II.2. Determining the value of a physical parameter.

• II.2.1. Direct measurement of length.
  • II.2.1.1. The dimensions of a body in space.
  • II.2.1.2. Measurement error.
  • II.2.1.3. Apply what you have learned about Direct Length Measurement.
• II.2.2. Direct area measurement.
• II.2.3. Indirect determination of the area.
  • II.2.3.1 Apply what you have learned about Indirect Area Determination.
• II.2.4. Direct measurement of the volume.
• II.2.5. Indirect measurement of the volume.
  • II.2.5.1 Apply what you have learned about indirect measurement of the volume.
• II.2.6. Direct measurement of the time interval.
  • II.2.6.1 Apply what you have learned about the direct measurement of the time interval.

II.2.1. Direct measurement of length.

Characterization of length as physical parameter:

Since ancient times, man has measured different lengths using as units of measure the step, elbow, palm etc.

Today, in order to have a valid unit of international length, the meter is used.

The standard meter is made of a platinum and iridium ruler, kept in special conditions, protected from moisture and temperature differences, at the International Bureau of Measures and Weights located in Sèvres (near Paris). Each country has a copy of the standard meter.

Instruments used for measuring lengths:

1 – Ruler

2 – Tape line

3 – Callipers

4 – Tailoring centimeter

Measuring the distance with the ruler:

When we want to measure with the ruler, we place it above the length to be measured so that the zero division is at one end of this length. To make the reading accurate, you need to make sure that the gaze falls perpendicular to the ruler, in front of the respective division.

Indirect measurement of lengths can be done for objects we cannot reach, for rugged, swampy or water-covered terrain.

Some methods used for indirect measurement of lengths are optical methods, related to the rectilinear propagation of light. Distance reading is done on a digital screen: rangefinder (laser beam), GPS.

II.2.1.1. The dimensions of a body in space.

A body in space has three dimensions:

Length - denoted by L = AB = CD = A'B '= C'D',

Width - denoted by l = BC = AD = B'C '= A'D'

Height - vertical dimension denoted by h = AA '= BB' = CC '= DD'.

II.2.1.2. Measurement error.

Any measurement has limited accuracy and therefore the notion of measurement error appears.

Sources of error can be:

• lack of accuracy of the measurement instrument;
• incorrect reading of the instrument indications;
• the lack of attention or skill of the person making the measurements;
• unfavorable environmental conditions (inadequate lighting, too hot or too cold, discomfort, etc.).

If we perform length measurements with a ruler, the measurement accuracy cannot exceed the smallest gradation of the ruler - respectively 1 mm.

So the measurement error due to the instrument used is equal to the smallest division of the instrument.

Example:

At the micrometer the measurement error decreases to 1 micron, ie one millionth of a meter (1 / 1000000).

The existence of measurement errors in the case of experimental determinations is normal and in order to obtain a result as close as possible to the true value of the measured quantity, the measurements are repeated several times and the experimental data are processed as I will show you in the next experiment.

👀 Experiment no. 1: Determining the actual length of a body.

Required materials:
Graduated ruler, pencil.

Experiment description:

• Use the graduated ruler to measure the length of the physics book.
• Measure the length of the physics book several times (at least 3 times), taking care to measure correctly each time.
• Fill in the following table of experimental data: you will pass your determinations and follow the steps according to the following model:

• Processes experimental data.
  • l is the length measured at least three times. I measured it four times, but the last value of 20 cm I excluded it, as it is far from the other values, being a rough measurement.
  • l_m is the average length, ie the arithmetic average of the three measured lengths. If some values appear very different from the others, they are written in the table, but they are cut, they represent gross errors. They are not taken into account in the calculation of the average length. The arithmetic average is equal to the ratio of the sum of all lengths to the number of determinations.
  • Δl is the absolute error, which is calculated by the difference between the measured length and the average length (the largest minus the smallest): Δl = l₁ – l_m sau Δl = l_m – l₁.
  • Δl_m is the absolute average error, which is calculated by making the arithmetic average of the absolute errors.
• After completing the table with experimental data, you must write the result of the determination, using the same number of decimals for all numbers. We will write the values with two decimal places, by rounding.

