<span>Charging by friction occurs, Electrons are transferred when one object rubs against another.
Another example of this would be socks on carpet.
Hope this helps!</span>
Many fundamental laws of physics and chemistry can be formulated when doing this.
Answer:
Explanation:
This problem relates to interference of light in thin films .
The condition of bright fringe in thin films which is sandwitched by two layers of medium having lesser refractive index is as follows.
2nt = (2n+1) λ / 2 , n is refractive index of thin layer , t is its thickness , λ is wavelength of light .
2 x 1.5 t = λ / 2 , if n = 0 for minimum thickness.
2 x 1.5 t = 600 / 2 nm
t = 100 nm .
The short answer is that the displacement is equal tothe area under the curve in the velocity-time graph. The region under the curve in the first 4.0 s is a triangle with height 10.0 m/s and length 4.0 s, so its area - and hence the displacement - is
1/2 • (10.0 m/s) • (4.0 s) = 20.00 m
Another way to derive this: since velocity is linear over the first 4.0 s, that means acceleration is constant. Recall that average velocity is defined as
<em>v</em> (ave) = ∆<em>x</em> / ∆<em>t</em>
and under constant acceleration,
<em>v</em> (ave) = (<em>v</em> (final) + <em>v</em> (initial)) / 2
According to the plot, with ∆<em>t</em> = 4.0 s, we have <em>v</em> (initial) = 0 and <em>v</em> (final) = 10.0 m/s, so
∆<em>x</em> / (4.0 s) = (10.0 m/s) / 2
∆<em>x</em> = ((4.0 s) • (10.0 m/s)) / 2
∆<em>x</em> = 20.00 m
Answer:
(a) 6.567 * 10^15 rev/s or hertz
(b) 8.21 * 10^14 rev/s or hertz
Explanation:
Fn= 4π^2k^2e^4m * z^2/(h^3*n^3)
Where Fn is frequency at all levels of n.
Z = 1 (nucleus)
e = 1.6 * 10^-19c
m = 9.1 * 10^-31 kg
h = 6.62 * 10-34
K = 9 * 10^9 Nm2/c2
(a) for groundstate n = 1
Fn = 4 * π^2 * (9*10^9)^2*(1.6*10^-19)^4* (9.1 * 10^-31) * 1 / (6.62 * 10^-31)^3 = 6.567 * 10^15 rev/s
(b) first excited state
n = 1
We multiple the groundstate answer by 1/n^3
6.567 * 10^15 rev/s/ 2^3
F2 = 8.2 * 10^ 14 rev/s
