We divide the thin rectangular sheet in small parts of height b and length dr. All these sheets are parallel to b. The infinitesimal moment of inertia of one of these small parts is

where

Now we find the moment of inertia by integrating from

to

The moment of inertia is

(from (-a/2) to

(a/2))
Answer:
0.010 m
Explanation:
So the equation for a pendulum period is:
where L is the length of the pendulum. In this case I'll use the approximation of pi as 3.14, and g=9.8 m\s. So given that it oscillates once every 1.99 seconds. you have the equation:

Evaluate the multiplication in front

Divide both sides by 6.28

Square both sides

Multiply both sides by m/s^2 (the s^2 will cancel out)
Now now let's find the length when it's two seconds

Divide both sides by 6.28

Square both sides

Multiply both sides by 9.8 m/s^2 (s^2 will cancel out)

So to find the difference you simply subtract
0.984 - 0.994 = 0.010 m
Use conservation of momentum ;
m1u1 + m2u2 = m1v1 + m2v2
1200×15.6 + 0 = 2700v
v = 18720/2700
v = 6.933 or ~ 7 m/s
Other than for the chemical symbol, the electron dot diagram for silicon would be the same as it was for carbon.
The reason for this is because electron dot diagrams are used to represent the electrons in the outermost, or valence, shell of an atom. In a group of the periodic table, all of the elements have the same number of valence shell electrons. This means that all elements belonging to the same group have the same electron dot diagram, except for the symbol of the element that is within the diagram.