a 10 kg block reaches a point with a velocity of 15 m per second and slides down a rough track my the coefficient of the kinetic energy between the two surface ab and the block iis0.52

5 0

11 months ago

A ruler of length 0.30m is pivoted at its centre. Equal and opposite forces of magnitude 2.0N are applied to the ends of the rul

kow [346]

Answer:

0.3858 Nm

Explanation:

The torque of the couple is the dot product of the force vector and the couple vector from 1 end of the ruler to the center. This equals to the product of their magnitude times the cosine() of the angle made by their direction:

$T = \vec{F} \cdot \vec{s} = Fscos(50^0) = 2 * 0.3 * 0.643 = 0.3858 Nm$

6 0

2 years ago