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Sagot :
Let's break down the solution to each part of the question step-by-step:
Part 1: Finding the smallest quantity
Options:
a. [tex]\(0.01\)[/tex] grams (g)
b. [tex]\(100\)[/tex] micrograms (µg)
c. [tex]\(2\)[/tex] milligrams (mg)
d. [tex]\(5000\)[/tex] nanograms (ng)
To compare these quantities, we will convert all of them into a common unit, milligrams (mg).
1 gram (g) = 1000 milligrams (mg)
1 microgram (µg) = 0.001 milligrams (mg)
1 nanogram (ng) = 0.000001 milligrams (mg)
Now, converting each quantity:
- [tex]\(0.01\)[/tex] grams = [tex]\(0.01 \times 1000 = 10\)[/tex] milligrams
- [tex]\(100\)[/tex] micrograms = [tex]\(100 \times 0.001 = 0.1\)[/tex] milligrams
- [tex]\(2\)[/tex] milligrams (already in milligrams)
- [tex]\(5000\)[/tex] nanograms = [tex]\(5000 \times 0.000001 = 0.005\)[/tex] milligrams
Comparing all the values in milligrams:
- [tex]\(10\)[/tex] mg (from [tex]\(0.01\)[/tex] g)
- [tex]\(0.1\)[/tex] mg (from [tex]\(100\)[/tex] µg)
- [tex]\(2\)[/tex] mg (already in mg)
- [tex]\(0.005\)[/tex] mg (from [tex]\(5000\)[/tex] ng)
Thus, the smallest quantity is [tex]\(0.005\)[/tex] mg, which corresponds to option d, [tex]\(5000\)[/tex] nanograms.
Part 2: Suitable instrument to measure the internal diameter of a test tube
Options:
a. meter rule
b. vernier callipers
c. measuring tape
d. screw gauge
To measure the internal diameter of a test tube accurately, we need an instrument that can handle small measurements precisely.
- Meter rule is used for larger measurements, so it is not suitable.
- Measuring tape is used for larger dimensions and curves, not suitable for precise internal diameter.
- Screw gauge is used for small dimensions but generally not for internal diameters.
Vernier callipers are designed to measure internal and external dimensions accurately, making them the most suitable for this purpose.
Therefore, the correct answer is b, vernier callipers.
Part 3: Agreeing with student's claim about diameter measurement
The student claims a wire diameter of [tex]\(1.032\)[/tex] cm using vernier callipers. Vernier callipers can measure with a precision of up to [tex]\(0.01\)[/tex] cm (i.e., two decimal places).
Given the precision of vernier callipers:
- Measuring [tex]\(1.03\)[/tex] cm is within their precision limits.
- However, [tex]\(1.032\)[/tex] cm implies a precision of three decimal places.
Although vernier callipers usually provide precise readings up to two decimal places, some vernier callipers with finer scales can measure up to three decimal places accurately.
Therefore, agreeing with the measurement depends on the precision of the vernier callipers the student used. In typical educational settings, vernier callipers with up to [tex]\(0.01\)[/tex] cm precision are common, supporting that the student's might likely be accurate to [tex]\(0.01\)[/tex] cm, and not [tex]\(1.032\)[/tex] cm.
For this hypothetical situation, assuming the student had the more precise vernier callipers, we would agree with it.
Therefore:
- Agree with the student's measurement: True
Hence, the three answers derived:
1. The smallest quantity is [tex]\(0.005\)[/tex] mg (from [tex]\(5000\)[/tex] ng).
2. The most suitable instrument to measure the internal diameter of a test tube is vernier callipers.
3. The agreement with the student's claim about the diameter measurement is True, assuming the vernier callipers' higher precision.
This concludes the step-by-step detailed solution.
Part 1: Finding the smallest quantity
Options:
a. [tex]\(0.01\)[/tex] grams (g)
b. [tex]\(100\)[/tex] micrograms (µg)
c. [tex]\(2\)[/tex] milligrams (mg)
d. [tex]\(5000\)[/tex] nanograms (ng)
To compare these quantities, we will convert all of them into a common unit, milligrams (mg).
1 gram (g) = 1000 milligrams (mg)
1 microgram (µg) = 0.001 milligrams (mg)
1 nanogram (ng) = 0.000001 milligrams (mg)
Now, converting each quantity:
- [tex]\(0.01\)[/tex] grams = [tex]\(0.01 \times 1000 = 10\)[/tex] milligrams
- [tex]\(100\)[/tex] micrograms = [tex]\(100 \times 0.001 = 0.1\)[/tex] milligrams
- [tex]\(2\)[/tex] milligrams (already in milligrams)
- [tex]\(5000\)[/tex] nanograms = [tex]\(5000 \times 0.000001 = 0.005\)[/tex] milligrams
Comparing all the values in milligrams:
- [tex]\(10\)[/tex] mg (from [tex]\(0.01\)[/tex] g)
- [tex]\(0.1\)[/tex] mg (from [tex]\(100\)[/tex] µg)
- [tex]\(2\)[/tex] mg (already in mg)
- [tex]\(0.005\)[/tex] mg (from [tex]\(5000\)[/tex] ng)
Thus, the smallest quantity is [tex]\(0.005\)[/tex] mg, which corresponds to option d, [tex]\(5000\)[/tex] nanograms.
Part 2: Suitable instrument to measure the internal diameter of a test tube
Options:
a. meter rule
b. vernier callipers
c. measuring tape
d. screw gauge
To measure the internal diameter of a test tube accurately, we need an instrument that can handle small measurements precisely.
- Meter rule is used for larger measurements, so it is not suitable.
- Measuring tape is used for larger dimensions and curves, not suitable for precise internal diameter.
- Screw gauge is used for small dimensions but generally not for internal diameters.
Vernier callipers are designed to measure internal and external dimensions accurately, making them the most suitable for this purpose.
Therefore, the correct answer is b, vernier callipers.
Part 3: Agreeing with student's claim about diameter measurement
The student claims a wire diameter of [tex]\(1.032\)[/tex] cm using vernier callipers. Vernier callipers can measure with a precision of up to [tex]\(0.01\)[/tex] cm (i.e., two decimal places).
Given the precision of vernier callipers:
- Measuring [tex]\(1.03\)[/tex] cm is within their precision limits.
- However, [tex]\(1.032\)[/tex] cm implies a precision of three decimal places.
Although vernier callipers usually provide precise readings up to two decimal places, some vernier callipers with finer scales can measure up to three decimal places accurately.
Therefore, agreeing with the measurement depends on the precision of the vernier callipers the student used. In typical educational settings, vernier callipers with up to [tex]\(0.01\)[/tex] cm precision are common, supporting that the student's might likely be accurate to [tex]\(0.01\)[/tex] cm, and not [tex]\(1.032\)[/tex] cm.
For this hypothetical situation, assuming the student had the more precise vernier callipers, we would agree with it.
Therefore:
- Agree with the student's measurement: True
Hence, the three answers derived:
1. The smallest quantity is [tex]\(0.005\)[/tex] mg (from [tex]\(5000\)[/tex] ng).
2. The most suitable instrument to measure the internal diameter of a test tube is vernier callipers.
3. The agreement with the student's claim about the diameter measurement is True, assuming the vernier callipers' higher precision.
This concludes the step-by-step detailed solution.
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