(b) Fig. 2.1 shows a uniform metre rule of mass 0.15 kg pivoted at the 20.0 cm mark. The rule is balanced by the tension in a string at the 80.0 cm mark as shown. [1] ? Fig. 2.1 Calculate (i) the weight \( W \) of the rule, ( \( g=10 \mathrm{~N} / \mathrm{kg} \) ) \[ W= \] 2 (ii) the tension on the string. ?
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- The mass of the rule is given as \( 0.15 \, \text{kg} \). - The acceleration due to gravity \( g \) is \( 10 \, \text{N/kg} \). \[ W = \text{mass} \times g = 0.15 \, \text{kg} \times 10 \, \text{N/kg} = 1.5 \, \text{N} \] Show more…
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