Mach line is:
θr = μ₂ - θ = 16 - 10.79 = 5.21°
