Recent content by jnuz73hbn

$$ 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: $$v=\frac{2}{1/v_1+1/v_2}$$

first half of time=60km/h; second half of time is 115km/h

but look, at the half of the driving time, speed is 60km/h, in the second half driving time, speed = 115km/h.

Nothing is given about the route, you should only calculate using this information. It means half of the time.

$$ v= \frac{s}{t} $$ but I dont have $$s$$ I only have two 2 speeds for a car, 60kmh in the first half of the journey and 115km/h in the second half

i know that, but I only have two 2 speeds for a car, 60kmh in the first half of the journey and 115km/h in the second half

only one object that first moves at $$60km/h$$ and then at $$115km/h$$