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:
21.35 cm^3
Explanation:
let the volume at the surface of fresh water is V.
The volume at a depth of 100 m is V' = 2 cm^3
temperature remains constant.
density of water, d = 1000 kg/m^3
Pressure at the surface of fresh water is atmospheric pressure,
P = Po = 1.013 x 10^5 N/m^2
The pressure at depth 100 m is P' = Po + hdg
P' = 
P' = 10.813 x 10^5 N/m^2
Use the Boyle's law
P V = P' V'
V = 21.35 cm^3
Thus, the volume of air bubble at the surface of fresh water is 21.35 cm^3.
Answer:
Chicken and peanut butter lol
Explanation:
Answer:
Add an arrow above the symbol p to show it is a vector. Sometimes it is italicized in textbooks.
Explanation:
