Answer:
Explanation:
Let the velocity be v
Total energy at the bottom
= rotational + linear kinetic energy
= 1/2 Iω² + 1/2 mv² ( I moment of inertia of shell = mr² )
= 1/2 mr²ω² + 1/2 mv² ( v = ω r )
= 1/2 mv² +1/2 mv²
= mv²
mv² = mgh ( conservation of energy )
v² = gh
v = √gh
= √9.8 x 1.8
= 4.2 m /s
The weight of the meterstick is:

and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance

from the pivot.
The torque generated by the weight of the meterstick around the pivot is:

To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:

from which we find the value of d2:

So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
Option 3 is the most reasonable
I hope this helped <3
Please give brainliest :)
Answer:
The units of the orbital period P is <em>years </em> and the units of the semimajor axis a is <em>astronomical units</em>.
Explanation:
P² = a³ is the simplified version of Kepler's third law which governs the orbital motion of large bodies that orbit around a star. The orbit of each planet is an ellipse with the star at the focal point.
Therefore, if you square the year of each planet and divide it by the distance that it is from the star, you will get the same number for all the other planets.
Thus, the units of the orbital period P is <em>years </em> and the units of the semimajor axis a is <em>astronomical units</em>.