Answer:
Atomic Size and Mass:
convert given density to kg/m^3 = 8900kg/m^3 2) convert to moles/m^3 (kg/m^3 * mol/kg) = 150847 mol/m^3 (not rounding in my actual calculations) 3) convert to atoms/m^3 (6.022^23 atoms/mol) = 9.084e28 atoms/m^3 4) take the cube root to get the number of atoms per meter, = 4495309334 atoms/m 5) take the reciprocal to get the diameter of an atom, = 2.2245e-10 m/atom 6) find the mass of one atom (kg/mol * mol/atoms) = 9.7974e-26 kg/atom Young's Modulus: Y=(F/A)/(dL/L) 1) F=mg = (45kg)(9.8N/kg) = 441 N 2) A = (0.0018m)^2 = 3.5344e-6 m^2 3) dL = 0.0016m 4) L = 2.44m 5) Y = 1.834e11 N/m^2 Interatomic Spring Stiffness: Ks,i = dY 1) From above, diameter of one atom = 2.2245e-10 m 2) From above, Y = 1.834e11 N/m^2 3) Ks,i = 40.799 N/m (not rounding in my actual calculations) Speed of Sound: v = ωd 1) ω = √(Ks,i / m,a) 2) From above, Ks,i = 40.799 N/m 3) From above, m,a = 9.7974e-26 kg 4) ω=2.0406e13 N/m*kg 5) From above, d=2.2245e-10 m 6) v=ωd = 4539 m/s (not rounding in actual calculations) Time Elapsed: 1) length sound traveled = L+dL = 2.44166 m 2) From above, speed of sound = 4539 m/s 3) T = (L+dL)/v = 0.000537505 s
Answer:
11300 kgm3
Hope this helps
Explanation:
The left side of the periodic table has elements that have less number of electrons in the valence shell.
These elements loose electrons easily.These elements appear as metals or metalloids in nature.These are hard solids.Their inter molecular forces are very strong.
The right side of the periodic table has elements that have more number of electrons in the valence shell.
These elements gain electrons easily.These elements appear as non metals most of which are gases.Their inter molecular forces are weak.
As a reference, consider the line from the point perpendicular to the mirror.
That direction is called 'normal' to the mirror.
The ray on the right leaves the point traveling 5° to the right of the normal,
and leaves the mirror on a path that's 10° to the right of the normal.
The ray on the left leaves the point traveling 5° to the left of the normal,
and leaves the mirror on a path that's 10° to the left of the normal.
The angle between the two rays after they leave the mirror is 20° .
Frankly, Charlotte, if there were more than 5 points available for this answer,
I'd seriously consider giving you a drawing too.
