This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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Step 1: Answer Question 9. A uniform metre rule has a mass of 145 g and a length of 100 cm. Its center of mass is at the 50 cm mark from end A. A mass of 300 g is hung at end A (0 cm mark). Let the string be positioned at a distance x from end A. For the rule to balance, the sum of clockwise moments about the pivot (string) must equal the sum of anticlockwise moments. Mass of rule M_r = 145 g. Mass at end A M_A = 300 g. Length of rule L = 100 cm. Position of rule's center of mass from A = 50 cm. Taking moments about the pivot (string at position x from A): Anticlockwise moment due to mass at A: M_A g × x Clockwise moment due to rule's mass: M_r g × (50 cm - x) Equating moments: M_A g x = M_r g (50 cm - x) 300 g × x = 145 g × (50 cm - x) 300x = 145 × 50 - 145x 300x = 7250 - 145x 300x + 145x = 7250 445x = 7250 x = (7250)/(445) ≈ 16.292 cm This is the position from end A. The question asks for the position from end B. Position from end B = 100 cm - x Position from end B = 100 cm - 16.292 cm = 83.708 cm. The position from end B is 83.71 cm. Step 2: Answer Question 10. Given: Mass m = 56 g = 0.056 kg. Radius r = 0.8 m. Frequency $f =