For our example: L = 25,9 cm ± 0,06 cm.

Experiment conclusion:
This result indicates that the actual value is in a range: 25,9 cm – 0,06 cm ≤ l ≤ 25,9 cm + 0,06 cm
So the real length of the physics book is: 25,84 cm ≤ l ≤ 25,96 cm.

Result of the determination = The average value ± The absolute average error

l = l_average ± Δl_average

🔓 Solved problem

1. Florin wants to determine the real value of the length of the kitchen table.

Following the measurements he found the following values: 1,5 m; 1,46 m; 1,6 m ; 1,2 m; 1,56 m. How did he proceed?

Solution:

He calculated l_m = The average length, ie the arithmetic average of the four measured lengths. The value of 1.2 m is very different, it is cut and not taken into account in the calculation of the average length, being a rough measurement.

He calculated for each measurement Δl = absolute error, which is calculated by the difference between the measured length and the average length (the largest minus the smallest); Δl = l₁ – l_m sau Δl = l_m – l₁.

Δl₁ = 1,53 - 1,5 = 0,03 m
Δl₂ = 1,53 - 1,46 = 0,07 m
Δl₃ = 1,6 - 1,53 = 0,07 m
Δl₄ = 1,56 - 1,53 = 0,03 m

He calculated Δl_m = the absolute average error, which is calculated by making the arithmetic average of the four absolute errors.

Florin wrote the result of the determination, using the same number of decimals for all numbers (values with two decimals, by rounding).
Result of the determination = The average value ± The absolute average error
l = l_average ± Δl_average = 1,53 m ± 0,05 m.

II.2.1.3. Apply what you have learned about Direct Length Measurement.

🔐 Homework

1. Which of the following statements are true or false and why?

a) 278 mm > 27,8 cm

b) 0,066 km = 66 m

c) 0,45 hm ≤ 7,9 dam

d) 562 dam ≥ 3495 dm

2. According to the model of Experiment no. 1, determines the width of the physics book.

3. Mary measured the diameter of the mouth of her favorite glass and found the values: 6,5 cm; 6,4 cm; 6,6 cm; 6,3 cm; 5,2 cm; 6,9 cm.

Use these values and determine the actual size of the glass diameter.

II.2.2. Direct area measurement.

The area of a surface shows how large that surface is.

Characterization of the area as physical parameter:

Square meter multiples and submultiples are also used.

🔦 Remark

Here's how to make these transformations from multiples and submultiples of m² to m²:

• Write the numerical value and open a parenthesis, in which the value of the given multiple or submultiple is written, close the parenthesis and raise everything to the respective power, ie to the square.
• Copy the given value again and multiply by the values in parentheses raised to the second power.
• The corresponding mathematical calculations are made and the result is given.

Examples of transformations from multiples and submultiples of m² into m²:

Direct measurement of the area is done using graph paper. On it, thin vertical and horizontal lines are drawn, delimiting squares with a side of 1 mm and an area of 1 mm², and thicker lines, delimiting squares with a side of 1 cm and an area of 1 cm².

👀 Experiment no. 2: Direct measurement of the area of a leaf with graph paper.

Required materials:
Graph paper, leaf, pencil.

Experiment description:

• Draw on the graph paper (you can also use a math sheet that has a side of 0,5 cm and an area of 0,25 cm², but the result will not be too precise) the contour of the leaf (you can choose any shape you want, not necessarily a leaf).
• Count the whole squares with an area of 1 cm² (those outlined with a blue marker), then those with an area of 0.25 cm² (those outlined in yellow), and group the whole ones and approximate them as whole squares with area of 0.25 cm². If you have patience, you can stop approximating the whole squares and count the small squares with the area of 1 mm² and their number multiplied by 1 mm² to turn it into cm², by dividing the result by 100.
• To calculate the area of the leaf (S), apply the formula: S = n ∙ Su, where n = no. squares and Su = area of the chosen unit (either cm² - the largest, or 0.25 cm² - the smallest).
• Draw the outline of the leaf on another graph paper and repeat the above operations so that you have at least three values of the area of the leaf of your choice.

