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The Position Vector of an Object of Mass 0

question 14

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The position vector of an object of mass 0.50 kg subject to a constant force is given by The position vector of an object of mass 0.50 kg subject to a constant force is given by   = (at<sup>2</sup>+bt)    + (ct<sup>2</sup>+dt)    + (et<sup>2</sup>+ft)    , where a = 2.0 m/s<sup>2</sup>, b = 3.0 m/s, c = 2.5 m/s<sup>2</sup>, d = -2.0 m/s, e = 1.0 m/s<sup>2</sup>, and f = 4.0 m/s. What is the angular momentum of the object about the origin at t = 2.0 s? A) (24   - 122   - 89   )  kg∙m<sup>2</sup>/s B) (-24   + 10   + 23   )  kg∙m<sup>2</sup>/s C) (25   - 14   + 20   )  kg∙m<sup>2</sup>/s D) (25   + 14   + 20   )  kg∙m<sup>2</sup>/s E) (24   + 122   + 23   )  kg∙m<sup>2</sup>/s = (at2+bt) The position vector of an object of mass 0.50 kg subject to a constant force is given by   = (at<sup>2</sup>+bt)    + (ct<sup>2</sup>+dt)    + (et<sup>2</sup>+ft)    , where a = 2.0 m/s<sup>2</sup>, b = 3.0 m/s, c = 2.5 m/s<sup>2</sup>, d = -2.0 m/s, e = 1.0 m/s<sup>2</sup>, and f = 4.0 m/s. What is the angular momentum of the object about the origin at t = 2.0 s? A) (24   - 122   - 89   )  kg∙m<sup>2</sup>/s B) (-24   + 10   + 23   )  kg∙m<sup>2</sup>/s C) (25   - 14   + 20   )  kg∙m<sup>2</sup>/s D) (25   + 14   + 20   )  kg∙m<sup>2</sup>/s E) (24   + 122   + 23   )  kg∙m<sup>2</sup>/s + (ct2+dt) The position vector of an object of mass 0.50 kg subject to a constant force is given by   = (at<sup>2</sup>+bt)    + (ct<sup>2</sup>+dt)    + (et<sup>2</sup>+ft)    , where a = 2.0 m/s<sup>2</sup>, b = 3.0 m/s, c = 2.5 m/s<sup>2</sup>, d = -2.0 m/s, e = 1.0 m/s<sup>2</sup>, and f = 4.0 m/s. What is the angular momentum of the object about the origin at t = 2.0 s? A) (24   - 122   - 89   )  kg∙m<sup>2</sup>/s B) (-24   + 10   + 23   )  kg∙m<sup>2</sup>/s C) (25   - 14   + 20   )  kg∙m<sup>2</sup>/s D) (25   + 14   + 20   )  kg∙m<sup>2</sup>/s E) (24   + 122   + 23   )  kg∙m<sup>2</sup>/s + (et2+ft) The position vector of an object of mass 0.50 kg subject to a constant force is given by   = (at<sup>2</sup>+bt)    + (ct<sup>2</sup>+dt)    + (et<sup>2</sup>+ft)    , where a = 2.0 m/s<sup>2</sup>, b = 3.0 m/s, c = 2.5 m/s<sup>2</sup>, d = -2.0 m/s, e = 1.0 m/s<sup>2</sup>, and f = 4.0 m/s. What is the angular momentum of the object about the origin at t = 2.0 s? A) (24   - 122   - 89   )  kg∙m<sup>2</sup>/s B) (-24   + 10   + 23   )  kg∙m<sup>2</sup>/s C) (25   - 14   + 20   )  kg∙m<sup>2</sup>/s D) (25   + 14   + 20   )  kg∙m<sup>2</sup>/s E) (24   + 122   + 23   )  kg∙m<sup>2</sup>/s , where a = 2.0 m/s2, b = 3.0 m/s, c = 2.5 m/s2, d = -2.0 m/s, e = 1.0 m/s2, and f = 4.0 m/s. What is the angular momentum of the object about the origin at t = 2.0 s?


Definitions:

Marginal Product

The additional output that is produced by employing one more unit of a particular input, while holding other inputs constant.

Fixed Cost

A financial outlay that is unaffected by variations in the production or sales levels of goods and services.

Marginal Cost

The expense associated with creating an extra unit of a product or service.

Total Cost

The complete cost of producing a specific quantity of output, including both fixed and variable costs.

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