Answer: A projectile which is fired horizontally is being constantly acted upon by acceleration due to gravity, acting vertically downwards. Hence, it does not follow a straight line path. Also Why a projectile fixed along the horizontal not follow a straight line path? Because the projectile fired horizontally is constantly acts upon by acceleration due to gravity acting vertically downwards.
Explanation:
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Answer:
n1 sin θ1 = n2 sin θ2 Snell's Law (θ1 is the angle of incidence)
sin θ2 = n1 / n2 * sin θ1
sin θ2 = 2.4 / 1.33 * sin θ1
sin θ2 = 1.80 * .407 = .734
θ2 = 47.2 deg
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
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