I believe the correct answer from the choices listed above is the last option. If the volatility of X is higher than that of Y, then <span>Y’s molecules experience stronger London dispersion forces than X’s molecules. All molecules has london dispersion forces. Also, the stronger the bond, the harder it is to volatilize. Hope this answers the question.</span>
<h2>You input potential (stored) energy into the rubber band system when you stretched the rubber band back. Because it is an elastic system, this kind of potential energy is specifically called elastic potential energy.</h2><h2>Hope it helps..</h2>
Answer:
It corresponds to a distance of 100 parsecs away from Earth.
Explanation:
The angle due to the change in position of a nearby object against the background stars it is known as parallax.
It is defined in a analytic way as it follows:

Where d is the distance to the star.
(1)
Equation (1) can be rewritten in terms of d:
(2)
Equation (2) represents the distance in a unit known as parsec (pc).
The parallax angle can be used to find out the distance by means of triangulation. Making a triangle between the nearby star, the Sun and the Earth (as is shown in the image below), knowing that the distance between the Earth and the Sun (150000000 Km), is defined as 1 astronomical unit (1AU).
For the case of (
):


Hence, it corresponds to a distance of 100 parsecs away from Earth.
<em>Summary:</em>
Notice how a small parallax angle means that the object is farther away.
Key terms:
Parsec: Parallax of arc second
Answer:
Normal stress = 66/62.84 = 1.05kips/in²
shearing stress = T/2 = 0.952/2 = 0.476 kips/in²
Explanation:
A steel pipe of 12-in. outer diameter d₂ =12in d₁= 12 -4in = 8in
4 -in.-thick
angle of 25°
Axial force P = 66 kip axial force
determine the normal and shearing stresses
Normal stress б = force/area = P/A
= 66/ (П* (d₂²-d₁²)/4
=66/ (3.142* (12²-8²)/4
= 66/62.84 = 1.05kips/in²
Tangential stress T = force* cos ∅/area = P/A
= 66* cos 25/ (П* (d₂²-d₁²)/4
=59.82/ (3.142* (12²-8²)/4
= 59.82/62.84 = 0.952kips/in²
shearing stress = tangential stress /2
= T/2 = 0.952/2 = 0.476 kips/in²