Answer:

Explanation:
As we know that the acceleration of a point on the rim of the disc is in two directions
1) tangential acceleration which is given as

2) Centripetal acceleration

here we know that


now we know that net linear acceleration is given as

so we have


The net force will point towards the acceleration of the object, as supported by Newton's second law.
Answer:
A. 
B. 
C. 
Explanation:
Given:
- spring constant,

- mass attached,

A)
for a spring-mass system the frequency is given as:



B)
frequency is given as:



C)
Time period of a simple harmonic motion is given as:


Answer:
Force is repulsive hence direction of force is away from wire
Explanation:
The first thing will be to draw a figure showing the condition,
Lets takeI attractive force as +ve and repulsive force as - ve and thereafter calculating net force on outer left wire due to other wires, net force comes out to be - ve which tells us that force is repulsive, hence direction of force is away from wire as shown in figure in the attachment.
1) A negatively charged ion is chloride
2) Moving from left to right, valence electrons increase by one.
3) The period number gives information about how many energy levels it has
4) Fluorine has a charge of 1–
5) Potassium and iodine form an ionic bond
The periodic table is an arrangement of elements into groups and periods based on their periodic properties.
In the periodic table, elements are arranged in groups and periods. There are 18 groups and 8 periods.
Chlorine is in group 17, there have seven outermost electrons hence the chlorine atom needs only one more electron in order to attain a stable octet. This is done by accepting one electron to form the negatively charged chloride ion.
As we move from one period to another, one extra electron is added to the outermost shell of elements. Hence, the valence electrons increases by one.
The period to which an element belongs shows you the number of shells or energy levels in the atom of that element.
Fluorine is in group 17. One electron is needed to achieve a stable octet. When an atom accepts one electron, its charge is 1–.
Bonding based on ionic charges occurs between metals and nonmetals. Potassium is a metal of group 1 and iodine is a non metal of group 17 hence they can bond together based on their ionic charges.
Learn more:brainly.com/question/23277186