Answer:
Angular velocity is same as frequency of oscillation in this case.
ω = 
x ![[\frac{L^{2}}{mK}]^{3/14}](https://tex.z-dn.net/?f=%5B%5Cfrac%7BL%5E%7B2%7D%7D%7BmK%7D%5D%5E%7B3%2F14%7D)
Explanation:
- write the equation F(r) = -K
with angular momentum <em>L</em>
- Get the necessary centripetal acceleration with radius r₀ and make r₀ the subject.
- Write the energy of the orbit in relative to r = 0, and solve for "E".
- Find the second derivative of effective potential to calculate the frequency of small radial oscillations. This is the effective spring constant.
- Solve for effective potential
- ω = x
Answer:
a) 2.41 km
b) 38.8°
Questions c and d are illegible.
Explanation:
We can express the displacements as vectors with origin on the point he started (0, 0).
When he traveled south he moved to (-3, 0).
When he moved east he moved to (-3, x)
The magnitude of the total displacement is found with Pythagoras theorem:
d^2 = dx^2 + dy^2
Rearranging:
dy^2 = d^2 - dx^2
The angle of the displacement vector is:
cos(a) = dx/d
a = arccos(dx/d)
a = arccos(3/3.85) = 38.8°
The distance travel is 69.5 meters.
<u>Explanation:</u>
Given datas are as follows
Speed = 27.8 meters / second
Time = 2.5 seconds
The formula to calculate the speed using distance and time is
Speed = Distance ÷ Time (units)
Then Distance = Speed × Time (units)
Distance = (27.8 × 2.5) meters
Distance = 69.50 meters
Therefore the distance travelled is 69.50 meters.
Answer:
Sagittarius
Explanation:
The center of the Milky Way galaxy lies in the direction of the Sagittarius constellation, about 26,000 light-years away.
