Answer:
C. At the bottom of the circle.
Explanation:
Lets take
Radius of the circle = r
Mass = m
Tension = T
Angular speed = ω
The radial acceleration towards = a
a= ω² r
Weight due to gravity = mg
<h3>At the bottom condition</h3>
T - m g = m a
T = m ω² r + m g
<h3>At the top condition</h3>
T + m g = m a
T= m ω² r -m g
From above equation we can say that tension is grater when ball at bottom of the vertical circle.
Therefore the answer is C.
C. At the bottom of the circle.
Let and be the vectors, and let be their common magnitude.
The resultant is times larger in magnitude than either vector alone, so .
Recall the dot product identity
where is the angle between the vectors and . In the special case of , we get
Now, to get the angle between and , we have
To compute the dot product, we take the dot product of the resultant with itself.
Solve for .
Since their dot product is zero, and are perpendicular, so .
On the periodic table, the element “K’ represents Potassium. Potassium has an atomic number of 19. It has an atomic mass of 39.098. The number of protons is equal to the atomic number, so Potassium has 19 protons. To find the number of neutrons in an element, subtract the atomic mass from the atomic number. (39-19). This means that the element K has 20 neutrons. Finally, to find the number of electrons, you just take the atomic number, and you have your number of electrons. Thus, Potassium has 19 electrons.
The tricks above work for any element on the Periodic Table, so feel free to use them! Here is a little way to remen then by:
A = P = E M - A = N
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T R L M T E
O O E A O U
M T C S M T
I O T S I R
C N R C O
S O N
# N # S
S
I really hope this helps :)
A more ethical approach to athletics is sportsmanship. Under a sportsmanship model, healthy competition is seen as a means of cultivating personal honor, virtue, and character. It contributes to a community of respect and trust between competitors and in society. The goal in sportsmanship is not simply to win, but to pursue victory with honor by giving one's best effort.