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Smallest Division = 0.002 cm Estimated uncertainty = ± 0.001_ cm Length, Diameter, Mass, L (cm) Calculated Cylindrical Sample V Standard % Error D (cm) m (g) (em) p (g/cm³) Value (g/cm³) Gray: 5.769 White: 7.365 Tin 3.245 1.266 28.88 8.40 – 8.75 Brass 5.010 1.266 53.35 Al 8.590 1.276 28.85 2.70

Smallest Division = 0.002 cm Estimated uncertainty = ± 0.001_ cm Length, Diameter, Mass, L (cm) Calculated Cylindrical Sample V Standard % Error D (cm) m (g) (em) p (g/cm³) Value (g/cm³) Gray: 5.769 White: 7.365 Tin 3.245 1.266 28.88 8.40 – 8.75 Brass 5.010 1.266 53.35 Al 8.590 1.276 28.85 2.70

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Complete this table using vernier caliper
Parts A, C, and D: Vernier Caliper Smallest Division = 0.002__ cm Estimated uncertainty + 0.001__ cm Cylindrical Length, Sample Diameter, Mass, D (cm) V Calculated Standard % Error L (cm) m (g) (cm³) P (g/cm³) Value (g/cm³) Gray: 5.769 White: 7.365 Tin 3.245 1.266 28.88 8.40 – 8.75 Brass 5.010 1.266 53.35 Al 8.590 1.276 28.85 2.70
Transcribed Image Text:Parts A, C, and D: Vernier Caliper Smallest Division = 0.002__ cm Estimated uncertainty + 0.001__ cm Cylindrical Length, Sample Diameter, Mass, D (cm) V Calculated Standard % Error L (cm) m (g) (cm³) P (g/cm³) Value (g/cm³) Gray: 5.769 White: 7.365 Tin 3.245 1.266 28.88 8.40 – 8.75 Brass 5.010 1.266 53.35 Al 8.590 1.276 28.85 2.70
Expert Solution
Step 1

Cylindrical sample : TIN

Length, L=3.245 cm

Diameter, D=1.266 cm

Volume: V=πD2L4=3.14×1.266×1.266×3.2454=4.08 cm3Mass, m=28.88 gdensity, ρ=mV                 =28.88 g4.08 cm3                 =7.078 g/cm3ρtheoretical=5.769 to 7.365% Error : 7.365-7.0787.078×100 to 7.078-5.7697.078×100              =4.05% to 18.499 %

Answer: Volume, V=4.08 cm3ρ=7.078 g/cm3%Error=4.05% to 18.499%

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