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guapka [62]
2 years ago
12

a student measures the mass of a metal weight using three separate balances and collects the following data: 38.7g, 38.9g and 38

.6g. The actual mass of the weight is 34.2 g. Evaluate the accuracy and precision of this student's measurements.
Physics
1 answer:
Lesechka [4]2 years ago
6 0

The values are precise because the values are close together but they are not accurate because they are all far from the true value.

<h3>What is the accuracy and precision?</h3>

The term accuracy refers to how much close to the true or the actual value that a measurement is. The precision refer to how close together the values that are obtained in a series of measurements are.

Now, we have the values that were obtained as; 38.7g, 38.9g and 38.6g. We can see that values are precise because the values are close together but they are not accurate because they are all far from the true value.

Learn more about precision:brainly.com/question/28336863

#SPJ1

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Take into account that density and relative density are given by:

\begin{gathered} \text{density}=\text{ mass/volume} \\ \text{relative density = density/density of water} \end{gathered}

Take into account that the volume associated to each of the given sustances in the table is determined by the Level Difference (because it is the change in the volume of the water of the recipient in which the substance is immersed).

The density of water in kg/m^3 is 1000 kg/m^3.

Due to the density must be given in kg/m^3, it is necessary to express the volumes of the table in m^3 and mass in kg, then, consider the following conversion factor:

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\begin{gathered} 53.2g\cdot\frac{1kg}{1000g}=0.0532kg \\ 6ml\cdot\frac{1m^3}{1000000ml}=0.000006m^3 \\ \text{density}=\frac{0.0532kg}{0.000006m^3}\approx8866.67\frac{kg}{m^3} \\ \text{relative density=}\frac{(\frac{8866.66kg}{m^3})}{(1000\frac{kg}{m^3})}\approx8.87 \end{gathered}

Cooper:

\begin{gathered} 57.4g=0.0574kg \\ 6ml=0.000006m^3 \\ \text{density}=\frac{0.0574kg}{0.000006m^3}\approx9566.67\frac{kg}{m^3} \\ \text{relative density=}\frac{\frac{9566.67kg}{m^3}}{1000kg}=9.57 \end{gathered}

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