Recent content by jnuz73hbn

My question is whether these formulas are sufficient to answer the question or if something is missing.

Is it correct for the problem? $$ r_{\max} = \sqrt{\frac{q^2}{16\pi\varepsilon_0mg}} = \sqrt{ \frac{(3.9\times10^{-7}\,\mathrm C)^2} {16\pi(8.854\times10^{-12}\,\mathrm{F/m})(2.8\times10^{-3}\,\mathrm{kg})(9.81\,\mathrm{m/s^2})} } \approx 0.1115\,\mathrm m = 11.15\,\mathrm{cm} $$ $$ r_{\min} =...

Ceiling = conducting and grounded (with a thin insulating surface layer). Sphere = insulating spherical shell, with homogeneous fixed charge q=+3,8 \times10^{-7}\,\mathrm C. No charge transfer between sphere and ceiling. The method of image charges is to be considered.

Given values are: q= 3.9 * 10^-7 C , m= 2,8 *10^-3 kg $$ r_{\max} = \sqrt{ \frac{(3.9\times10^{-7})^2} {16\pi(8.854\times10^{-12})(2.8\times10^{-3})(9.81)} } \approx 0.1115\,\mathrm{m} $$

for the first I calculated this. Is that formula correct for r_max? Because 0,7m for a radius of a sphere is very big. $$ r_{\max} = \sqrt{ \frac{ (8.99\times10^9\,\mathrm{N\,m^2/C^2}) (3.9\times10^{-7}\,\mathrm{C})^2 }{ 4(2.8\times10^{-3}\,\mathrm{kg})(9.81\,\mathrm{m/s^2}) } } \approx...

$$ r_{\max}=\sqrt{\frac{kq^2}{4mg}} $$ $$ r_{\min}=\sqrt{\frac{k|q|}{E_{\mathrm{crit}}}} $$

$$ E = \frac{5.0\;\mathrm{V}}{0.020\;\mathrm{m}} = 250\;\mathrm{V/m} F = q\,E = (-e)\times E = -(1.602\times10^{-19}\;\mathrm{C})\times250\;\mathrm{V/m} = -4.005\times10^{-17}\;\mathrm{N} |F| = 4.005\times10^{-17}\;\mathrm{N} a = \frac{|F|}{m_e} =...

got it, thank you , now I have a good density

well, $$ F_{\mathrm{buoyancy}} = (m_3 - m_2) \cdot g = (843.2 - 817.0)\,\mathrm{g} \cdot 9.81 = 0.0262\,\mathrm{kg} \cdot 9.81 \approx 0.257\,\mathrm{N} $$ The displaced volume of water is: $$ V = \frac{0.257}{1000 \cdot 9.81} \approx 2.62 \times 10^{-5}\,\mathrm{m^3} = 26.2\,\mathrm{cm^3} $$...

We assume the sample is shaped like a perfect cylinder. The volume is given by: $$V = \pi \cdot \left( \frac{7,2}{2} \right)^2 \cdot 1,8 = \pi \cdot 3,6^2 \cdot 1,8 = 73,2\,\mathrm{cm^3}$$ Then the approximate density is: $$ \rho = \frac{98,7}{73,2} = 1,348\,\mathrm{g/cm^3} =...

I just want to know where the formula comes from using the unit circle or how it relates to sin cos in the unit circle

however, i wanted to go via the unit circle with sinus and cosine to derive exactly this definition

$$ \ ω = \frac{Δα}{Δt} \ $$

is this new formula correct?

$$t_1=\frac{D/2}{v_1}$$ that means: $$v=\frac{2}{1/v_1+1/v_2}$$