Answer:
this is a no brainer
Explanation:
As air pressure in an area increases, the density of the gas particles in that area increases.
Electron configurations:
Ge: [Ar] 3d10 4s2 4p2 => 6 electrons in the outer shell
Br: [Ar] 3d10 4s2 4p5 => 7 electrons in the outer shell
Kr: [Ar] 3d10 4s2 4p6 => 8 electrons in the outer shell
The electron affinity or propension to attract electrons is given by the electronic configuration. Remember that the most stable configuration is that were the last shell is full, i.e. it has 8 electrons.
The closer an atom is to reach the 8 electrons in the outer shell the bigger the electron affinity.
Of the three elements, Br needs only 1 electron to have 8 electrons in the outer shell, so it has the biggest electron affinity (the least negative).
Ge: needs 2 electrons to have 8 electrons in the outer shell, so it has a smaller (more negative) electron affinity than Br.
Kr, which is a noble gas, has 8 electrons and is not willing to attract more electrons at all, the it has the lowest (more negative) electron affinity of all three to the extension that really the ion is so unstable that it does not make sense to talk about a number for the electron affinity of this atom.
Answer:
56.7°
Explanation:
Imagine a rectangle triangle.
The triangle has 3 sides.
One side is the height of the tower, let's name it A.
Another side is the distance from the base of the tower to the point where the waire touches the ground. Let's name that B.
Sides A and B are perpendicular.
The other side is the length of the wire. Let's name it C.
From trigonometry we know that:
cos(a) = B / C
Where a is the angle between B anc C, between the wire and the ground.
Therefore
a = arccos(B/C)
a = arccos(552/1005) = 56.7°
Answer:
See Explanation
Explanation:
The relationship between angle of an incline and the acceleration of an object moving down the incline.
As the angle of an incline increases, so does the acceleration of the body moving down the incline increases, resolving the force acting on an inclined object
Parallel force = mgsin, perpendicular = mgcosΘ
With th weigh component 'mg' of the parallel force accounting for the acceleration of the body down the incline.
mgsinΘ = ma
Fnet = ma
B.) From Fnet = ma
Fnet = ma
a = Fnet / m
Where Fnet = Net force = mgsinΘ, a = acceleration
