Answer:
A negatively charged particle -q is placed at the center of a uniformly charged ring, where the ring has a total positive charge Q as shown in the following figure. The particle, confined to move along the x axis, is moved a small distance x along the axis ( where x << a) and released. Show that the particle oscillates in simple harmonic motion with a frequency given by,
Answer:
359 m/s
Explanation:
Highest frequency a person can hear is 20,000Hz.
f = c × λ
Frequency = Wave speed × Wavelength
Wavelength = 20,000 ÷ 343 = 58.31m
Wavespeed = Frequency ÷ Wavelength
c = 20,950 ÷ 58.31 = 359.3 m/s
= 359 m/s (3 s.f.)
Answer:
0.546 ohm / μm
Explanation:
Given that :
N = 1.015 * 10^17
Electron mobility, u = 3900
Hole mobility, h = 1900
Ng = 4.42 x10^22
q = 1.6*10^-19
Resistivity = 1/qNu
Resistivsity (R) = 1/(1.6*10^-19 * 1.015 * 10^17 * 3900)
= 0.01578880889 ohm /cm
Resistivity of germanium :
R = 1 / 2q * sqrt(Ng) * sqrt(u*h)
R = 1 / 2 * 1.6*10^-19 * sqrt(4.42 x10^22) * sqrt(3900*1900)
R = 1 /0.0001831
R = 5461.4964 ohm /cm
5461.4964 / 10000
0.546 ohm / μm
Answer:
and 20.86 seconds are the values of the rate constant and the half-life for this process respectively..
Explanation:
Expression for rate law for first order kinetics is given by:
where,
k = rate constant
t = age of sample
= let initial amount of the reactant
a = amount left after decay process
We have :
t = 95 s
Half life is given by for first order kinetics::
and 20.86 seconds are the values of the rate constant and the half-life for this process respectively..
