30. (a) The acceleration vector is obtained by dividing the force vector by the (scalar) mass: a→= F→/m= (3.00 m/s2)i^– (4.00 m/s2)j^+ (2.00 m/s2)k^.(b) Use of Eq. 11-18 leads directly toL→= (42.0 kg.m2/s)i^+ (24.0 kg.m2/s)j^+ (60.0 kg.m2/s)k^.(c) Similarly, the torque is rFτ=×GGG= (–8.00 N.m)i^– (26.0 N.m)j^– (40.0 N.m)k^.(d) We note (using the Pythagorean theorem) that the magnitude of the velocity vector is
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