• Fill in the experimental data table and process the data in the table.

Experiment conclusion:
So the real area of the leaf is: S = S_average ± ΔS_average = 14,58 cm² ± 0,11cm².

The real result of the leaf area: S = S_average ± ΔS_average.

II.2.3. Indirect determination of the area.

The measurement of the area by indirect methods, in the case of surfaces with regular geometric shape, is done by measuring the linear dimensions and using calculation formulas:

• For a rectangle (The rectangle is a particular case of a parallelogram, which has all right angles), measure the length of the rectangle (L = AB = CD) and the width of the rectangle (l = AD = BC) and then apply the calculation formula:

• For a square (The square is a particular case of rectangle, which has all right angles and all four equal sides, denoted by l = side of the square = AB = BC = CD = AD) measure its side and then apply the calculation formula:

• For any triangle (The triangle is a polygon consisting of three sides that meet two by two, forming three internal angles) measure one side of it (a) and its height (h) and then apply the calculation formula:

• For a parallelogram (The parallelogram represents a quadrilateral that has opposite and parallel sides) measure the length of the parallelogram (b = L = AB = CD) and the height of the parallelogram (h = AF) and then apply the calculation formula:

🔦 Remark

Measurement units for the area of land areas in agriculture often used are 1 ar = 100 m² (equivalent to the area of a square with a side of 10 m) and 1 hectare = 1 ha = 100 ar.

🔓 Solved problems

1. A tennis court (rectangular) has a length of 2,377 dam and a width of 8230 mm for single play. Calculate the area of the rectangle in m².

Solution:

We note the data of the problem and transform the given parameters into SI:

We apply the formula for calculating the area of a rectangle and replace the problem data. Always add the measurement unit to the obtained result.

S = L ∙ l = 23,77 m ∙ 8,23 m = 195,62 m²

2. A certain triangle has a side of 0,008 km and the height corresponding to this side is 670 cm. Find the area of the surface of this triangle.

Solution:

We note the data of the problem and transform the given parameters into SI:

We apply the formula for calculating the area of a triangle and replace the problem data. Always add to the result obtained the measurement unit m².

3. The floor of a room is covered with square tiles, measuring 40 cm. If you count 10 tiles on the length of the room and 8 tiles on the width, what is the area of the floor surface?

Solution:

We note the data of the problem and transform the given parameters into SI:

We apply the formula for calculating the area of a rectangle and replace the problem data. Always add the measurement unit to the obtained result.

S = L ∙ l = 4 m ∙ 3,2 m = 12,8 m²

4. Calculate the surface areas of a box that has the following dimensions:

L = 22,5 cm

l = 11,3 cm

h = 7 cm

Solution:

We transform the given parameters into SI:

• A box (parallelepiped) has six surfaces, two by two equal. So we calculate the areas of rectangular surfaces that are different, that is, three areas. We apply the formula for calculating the area of a rectangle and replace the problem data. The unit of measure is always added to the obtained result.

S₁ = L ∙ l = 0,225m ∙ 0,113 m = 0,025425 m²

S₂ = L ∙ h = 0,225m ∙ 0,07 m = 0,01575 m²

S₃ = l ∙ h = 0,113m ∙ 0,07 m = 0,00791 m²

II.2.3.1 Apply what you have learned about Indirect Area Determination.

🔐 Homework

1. Determine the areas of the three surfaces of the physics textbook by measuring its length, width, and height.

2. Transforms:

a) 520 cm² = ? m²

b) 4,9 dam² = ? m²

3. Determine the area of the following contour:

II.2.4. Direct measurement of the volume.

The volume of a body is the place occupied by a body in space.

Characterization of the volume as physical parameter:

Another measurement unit for volume (capacity) is the liter: 1 L = 1dm³.

