By copying their genomes, they retain the tool kit and at the same time generate a garage full of spare parts. Gene duplication can provide the raw material for expression changes to occur, and polyploidy itself can trigger epigenetic changes
As the core collapses, the outer layers of the star are expelled. A planetary nebula is formed by the outer layers. The core remains as a white dwarf and eventually cools to become a black dwarf. ... Like low-mass stars, high-mass stars are born in nebulae and evolve and live in the Main Sequence
hydrogen shell burning - outer layers swell. Red Giant Branch - helium ash core compresses - increased hydrogen shell burning. First Dredge Up - expanding atmosphere cools star - stirs carbon, nitrogen and oxygen upward - star heats up.
To solve this exercise it is necessary to take into account the concepts related to Tensile Strength and Shear Strenght.
In Materials Mechanics, generally the bodies under certain loads are subject to both Tensile and shear strenghts.
By definition we know that the tensile strength is defined as

Where,
Tensile strength
F = Tensile Force
A = Cross-sectional Area
In the other hand we have that the shear strength is defined as

where,
Shear strength
Shear Force
Parallel Area
PART A) Replacing with our values in the equation of tensile strenght, then

Resolving for F,

PART B) We need here to apply the shear strength equation, then



In such a way that the material is more resistant to tensile strength than shear force.
Answer:
d = 68.5 x 10⁻⁶ m = 68.5 μm
Explanation:
The complete question is as follows:
An optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is 1.70m from the slits. If the slits are illuminated with coherent light of wavelength 633 nm, how far apart should the slits be?
The answer can be given by using the formula derived from Young's Double Slit Experiment:

where,
d = slit separation = ?
λ = wavelength = 633 nm = 6.33 x 10⁻⁷ m
L = distance from screen (detector) = 1.7 m
y = distance between bright fringes = 15.7 mm = 0.0157 m
Therefore,

<u>d = 68.5 x 10⁻⁶ m = 68.5 μm</u>
Answer:
The distance is 55.636 billion miles, or 528.2 AU.
Explanation:
Since the distance from the Sun to Neptune is 2.7818 billion miles, the distance from the Sun to Planet Nine would be 20 times that, which is:

or 55.636 billion miles.
Since 1 astronomical unit (AU) is 93 million miles, that distance is also:
