Explanation:
The blocks are spinning around the center of mass, but they're also sliding horizontally. So they still have a positive displacement.
You can download
the answer here
bit.
ly/3gVQKw3
Answer:
F = 2.26 × 10⁻³ N
Explanation:
given,
length of rod = 11 cm
charge = 19 nC
linear charge density = 3.9 x 10⁻⁷ C/m
electric force at 2 cm away.

F = E q

integrating from 0.02 to 0.02 + L
![F= \dfrac{2K\lambda\ q}{L}[ln(0.02+L)-ln(0.002)]](https://tex.z-dn.net/?f=F%3D%20%5Cdfrac%7B2K%5Clambda%5C%20q%7D%7BL%7D%5Bln%280.02%2BL%29-ln%280.002%29%5D)
![F= \dfrac{2\times 9 \times 10^9\times 3.9\times 10^{-7}\times 19 \times 10^{-9}}{0.11}[ln(0.02+0.11)-ln(0.002)]](https://tex.z-dn.net/?f=F%3D%20%5Cdfrac%7B2%5Ctimes%209%20%5Ctimes%2010%5E9%5Ctimes%203.9%5Ctimes%2010%5E%7B-7%7D%5Ctimes%2019%20%5Ctimes%2010%5E%7B-9%7D%7D%7B0.11%7D%5Bln%280.02%2B0.11%29-ln%280.002%29%5D)
F = 2.26 × 10⁻³ N
Molecular formulas:
- CH₂O;
- C₂H₄O₂;
- C₆H₁₂O₆.
<h3>Explanation</h3>
The empirical formula of a compound tells only the ratio between atoms of each element. The empirical formula CH₂O indicates that in this compound,
- for each C atom, there are
- two H atoms, and
- one O atom.
The molecular weight (molar mass) of the molecule depends on how many such sets of atoms in each molecule. The empirical formula doesn't tell anything about that number.
It's possible to <em>add</em> more of those sets of atoms to a molecular formula to increase its molar mass. For every extra set of those atoms added, the molar mass increase by the mass of that set of atoms. The mass of one mole of C atoms, two mole of H atoms, and one mole of O atoms is
.
- CH₂O- 30.0 g/mol;
- C₂H₄O₂- 30.0 + 30.0 = 2 × 30.0 = 60.0 g/mol;
- C₃H₆O₃- 30.0 + 30.0 + 30.0 = 3 × 30.0 = 90.0 g/mol.
It takes one set of those atoms to achieve a molar mass of 30.0 g/mol. Hence the molecular formula CH₂O.
It takes two sets of those atoms to achieve a molar mass of 60.0 g/mol. Hence the molecular formula C₂H₄O₂.
It takes
sets of those atoms to achieve a molar mass of 180.0 g/mol. Hence the molecular formula C₆H₁₂O₆.

In order to find it's square root, we could make it into two square roots.

Let us find the square roots of both radicals seprately.

Each pair of a number inside square root gives a number out .



Therefore,

