Answer:
Between 120 and 180 seconds
Answer:
1.17 m
Explanation:
From the question,
s₁ = vt₁/2................ Equation 1
Where s₁ = distance of the reflecting object for the first echo, v = speed of the sound in air, t₁ = time to dectect the first echo.
Given: v = 343 m/s, t = 0.0115 s
Substitute into equation 1
s₁ = (343×0.0115)/2
s₁ = 1.97 m.
Similarly,
s₂ = vt₂/2.................. Equation 2
Where s₂ = distance of the reflecting object for the second echo, t₂ = Time taken to detect the second echo
Given: v = 343 m/s, t₂ = 0.0183 s
Substitute into equation 2
s₂ = (343×0.0183)/2
s₂ = 3.14 m
The distance moved by the reflecting object from s₁ to s₂ = s₂-s₁
s₂-s₁ = (3.14-1.97) m = 1.17 m
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:
Symptoms of excessive stress include all of the following EXCEPT: increased energy level.
