Answer:
Tendons connect muscle to bone. These tough, yet flexible, bands of fibrous tissue attach the skeletal muscles to the bones they move. Essentially, tendons enable you to move; think of them as intermediaries between muscles and bones.
Hope this helps! (:
I believe it would be a musical note
To solve this problem it is necessary to apply the concepts related to the Third Law of Kepler.
Kepler's third law tells us that the period is defined as

The given data are given with respect to known constants, for example the mass of the sun is

The radius between the earth and the sun is given by

From the mentioned star it is known that this is 8.2 time mass of sun and it is 6.2 times the distance between earth and the sun
Therefore:


Substituting in Kepler's third law:






Therefore the period of this star is 3.8years
Answer:
1742.24106 revolutions per day
Explanation:
v = Velocity
d = Diameter = 1.1 km
r = Radius = 
g = Acceleration due to gravity = 9.81 m/s²
g = 0.9 g
The centrifugal force will balance the gravitational force


The rotation speed is 1742.24106 revolutions per day