Answer:
The frequency of oscillation of the simple pendulum is 0.49 Hz.
Explanation:
Given that,
Mass of the simple pendulum, m = 0.35 kg
Length of the string to which it is attached, l = 1 m
We need to find the frequency of oscillation. The frequency of oscillation of the simple pendulum is given by :

So, the frequency of oscillation of the simple pendulum is 0.49 Hz. Hence, this is the required solution.
Answer:
Explanation:
Given that,
Current in loops are
i1 = 12A
i2 = 20A
The loops are 3.4cm apart
The magnetic field at the center is found to be zero, so when want to find the radius of bigger loop
Magnetic Field is given as
B= μoi/2πr
Where,
μo is a constant = 4π×10^-7 Tm/A
r is the distance between the two wires
i is the current in the wires
B is the magnetic field
NOTE
Field due to large loop should be equal to the smaller loop.
B1 = B2
μo•i1 / 2π•r1 = μo•i2 / 2π•r2
Then, μo, 2π cancels out, so we have
i1 / r1 = i2 / r2
Make r2 subject of formula
i1•r2 = i2•r1
r2 = i2•r1 / i2
r2 = 20×3.4/12
r2 = 5.67cm
The radius of the bigger loop is 5.67cm.
I dont know what the statements are but concave lens are thinner in the middle which cause light to diverge or scatter
Answer:
C. Code of ethics
Explanation:
The code of ethics of a company is the document where all the values that the members of the company should live and obbey, they are often guideliness on how to act and beheave and how to treat customers, providers and co-workers. IT also has all of the standards for all managers ethical responsabilities, and ethical responsabilities of the general of the employees of the company.
Answer:
at point A the ball possess pontetial energy , point B kinetic energy then point C pontetial energy