Cubic meter multiples and submultiples are also used.

🔦 Remark

Here is how to perform these transformations from multiples and submultiples of m³ into m³:

• Write the numerical value and open a parenthesis, in which the value of the given multiple or submultiple is written, close the parenthesis and raise everything to the respective power, ie to the cube (3rd power).
• Copy the given value again and multiply by the values in parentheses raised to the third power.
• The corresponding mathematical calculations are made and the result is given.

Examples of transformations from multiples and submultiples of m³ or liter (L) into m³:

To perform calculations with transformations you need to know the values of multiples and submultiples.

You also need to know very well the operations with fractions:

• Multiplying two fractions is done by multiplying numerator (number on the fraction line) by numerator and denominator (number below the fraction line) by denominator.
• The division of two fractions is done by multiplying the fraction from the numerator by the inverse (inverted) of the fraction from the denominator.

👀 Experiment no. 3: Measuring the volume of a body with the graduated cylinder.

Required materials:
Graduated cylinder, water, string, a body.

Experiment description:

1. The first step is to determine the volume of a division = 1 div = the minimum volume between two consecutive lines. Look carefully at the measurement and find the measurement unit of the cylinder used. How do we proceed?

• Note two consecutive graduations (one after the other) of the cylinder and subtract them: 50 ml - 40 ml = 10 ml
• Count how many divisions there are between these notations: 10 divisions = 10 ml
• With the simple rule of three, we find out the volume of one division:

2. Put water in the cylinder and measure its volume, denoted by V₁ = 35 ml.

3. Insert the body into the water in the cylinder. The fluid level has risen. We will note the new volume read V₂ = 39 ml.

4. The volume of the body is the difference between V₂ (water+body volume) and V₁ (water volume), ie:

V_body = V₂ – V₁

Remark:

The graduated cylinder must be on a horizontal surface (on the table). The free surface of the liquid is slightly curved (called the meniscus) - higher at the contact of the liquid with the walls of the graduated cylinder. Position the eyes at the free surface of the liquid and read the volume at its base.

Experiment conclusion:

The volume of the body is the difference between V₂ (water volume + body) and V₁ (water volume), ie:

V_body = V₂ – V₁

V_body = 39 ml – 35 ml = 4 ml.

II.2.5. Indirect measurement of the volume.

Volume measurement by indirect methods, in the case of bodies with regular geometric shape, is done by measuring linear dimensions and using calculation formulas (in 8th Grade you will learn volume formulas for other geometric bodies):

• For parallelepiped we have the volume formula:

• For the cube we have the volume formula:

The cube is the rectangular parallelepiped with all edges equal. The faces of a cube are square and congruent.

🔓 Solved problems

1. A room has a length of 0.06 hm, a width of 40 dm and a height of 330 cm. Calculate the volume of air in the room in m³.

Solution:

We note the data of the problem and transform the given parameters into SI:

The air being a gas occupies the entire volume of the room. We apply the formula for calculating the volume of a parallelepiped and replace the problem data. Always add the measurement unit to the obtained result.

V = L ∙ l ∙ h = 6 m ∙ 4 m ∙ 3,3 m = 79,2 m³.

2. Pour 500 cm³ of water into a coffee maker to make the coffee. Knowing that a cup of coffee has 150 mL, how many coffees have you made?

Solution:

We note the data of the problem and transform the given parameters into SI:

We divide the volume of the coffee maker by the volume of the cup:

II.2.5.1 Apply what you have learned about indirect measurement of the volume.

🔓 Solved problem

3. Determine the volume of the body in the following image, knowing that only water is placed in the first cylinder, and in the second was added to the water from the first cylinder, the body whose volume you need to determine.

Solution:

🔐 Homework

1. A cube has a side of 5 dm, and a parallelepiped has the following dimensions 800 mm; 0,04 hm and 0,3 dam. Which of the two bodies has the largest volume?

2. Transform into m³:

a) 4.800 dm³ = ? m³

b) 0,06 hm³ = ? m³

c) 53.000 mm³ = ? m³

II.2.6. Direct measurement of the time interval.

