Answer:
2.Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and Neptune
Explanation:
1. I do not understand what is being asked.
If you could comment with more understandable question ill do my best to help. (:
Answer:
"*The incident ray, the reflected ray and the normal to the reflection surface at the point of the incidence lie in the same plane.
*The angle which the incident ray makes with the normal is equal to the angle which the reflected ray makes to the same normal.
*The reflected ray and the incident ray are on the opposite sides of the normal."
Explanation:
![\huge\mathfrak\red{✔Answer:-}](https://tex.z-dn.net/?f=%5Chuge%5Cmathfrak%5Cred%7B%E2%9C%94Answer%3A-%7D)
Strength: able to detect planets in a wide range of orbits, as long as orbits aren't face on
Limitations: yield only planet's mass and orbital properties
Answer:
The correct answer to the following question will be Option A (I1 > I2).
Explanation:
Method for moment of inertia because of it's viewpoint including object at a mean distance "r" from the axis is,
⇒ mr²
<u>For Case 1:</u>
Let the length of a rod be "r".
The axis passes via the middle of that same rod so that the range from either the axis within each dumbbell becomes "
".
Now,
Now total moment of inertia = sum of inertial moment due to all of the dumbbell
⇒ ![l1=m(\frac{r}{2})^2+m(\frac{r}{2})^2](https://tex.z-dn.net/?f=l1%3Dm%28%5Cfrac%7Br%7D%7B2%7D%29%5E2%2Bm%28%5Cfrac%7Br%7D%7B2%7D%29%5E2)
⇒ ![\frac{mr^2}{2}](https://tex.z-dn.net/?f=%5Cfrac%7Bmr%5E2%7D%7B2%7D)
<u>For Case 2:</u>
Axis moves via one dumbbell because its range from either the axis becomes zero (0) and its impact is zero only at inertia as well as other dumbbell seems to be at a range "r" from either the axis
Now,
Total moment of inertia = moment of inertia of dumbbell at distance "r".
![l2=mr^2](https://tex.z-dn.net/?f=l2%3Dmr%5E2)
And now we can infer from this one,
⇒ ![mr^2>\frac{mr^2}{2}](https://tex.z-dn.net/?f=mr%5E2%3E%5Cfrac%7Bmr%5E2%7D%7B2%7D)
So that "I1 > I2" is the right answer.