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kobusy [5.1K]
2 years ago
14

In separate experiments, a large number of particles (all with the same charge but with a wide variety of masses, speeds, and sp

in angular momentum) are fired into a region containing a uniform magnetic field. The velocities of the particles are all perpendicular to the field. What do the particles that move in orbits of the same radius have in common
Physics
1 answer:
melomori [17]2 years ago
4 0

The particles that move in orbits of the same radius have the same momentum.

<h3>Orbital angular momentum:</h3>

A point particle's three-dimensional angular momentum is traditionally represented by the pseudovector r p, which is the cross product of the particle's position vector r (relative to some origin) and momentum vector, which in Newtonian physics is denoted by p = mv.

L = mrV_{prep} = mr²w is the particle's orbital angular momentum in units of magnitude. The part of the particle's velocity that is here perpendicular to the axis of rotation is designated as V_{perp.} The right-hand rule indicates the direction of the angular momentum. In isolated systems, the angular momentum is conserved.

Learn more about momentum here:

brainly.com/question/25849204

#SPJ4

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julia-pushkina [17]

Hence the expression of ω in terms of m and k is

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Given the expressions;

T_s = 2 \pi \sqrt{\frac{m}{k} } \ and \ T_s = \frac{2 \pi}{\omega}

Equating both expressions we will have;

2 \pi \sqrt{\frac{m}{k} }  = \frac{2 \pi}{\omega}

Divide both equations by 2π

\frac{2 \pi\sqrt{\frac{m}{2 \pi} } }{2 \pi}=\frac{\frac{2 \pi}{\omega} }{2\pi}\\\sqrt{\frac{m}{2 \pi} } = \frac{1}{\omega}\\

Square both sides

(\sqrt{\frac{m}{k} } )^2 = (\frac{1}{\omega} )^2\\\frac{m}{k} = \frac{1}{\omega ^2} \\\omega ^2 = \frac{k}{m}

Take the square root of both sides

\sqrt{\omega ^2} =\sqrt{\frac{k}{m} } \\\omega = \sqrt{\frac{k}{m}

Hence the expression of ω in terms of m and k is

\omega = \sqrt{\frac{k}{m}

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Mark me Brianliest please:)
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