Question

if particle project with speed a and angle theta with horizontal then find locus of particle.
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Answer 1

Answer:

of the particle in the lowest quantumparticle is moving in a three dimensions potential V(x)=1/2mw^2(2x^2+y^2+4z^2). if the mass of the particle is m then what is the energy of the particle in the lowest quantumif particle project with speed a and angle theta with horizontal then find locus of particle.

Explanation:

particle is moving in a three dimensions potential V(x)=1/2mw^2(2x^2+y^2+4z^2). if the mass of the particle is m then what is the energy of the particle in the lowest quantumparticle is moving in a three dimensions potential V(x)=1/2mw^2(2x^2+y^2+4z^2). if the mass of the particle is m then what is the energy of the particle in the lowest quantumparticle is moving in a three dimensions potential V(x)=1/2mw^2(2x^2+y^2+4z^2). if the mass of the particle is m then what is the energy of the particle in the lowest quantumif particle project with speed a and angle theta with horizontal then find locus of particle.