If this case could ever happen, the speed would follow from this formula:
with f the frequency and lambda the wavelength. We are give a wavelength of 10m. The frequencies of the visible light can range between 400 to about 790 Terahertz, so let us pick a middle point of 600 THz ("green-ish") as a "representative."
The speed of such a wave would have to be 6e+15 m/s (which would be 7 orders of magnitude higher than the universal speed of light constant)
Answer:
Coefficient of friction = 0.836
Explanation:
If v be the speed after one quarter of the circular path
v² = 2as = 2 x 1.85 x 2πr/4 ; v²/r = 1.85 x 3.14 = 5.8
tangential acceleration = 5.8 m/s²
radial acceleration = v² /r = 5.8
total acceleration = √2 x 5.8
m x√2 x 5.8 = m x g xμ
μ = √2 x 5.8 / 9.8 = 0.836
A 5.00 A current runs through a 12 gauge copper wire (diameter 2.05 mm) and through a light bulb. Copper has 8.5*10^28 free electrons per cubic metre.
a) How many electrons pass through the light bulb each second?
b) What is the current density in the wire? (answer in A/m^2)
<span>c) At what speed does a typical electron pass by any given point in the wire? (answer in m/s)
</span>a) 5.0 A = 5.0 C/s
. Number of electrons in 5.0C = 5.0 / 1.60^-19 = 3.125^19
. 5.0 A ►= 3.125^19 electrons/s
b) A/m² = 5.0 / π(1.025^-3 m)² .. .. ►= 1.52^6 A/m²
c) Charge density (q/m³) = 8.50^28 e/m³ x 1.60^-19 = 1.36^10 C/m³
(q/m³)(m²)(m/s) = q/s (current i in C/s [A])
(m²) = Area
(m/s) = mean drift speed
(q/m³)(A)(v) = i
v = i.[(q/m³)A]ˉ¹
<span>v = 5.0 [1.36^10 * π(1.025^-3 m)²]ˉ¹.. .. ►v = 1.10^-4 m/s</span>
I think it’s proximity of the sun
Your fluid intake is enhanced, you feel refreshed, you get a sugar high for a while, and a lot if the rust is removed from your stomach and small intestine.
