Diurnal motion, apparent daily motion of the heavens from east to west in which celestial objects seem to rise and set, a phenomenon that results from the Earth's rotation from west to east. The axis of this apparent motion coincides with the Earth's axis of rotation. Retrograde motion is an APPARENT change in the movement of the planet through the sky. It is not REAL in that the planet does not physically start moving backwards in its orbit. It just appears to do so because of the relative positions of the planet and Earth and how they are moving around the Sun.

6 0

4 years ago

A slender rod 100.00cm long is used as a meter stick. Twoparallel axes which are perpendicular to the rod are considered.The fir

vredina [299]

Answer:

I / $I_{cm}$ = 1.48 , The correct answer is d

Explanation:

the moment of inertia is given by

I = ∫ r² dm

For figures with symmetry it is tabulated. In the case of a thin variation, the moment of inertia with respect to its center of mass is

$I_{cm}$ = 1/12 M L2

There is a widely used theorem, which is the parallel axis theorem, where the moment of inertia of any parallel axis, is the moment of mass inertia plus the moment of inertia of the body taken as a particle

I = $I_{cm}$ + M D²

Let's put these expressions to our case.

As the bar is one meter long its center of mass that this Enel midpoint corresponds to

$I_{cm}$ = 1/12 m L²

$I_{cm}$ = 1/12 m 1.00²

$I_{cm}$ = 8.33 10⁻² m

Let's use the parallel axes theorem for the axis that passes through x = 30 cm. The distance from the enrode masses to the axis is

D = $x_{cm}$ - 0.30

D = 0.50 - 0.30 = 0.20 m

I = $I_{cm}$ + m D²

I = 8.33 10⁻² m + m 0.2²

I = (8.33 10⁻² + 4 10⁻²) m

I = 12.33 10⁻² m

The relationship between these two moments of inertia

I / $I_{cm}$ = 12.333 10⁻² / 8.333 10⁻²

I / $I_{cm}$ = 1.48

The correct answer is d

3 0

3 years ago

The results from Rutherford's alpha particle scattering experiment led to

Brut [27]

Answer:

Explanation:

Can you help me in mathematics

7 0

3 years ago