WarSun

While constantly circling around a center, the rock will always have two vectors acting on it (forces in two direction). 1. towards the center 2. at tangent or perpendicular to the string The direction of these two forces keep changing as the rock travels along the circular path. Taking these two forces into account, the final or resultant force would be along the circumference of the circle. In your case, when the string is cut, the force acting towards the center is removed and the only force acting on the rock would be the force that is perpendicular to the string, ( which is upward if seen from above ). Another example would be- if the sun's gravitational pull suddenly disappears, then earth would continuously keep moving towards the tangent direction of its orbit.

BF3 has 6 normal modes of vibration (A1', A2'', 2E'), which are made up of SALCs of 3 stretches and 3 bends. The observed bands around 1450-1500 cm^-1 in the spectrum of BF3 are assigned to asymmetric B-F stretches (E') shows marked splitting because of 10-B and 11-B abundance in natural boron. The Raman band at 888 cm^-1 assigned to the A1' symmetric B-F stretch shows no isotopic effects. Explain why only the asymmetric stretches show isotopic effects. I can't figure out why only the asymmetric stretches show isotopic effects. I think that since the frequency of symmetric stretch is the same it wouldn't show splitting. During asymmetric stretch there are two different frequencies. Thus, showing splitting. But I don't know if the answer is logical and how do I relate it to isotopic effect. Thanks.