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A 1.40-kg particle moves in the xy plane with a velocity of v with arrow = (4.00 î − 3.30 ĵ) m/s. Determine the angular momentum of the particle about the origin when its position vector is r with arrow = (1.50 î + 2.20 ĵ) m.

Respuesta :

Karabo99

Answer:

the angular momentum of the paticle is 8.4i - 10.2j.

Explanation:

the angular momentum is given by:

L = m×v×r

  = (1.40)×(4.00i - 3.30j)×(1.50i + 2.20j)

   = (1.40)×[(4.00×1.50)i - (3.30×2.20)j]

   = (1.40)×(6i - 7.26j)

   = 8.4i - 10.2j

Therefore, the angular momentum of the paticle is 8.4i - 10.2j.

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