The time interval represents the duration of an event.

Characterization of time (duration of an event) as a physical parameter:

Other measurement units for time interval are:

• Minute = 1 min = 60 s
• Hour = 1 h = 60 min = 60 ∙ 60 s = 3.600 s
• Day = 24 h = 24 ∙ 3600 s = 86400 s
• Week = 7 days = 7 ∙ 86400 s = 604800 s
• Moon = 30 days = 30 ∙ 86400 s = 2592200 s
• Year = 365 days = 365 ∙ 86400 s = 31536000 s

🔦 Remark

The Romans called the hours before noon ante meridiem (before noon), and those in the afternoon, post meridiem. Today, they are abbreviated a.m. and p.m. and are used with the meaning of morning and afternoon.

If you use the electronic stopwatch on the phone, which also measures hundredths of seconds, then its accuracy has increased to 0.01s. On the other hand, if you use a ticker, its accuracy is lower and the error can be 1s .

👀 Experiment no. 4: Measuring the period of a pendulum with a stopwatch.

Required materials:
Ball or nut, thread, stopwatch (you can use the phone)

Experiment description:

• Take a small and heavy body (a ball, a nut, a key, etc.) and tie it to the end of a wire to get a pendulum. Attach the wire to a horizontal support (for example, a table).

• Remove the wire from the (vertical) equilibrium position and lift it to one side. Then release it and the pendulum will move to either side of this position, ie it will start to oscillate. The stopwatch starts when the body is released.

• Measure the time interval (t) in which the body returns a number of times "n" to point A ("n" can have different values: 3, 5, 8, 10 etc. - as much as you want).

• Repeat the operation at least 3 times, giving it to "n" different values.

• Calculate the period (T) = the time in which the body makes a complete oscillation, ie the time in which the pendulum climbed to the other side and returned to the initial position (round trip). If in a time "t" are performed "n" full oscillations, then the period (T) is calculated using the relation:

• Fill in an experimental data table using the below template.

Experiment conclusion:
Write the measurement result: T = T_mean ± ΔT_mean = 1,73 ± 0,03 (s)

🔓 Solved problem

1. A movie broadcast on a TV station started at 20:30 and ended at 22:20. If he was interrupted by four commercials, each lasting 8 minutes, how long did the film last, expressed in hours, minutes and seconds?

Solution:

We calculate the broadcast time of both the film and the advertisement, subtracting the end time from the start time of the film: to decrease the minutes, I borrow from the unit of hours by one hour, ie 60 minutes and add them to 20 minutes and say 80 min - 30 min = 50 min.
Then decrease 21 h - 20 h = 1h. 22:20 - 20:30 = 1 h 50 min = t₁.

We calculate the time of the advertising blocks (sequences): t₂ = 4 ∙ 8 min = 32 min

To find out the length of the movie only, subtract t₂ from t₁:

t_movie = t₁ – t₂ = 1 h 50 min – 32 min = 1 h 18 min
t_movie = 60 min + 18 min = 78 min = 78 ∙ 60 s = 4680 s

To find out the duration in hours we transform 18 min into h with the simple rule of 3:

II.2.6.1 Apply what you have learned about the direct measurement of the time interval.

🔐 Homework

1. Which of the following statements are true and false, respectively, and why?

a) 650 cs = 6,5 s

b) 1 h 5 min 40 s > 80 min 50 s

c) 5 days 3 h 30 min 20 s < 4 days 25 h 10 min 10 s

d) 4,6 hs = 460 s

2. Julia left home at 7:20 and arrived at school at 7:50. Knowing that she had 6 hours of class, every hour of 50 minutes with a break of 10 minutes and that on the way back she did 15 minutes more than on arrival, at what time did Julia get home?

3. Michael performed the experiment no. 4 to measure the oscillation period of a pendulum. He timed the time in which the pendulum made 10 oscillations each time and noted them in the following table:

Calculate the real value of the pendulum period in Michael's experiment.