What is the instrument uncertainty of the gram scale shown? (.2 pt) What is the least count of the Newton scale shown? (.2 pt) g NEWTONS grams Newtons I measure a small time ?t, taking four trials .0238, .0254, .0218 and .0226 seconds. Five people separately measure the distance x traveled in this time ?t, the results of their x distances in centimeters was 4.52, 4.34, 4.66, 4.45 and 4.58 centimeters. In the calculations below round off consistently considering the data and our rules. (.3 pt) a) Calculate the average and average deviation of ?t (.3 pt) b) Calculate the average and average deviation of x (.4 pt) c) Use the ?t and x averages and uncertainties to calculate the average velocity and its uncertainty using
in units of meters per second. Determine what that uncertainty is in units of meters per second. .3 pt) Two measured values with ranges are (42.4 minutes ± 4%) and (39.5 minutes ± 2%). Show each of these ranges as numerical values and determine and state whether or not these ranges overlap.
The Correct Answer and Explanation is:
Instrument Uncertainty and Least Count
1. What is the instrument uncertainty of the gram scale shown?
- Answer: 5 grams
- Explanation: The gram scale has numbered markings every 50 grams (0, 50, 100, 150). Between each numbered marking (e.g., between 50 and 100), there are five intervals, meaning each marked line represents 10 grams. This smallest marked division is the least count of the instrument. A common rule for estimating the uncertainty of an analog instrument is to take half of the least count.
- Least Count = 10 g
- Instrument Uncertainty = 0.5 × Least Count = 0.5 × 10 g = 5 g.
2. What is the least count of the Newton scale shown?
- Answer: 0.1 Newtons
- Explanation: The Newton scale has numbered markings for each whole Newton (0, 1, 2, 3, 4, 5). Between each whole Newton (e.g., between 0 and 1), there are 10 small divisions. The value of each small division is therefore 1 N / 10 = 0.1 N. The least count is the smallest value that can be read directly from the scale’s markings.
Data Analysis: Velocity and Uncertainty
Given time data (Δt): 0.0238, 0.0254, 0.0218, 0.0226 seconds.
Given distance data (x): 4.52, 4.34, 4.66, 4.45, 4.58 centimeters.
a) Calculate the average and average deviation of Δt
- Average Δt (Δt_avg):
(0.0238 + 0.0254 + 0.0218 + 0.0226) s / 4 = 0.0936 s / 4 = 0.0234 s - Average Deviation δ(Δt):
Deviations: |0.0238-0.0234|=0.0004, |0.0254-0.0234|=0.0020, |0.0218-0.0234|=0.0016, |0.0226-0.0234|=0.0008
Average Deviation = (0.0004 + 0.0020 + 0.0016 + 0.0008) s / 4 = 0.0048 s / 4 = 0.0012 s
Applying rounding rules (uncertainty to 1 significant figure, value to the same decimal place):
Δt = 0.023 ± 0.001 s.
- Final Answer a): Average Δt = 0.023 s, Average Deviation = 0.001 s
b) Calculate the average and average deviation of x
- Average x (x_avg):
(4.52 + 4.34 + 4.66 + 4.45 + 4.58) cm / 5 = 22.55 cm / 5 = 4.51 cm - Average Deviation (δx):
Deviations: |4.52-4.51|=0.01, |4.34-4.51|=0.17, |4.66-4.51|=0.15, |4.45-4.51|=0.06, |4.58-4.51|=0.07
Average Deviation = (0.01 + 0.17 + 0.15 + 0.06 + 0.07) cm / 5 = 0.46 cm / 5 = 0.092 cm
Applying rounding rules:
δx rounds to 0.09 cm (1 significant figure). The average is already at the same decimal place.
- Final Answer b): Average x = 4.51 cm, Average Deviation = 0.09 cm
c) Calculate the average velocity and its uncertainty in m/s
- Convert to SI units (m and s):
- x = 4.51 cm = 0.0451 m
- δx = 0.09 cm = 0.0009 m
- t = 0.023 s
- δt = 0.001 s
- Calculate Average Velocity (v_average):
v_average = x / t = 0.0451 m / 0.023 s ≈ 1.96 m/s - Calculate Uncertainty in Velocity (δv):
For division (v = x/t), we add the fractional uncertainties:
δv/v = δx/x + δt/t- Fractional uncertainty in x: 0.0009 m / 0.0451 m ≈ 0.01996
- Fractional uncertainty in t: 0.001 s / 0.023 s ≈ 0.04348
- Total fractional uncertainty: δv/v = 0.01996 + 0.04348 = 0.06344
- Absolute uncertainty (δv) = v * (δv/v) = 1.96 m/s * 0.06344 ≈ 0.124 m/s
- Round to Final Answer:
Round the uncertainty δv to one significant figure: δv ≈ 0.1 m/s.
Round the average velocity to the same decimal place (the tenths place): v_average ≈ 2.0 m/s.
- Final Answer c): The average velocity is 2.0 ± 0.1 m/s. The uncertainty is 0.1 m/s.
Range Overlap
Two measured values with ranges are (42.4 minutes ± 4%) and (39.5 minutes ± 2%). Show each of these ranges as numerical values and determine and state whether or not these ranges overlap.
1. Calculate the first numerical range:
- Value: 42.4 minutes
- Uncertainty: 4% of 42.4 = 0.04 × 42.4 = 1.696 minutes
- Range: 42.4 ± 1.696 minutes
- Numerical Range 1: [40.704, 44.096] minutes
2. Calculate the second numerical range:
- Value: 39.5 minutes
- Uncertainty: 2% of 39.5 = 0.02 × 39.5 = 0.79 minutes
- Range: 39.5 ± 0.79 minutes
- Numerical Range 2: [38.71, 40.29] minutes
3. Determine if the ranges overlap:
To check for overlap, we compare the highest value of the lower range to the lowest value of the higher range.
- The maximum value of the second range is 40.29 minutes.
- The minimum value of the first range is 40.704 minutes.
Since 40.29 is less than 40.704, there is a gap between the two ranges.
- Final Answer: The ranges do not overlap.
