Based on the calculations, the average velocity is equal to 360 m/s and the percent difference is equal to 4.72%.
<h3>What is average velocity?</h3>
An average velocity can be defined as the total distance covered by a physical object divided by the total time taken.
<h3>What is an
average?</h3>
An average is also referred to as mean and it can be defined as a ratio of the sum of the total number in a data set to the frequency of the data set.
<h3>How to calculate the
average velocity?</h3>
Mathematically, the average velocity for this data set would be calculated by using this formula:
Average = [F(v)]/n
Vavg = [v₁ + v₂ + v₃ + v₄ + v₅)/5
Since the values of the average velocity from the table are missing, we would assume the following values for the purpose of an explanation:
Substituting the parameters into the formula, we have:
Vavg = [300 + 450 + 500 + 250 + 300)/5
Vavg = 1800/5
Vavg = 360 m/s.
Next, we would calculate the percent difference by using this formula:
![Percent \;difference = \frac{[V_{avg}\;-\;V_{sound}]}{V_{sound}} \times 100](https://tex.z-dn.net/?f=Percent%20%5C%3Bdifference%20%3D%20%5Cfrac%7B%5BV_%7Bavg%7D%5C%3B-%5C%3BV_%7Bsound%7D%5D%7D%7BV_%7Bsound%7D%7D%20%5Ctimes%20100)
Percent difference = [360 - 343]/360 × 100
Percent difference = 17/360 × 100
Percent difference = 0.0472 × 100
Percent difference = 4.72%.
Read more on average here: brainly.com/question/9550536
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Answer:
im not too sure about that all i know is history
Answer:
Solution
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Correct option is
C
3 cm
RI=apparent depthreal depth
Substituting, 34=apparentdepth12
Therefore, apparent depth=412×3=9
The height by which it appears to be raised is 12−9=3cm
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SIMILAR QUESTIONS
A coin is placed at the bottom of a glass tumbler and then water is added. It appeared that the depth of the coin has been reduced because
Medium
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A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?
Moment about the pivot must be equal for the seesaw to balance. Initially, the first cat and the bowl are at 2 m from the pivot.
The moment due to cat = 5.3*2 = 10.6 kg.m
The moment due to bowl = 2.5*2 = 5 kg.m
The unbalanced moment = 10.6 - 5 = 5.6 kg.m
Therefore, the 3.7 kg cat should stand at a distance x from the pivot in left to balance the 5.6 kg.m.
That is,
3.7*x = 5.6 => x = 5.6/3.7 = 1.5134 m to the left (on the side of the bowl)