Use your periodic table to answer the following question.
1 answer:
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Answer:
a) 2.41 km
b) 38.8°
Questions c and d are illegible.
Explanation:
We can express the displacements as vectors with origin on the point he started (0, 0).
When he traveled south he moved to (-3, 0).
When he moved east he moved to (-3, x)
The magnitude of the total displacement is found with Pythagoras theorem:
d^2 = dx^2 + dy^2
Rearranging:
dy^2 = d^2 - dx^2
The angle of the displacement vector is:
cos(a) = dx/d
a = arccos(dx/d)
a = arccos(3/3.85) = 38.8°
(a) The angular acceleration of the wheel is given by
where
and 
are the initial and final angular speed of the wheel, and t the time.
In our problem, the initial angular speed is zero (the wheel starts from rest), so the angular acceleration is
(b) The wheel is moving by uniformly rotational accelerated motion, so the angle it covered after a time t is given by
where
is the initial angular speed. So, the angle covered after a time t=3.07 s is
where
In our problem, the initial angular speed is zero (the wheel starts from rest), so the angular acceleration is
(b) The wheel is moving by uniformly rotational accelerated motion, so the angle it covered after a time t is given by
where
The answer is 107 degrees. The geometric shape for ammonia is Trigonal Pyramidal, even though its electron geometry is “Tetrahedral”. This is because ammonia has a lone pair of electrons that occupy its space like the other 3 hydrogens in the geometric structure.
The answer 180 degrees. This is because of the linear geometric structure of carbon dioxide. The oxygen atom is on either side of the carbon atom, each is bound by a double covalent bond. All the atoms are involved in the bond and there are no one pair electrons.
The answer is tetrahedral geometry. This is because all the 4 valence electrons of the carbon are involved in a bond with a hydrogen atom. The angles in a tetrahedral geometric arrangement, such as in methane, is 109.5 degrees, where the hydrogen atoms are as far apart, from each other, as possible .
