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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?


A) (24 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 - 122
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 - 89
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 ) kg∙m2/s
B) (-24 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 + 10
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 + 23
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 ) kg∙m2/s
C) (25 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 - 14
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 + 20
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 ) kg∙m2/s
D) (25 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 + 14
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 + 20
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 ) kg∙m2/s
E) (24 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 + 122
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 + 23
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 ) kg∙m2/s

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