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
Explanation:
We know that the number of complete waves formed in 1 sec time is frequency and the distance between two consecutive crests or troughs is wavelength. And we have the formula that
Velocity = wavelength * frequency
or, frequency = velocity / wavelength
Here we can see frequency is directly proportional to velocity and indirectly proportional to wavelength.
So as the wavelength increases frequency decreases and as the wavelength decreases frequency increases.
Hope you understood
Answer:
Star A is brighter than Star B by a factor of 2754.22
Explanation:
Lets assume,
the magnitude of star A = m₁ = 1
the magnitude of star B = m₂ = 9.6
the apparent brightness of star A and star B are b₁ and b₂ respectively
Then, relation between the difference of magnitudes and apparent brightness of two stars are related as give below: 
The current magnitude scale followed was formalized by Sir Norman Pogson in 1856. On this scale a magnitude 1 star is 2.512 times brighter than magnitude 2 star. A magnitude 2 star is 2.512 time brighter than a magnitude 3 star. That means a magnitude 1 star is (2.512x2.512) brighter than magnitude 3 bright star.
We need to find the factor by which star A is brighter than star B. Using the equation given above,



Thus,

It means star A is 2754.22 time brighter than Star B.
Answer:
copper and aluminum have the highest thermal conductivity while steel and bronze have the lowest. Heat conductivity is a very important property when deciding which metal to use for a specific application.
Explanation:
brainiest pls
Answer:
PART A)
External force will be 75 N
PART B)
distance moved will be 1.125 m
Explanation:
PART A)
Given that net force on the mower is

now we also know that friction force due to ground is given as

now we have



so external force will be 75 N
PART B)
deceleration due to friction when external force is removed from it


now we can find the distance by kinematics



so the distance moved will be 1.125 m