You observe a ball that moves (33.6pm 0.2)cm in (11.49pm 0.02)s . What is the observed speed of the ball (best estimate and most probable uncertainty)? (speed=distance/time) (1.62pm 0.01)cm/s (1.739pm 0.005)cm/s (2.92pm 0.01)cm/s (2.92pm 0.02)cm/s (1.619pm 0.008)cm/s

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You observe a ball that moves (33.6pm 0.2)cm in (11.49pm 0.02)s . What is the observed speed of the ball (best estimate and most probable uncertainty)? (speed=distance/time) (1.62pm 0.01)cm/s (1.739pm 0.005)cm/s (2.92pm 0.01)cm/s (2.92pm 0.02)cm/s (1.619pm 0.008)cm/s

Answer 1

# Brief Explanations: To find the observed speed, we use the formula $ \text{speed} = \frac{\text{distance}}{\text{time}} $. The best estimate is calculated by dividing the distance by the time. To find the uncertainty, we use the propagation of uncertainties formula for division: $ \frac{\Delta v}{v} = \sqrt{\left(\frac{\Delta d}{d}\right)^2 + \left(\frac{\Delta t}{t}\right)^2} $. # Answer: The observed speed is $ \frac{33.6 \text{ cm}}{11.49 \text{ s}} \approx 2.92 \text{ cm/s} $. To find the uncertainty: \[ \frac{\Delta v}{v} = \sqrt{\left(\frac{0.2}{33.6}\right)^2 + \left(\frac{0.02}{11.49}\right)^2} \approx \sqrt{(0.00595)^2 + (0.00174)^2} \approx 0.0062 \] Thus, the uncertainty in speed is: \[ \Delta v \approx 2.92 \times 0.0062 \approx 0.0181 \text{ cm/s} \] Rounding to match the significant figures: \[ \Delta v \approx 0.02 \text{ cm/s} \] Therefore, the observed speed is: \[ (2.92 \pm 0.02) \text{ cm/s} \] # Answer: $(2.92\pm 0.02) \text{ cm/s}$

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You observe a ball that moves (33.6pm 0.2)cm in (11.49pm 0.02)s . What is the observed speed of the ball (best estimate and most probable uncertainty)? (speed=distance/time) (1.62pm 0.01)cm/s (1.739pm 0.005)cm/s (2.92pm 0.01)cm/s (2.92pm 0.02)cm/s (1.619pm 0.008)cm/s

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