C.) Laser. the light from the laser reflects off the shiny surface as the CD rotates
Answer: m∠P ≈ 46,42°
because using the law of sines in ΔPQR
=> sin 75°/ 4 = sin P/3
so ur friend is wrong due to confusion between edges
+) we have: sin 75°/4 = sin P/3
=> sin P = sin 75°/4 . 3 = (3√6 + 3√2)/16
=> m∠P ≈ 46,42°
Explanation:
Well according to Newton’s first law of motion, a body will remain in the state of rest or linear motion provided that an *external force* has been applied. So no, a force doesn’t need to keep a body to remain in linear motion, because F=ma, during uniform linear motion velocity is constant, hence acceleration is zero, so F=0
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