Name: Uncertainty in a Measurement
Uploaded: 2019-07-19
Duration: 7 min 59 s
Description: With error always present in measurement, uncertainty estimates the limits of an error in measurement. Learn the different sources of experimental uncertaint...

Question

Water is poured into a container that has a small leak. The mass of the water is given as a function of time $t$ by $m = 5.00 t^{0 .8} - 3 .00 t + 20 .00$ with $t \geq 0, m$ in grams and $t$ in seconds.

(a) At what time is the water mass greatest?

(b) What is the greatest mass?

In kilograms per minute, determine the rate of mass change at

(c) $t = 3 .00 s$

(d) $t = 5 .00 s$

Accepted Answer

Given:

Mass of water, $m = 5.00 t^{0.8} - 3.00 t + 20.00$

(a) If mass is maximum, then

$\frac{d m}{d t} = 0 \frac{d}{d t} (5.00 t^{0.8} - 3.00 t + 20) = 0 (5.00 \times 0.8 t^{- 0.2} - 3) = 0 4 t^{- 0.2} - 3 = 0 4 t^{- 0.2} = 3 t = 4.21 s$

(b) The greatest mass

$m_{\max} = 5 \times {(4.21)}^{0.8} - (3 \times 4.21) + 20 m_{\max} = 23.16 g$

(c)

${(\frac{d m}{d t})}_{t = 2 s} = 4 t^{- 0.2} - 3 (\frac{d m}{d t}) = 4 {(2)}^{- 0.2} - 3 (\frac{d m}{d t}) = 0.48 g s^{- 1} (\frac{d m}{d t}) = 0.0288 kg \min^{- 1}$

(d)

${(\frac{d m}{d t})}_{t = 5 s} = 4 t^{- 0.2} - 3 (\frac{d m}{d t}) = 4 {(5)}^{- 0.2} - 3 (\frac{d m}{d t}) = 0.10 g s^{- 1} (\frac{d m}{d t}) = 0.006 kg \min^{- 1}$

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