Question

1. use the correct values of mass and acceleration to determine the force from the fan.

unloaded mass | unloaded acceleration | unloaded force (n) | loaded mass | loaded acceleration | loaded force (n)
0.767 | 1.317 | 1010 | 0.6825 | 1.67 | 1.14

1. determine the precision of your forces. hint: the median of 2 values = the average value.

median force | uld = 1.1 x median | lld = 0.9 x median | precise? (yes/no)
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1. determine the accuracy of your forces. the actual value should be 0.23 n.

average force | ule = 1.1 x actual | lle = 0.9 x actual | accurate? (yes/no)
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Explanation:

Question 2

Step1: Recall Newton's Second Law

Force \( F = m \times a \). We have two cases: unloaded and loaded. Let's first check the unloaded case: mass \( m = 0.767 \) (assuming units are kg) and acceleration \( a = 1.317 \) (assuming units are \( \text{m/s}^2 \)). Wait, but the unloaded force is given as 1010 N, which seems inconsistent. Wait, maybe the loaded mass is 0.6825 kg, loaded acceleration 1.67 \( \text{m/s}^2 \), loaded force 1.14 N. Wait, maybe the unloaded mass and acceleration are for a different calculation? Wait, the problem says "use the correct values of mass and acceleration to determine the force from the fan". Let's recalculate both unloaded and loaded forces.

Step2: Calculate Unloaded Force

Using \( F = m \times a \), unloaded mass \( m_{unloaded} = 0.767 \) kg, unloaded acceleration \( a_{unloaded} = 1.317 \text{ m/s}^2 \). So \( F_{unloaded} = 0.767 \times 1.317 \approx 1.010 \) N (wait, the table has 1010 N, which is probably a typo, maybe 1.010 N).

Step3: Calculate Loaded Force

Loaded mass \( m_{loaded} = 0.6825 \) kg, loaded acceleration \( a_{loaded} = 1.67 \text{ m/s}^2 \). So \( F_{loaded} = 0.6825 \times 1.67 \approx 1.140 \) N (matches the table's 1.14 N). So the forces are approximately 1.01 N (unloaded) and 1.14 N (loaded).

Step1: Find Median Force

We have two forces: 1.01 N (unloaded) and 1.14 N (loaded). Median of two values is the average. So \( \text{Median Force} = \frac{1.01 + 1.14}{2} = \frac{2.15}{2} = 1.075 \) N.

Step2: Calculate ULD and LLD

\( \text{ULD} = 1.1 \times \text{Median} = 1.1 \times 1.075 = 1.1825 \) N.
\( \text{LLD} = 0.9 \times \text{Median} = 0.9 \times 1.075 = 0.9675 \) N.

Step3: Check Precision

Check if both forces (1.01 and 1.14) lie between LLD and ULD.
1.01 is between 0.9675 and 1.1825.
1.14 is between 0.9675 and 1.1825.
So Precise? YES.

Step1: Find Average Force

Average of 1.01 N and 1.14 N: \( \text{Average Force} = \frac{1.01 + 1.14}{2} = 1.075 \) N.

Step2: Calculate ULE and LLE

Actual value = 0.23 N.
\( \text{ULE} = 1.1 \times 0.23 = 0.253 \) N.
\( \text{LLE} = 0.9 \times 0.23 = 0.207 \) N.

Step3: Check Accuracy

Check if average force (1.075 N) lies between LLE and ULE.
1.075 N is greater than 0.253 N, so it does not lie between 0.207 and 0.253.
Thus, Accurate? NO.

Answer:

(for Question 2):
Unloaded Force: \( \approx 1.01 \) N, Loaded Force: \( \approx 1.14 \) N (assuming mass in kg and acceleration in \( \text{m/s}^2 \))

Question 3