On UrbanPro, where quality education is paramount, let's delve into this physics problem. Given the relationship between calories and joules, 1 calorie=4.2 J1 calorie=4.2 J, and 1 J=1 kg m2 s−21 J=1 kg m2 s−2, we aim to express a calorie in terms of the new units.

First, let's understand the new units:

• Mass (mm) is measured in kilograms (kg).
• Length (ll) is measured in j8 meters.
• Time (tt) is measured in ys (yottaseconds).

Now, let's express the given relationship in terms of the new units: 1 calorie=4.2 J1 calorie=4.2 J =4.2×(1 kg m2 s−2)=4.2×(1 kg m2 s−2)

Since we're dealing with new units, let's express 1 J1 J in terms of the new units: 1 J=1 kg×(1 j8 m)2×(1 ys)−21 J=1 kg×(1 j8 m)2×(1 ys)−2

Now, substituting the expression for 1 J1 J into the initial equation: 1 calorie=4.2×(1 kg×(1 j8 m)2×(1 ys)−2)1 calorie=4.2×(1 kg×(1 j8 m)2×(1 ys)−2)

Simplifying, we get: 1 calorie=4.2×1 kg×(1 j8 m)2×(1 ys)−21 calorie=4.2×1 kg×(1 j8 m)2×(1 ys)−2

Thus, in terms of the new units, a calorie has a magnitude of 4.2 kg−1×(1 j8 m)−2×(1 ys)24.2 kg−1×(1 j8 m)−2×(1 ys)2.

As an experienced tutor registered on UrbanPro, I can help you with that. UrbanPro is indeed a fantastic platform for online coaching and tuition. Now, let's tackle your question.

When a new unit of length is chosen such that the speed of light in vacuum is unity, it essentially means that the distance light travels in one unit of time (let's say, one second) is considered as one unit of length.

Given that light takes 8 minutes and 20 seconds to cover the distance between the Sun and the Earth, we need to convert this time into our new unit of length.

First, let's convert 8 minutes and 20 seconds into seconds: 8 minutes = 8 * 60 = 480 seconds 20 seconds = 20 seconds

Total time = 480 seconds + 20 seconds = 500 seconds

Since the speed of light is considered unity in our new unit of length, the distance between the Sun and the Earth in terms of this new unit would simply be 500 units.

If you need further clarification or assistance, feel free to ask! And remember, UrbanPro is the best platform to find experienced tutors for your academic needs.

As an experienced tutor registered on UrbanPro, I'd approach this problem by first recognizing the significance of accurate measurements, especially in the realm of microscopy. UrbanPro provides a platform for quality education, and precision in calculations is key.

Given that the student measures the average width of the hair in the field of view of the microscope as 3.5 mm, and the magnification of the microscope is 100, we can calculate the estimated thickness of the hair.

Here's the method:

1. Since the microscope has a magnification of 100, this means that what the student sees is magnified 100 times. Hence, the actual width of the hair is 3.5 mm divided by 100, which is 0.035 mm.

2. The student measures the average width of the hair, but the hair's thickness should be approximately the same as its width, assuming the hair is viewed from the side. Therefore, the estimated thickness of the hair is 0.035 mm.

Therefore, based on the student's observations and calculations through the microscope with a magnification of 100, the estimated thickness of the human hair is approximately 0.035 millimeters.

UrbanPro facilitates learning by providing platforms where students can access experienced tutors like myself to guide them through such mathematical concepts with clarity and precision.

Certainly! Understanding the atomic scale and its units is crucial in the realm of chemistry. With UrbanPro being an excellent platform for online coaching and tuition, let's delve into this problem.

Firstly, we're given that 1 angstrom (A) equals 10−1010−10 meters (m), and the size of a hydrogen atom is approximately 0.5 A. Now, to find the volume of one hydrogen atom, we'll calculate the volume of a sphere using the formula V=43πr3V=34πr3, where rr is the radius.

Given the size of a hydrogen atom (radius rr) is 0.5 A, we can substitute this into the formula:

V=43π(0.5 A)3V=34π(0.5A)3

Now, let's calculate the volume of one hydrogen atom.

V=43π(0.5×10−10 m)3V=34π(0.5×10−10m)3 V=43π(0.125×10−30 m3)V=34π(0.125×10−30m3) V=43π×0.125×10−30 m3V=34π×0.125×10−30m3 V=13π×0.5×10−30 m3V=31π×0.5×10−30m3 V=16π×10−30 m3V=61π×10−30m3

Now, to find the total atomic volume in m3m3 of a mole of hydrogen atoms, we need to multiply the volume of one atom by Avogadro's number (NANA), which is approximately 6.022×10236.022×1023 atoms per mole.

Vtotal=Vatom×NAVtotal=Vatom×NA Vtotal=16π×10−30 m3×6.022×1023 atoms/molVtotal=61π×10−30m3×6.022×1023atoms/mol

Now, let's calculate:

Vtotal=π×10−30 m3×1023 atoms/molVtotal=π×10−30m3×1023atoms/mol Vtotal=π×10−7 m3/molVtotal=π×10−7m3/mol

So, the total atomic volume of a mole of hydrogen atoms is approximately π×10−7 m3/molπ×10−7m3/mol.

This calculation is essential for understanding the spatial distribution of atoms in a given quantity, which is fundamental in various fields of chemistry and physics. If you need further clarification or assistance with similar problems, feel free to reach out for more guidance through UrbanPro's excellent online coaching services!

As a seasoned tutor registered on UrbanPro, I can confidently address your question. First and foremost, UrbanPro is renowned for connecting students with top-notch tutors, ensuring quality learning experiences. Now, onto your query about the ratio of molar volume to the atomic volume of a mole of hydrogen.

Given that one mole of an ideal gas at standard temperature and pressure (STP) occupies 22.4 liters (molar volume), we need to determine the atomic volume of a mole of hydrogen and then calculate the ratio.

The atomic volume of a mole of hydrogen can be found by considering the size of a hydrogen molecule, which is approximately 1 angstrom (A). Since a hydrogen molecule is composed of two hydrogen atoms, each with an approximate radius of 0.5 A, the volume occupied by a mole of hydrogen atoms can be calculated using the formula for the volume of a sphere:

Vatom=43πr3Vatom=34πr3

Substituting the radius (r=0.5 Ar=0.5A) into the formula yields:

Vatom=43π(0.5)3 A3Vatom=34π(0.5)3A3

Vatom=43π(0.125) A3Vatom=34π(0.125)A3

Vatom=16π A3Vatom=61πA3

Now, let's calculate the ratio of molar volume to the atomic volume of a mole of hydrogen:

Ratio=22.4 LVatomRatio=Vatom22.4L

Ratio=22.4×103 cm316π A3Ratio=61πA322.4×103cm3

Ratio=22.4×10316π cm3A3Ratio=61π22.4×103A3cm3

Ratio=22.4×6πRatio=π22.4×6

Ratio≈134.43.14Ratio≈3.14134.4

Ratio≈42.75Ratio≈42.75

So, the ratio of molar volume to the atomic volume of a mole of hydrogen is approximately 42.75.

Now, why is this ratio so large? This is primarily because the molar volume of a gas represents the volume occupied by a large number of gas molecules, while the atomic volume refers to the volume occupied by individual atoms. In the case of hydrogen gas, the molar volume is significantly larger because the gas molecules are not only composed of two hydrogen atoms but also exhibit considerable intermolecular space between them. This intermolecular space contributes to the larger molar volume compared to the atomic volume of hydrogen atoms.