The main requirement for a good conductor of electricity is to have a lot of valence electrons. Valence electrons are the electrons of the outer shells of atoms not bound with other atoms (for example through covalent bounds). These electrons are "free to escape" as soon as an electric field with enough intensity is applied to the material, and therefore these electrons will be free to move in the material producing an electric current.
Answer:
Spring's displacement, x = -0.04 meters.
Explanation:
Let the spring's displacement be x.
Given the following data;
Mass of each shrew, m = 2.0 g to kilograms = 2/1000 = 0.002 kg
Number of shrews, n = 49
Spring constant, k = 24 N/m
We know that acceleration due to gravity, g is equal to 9.8 m/s².
To find the spring's displacement;
At equilibrium position:
Fnet = Felastic + Fg = 0
But, Felastic = -kx
Total mass, Mt = nm
Fg = -Mt = -nmg
-kx -nmg = 0
Rearranging, we have;
kx = -nmg
Making x the subject of formula, we have;
![x = \frac {-nmg}{k}](https://tex.z-dn.net/?f=%20x%20%3D%20%5Cfrac%20%7B-nmg%7D%7Bk%7D%20)
Substituting into the formula, we have;
![x = \frac {-49*0.002*9.8}{24}](https://tex.z-dn.net/?f=%20x%20%3D%20%5Cfrac%20%7B-49%2A0.002%2A9.8%7D%7B24%7D%20)
![x = \frac {-0.9604}{24}](https://tex.z-dn.net/?f=%20x%20%3D%20%5Cfrac%20%7B-0.9604%7D%7B24%7D%20)
x = -0.04 m
Therefore, the spring's displacement is -0.04 meters.
Answer:
the frequency of the second harmonic of the pipe is 425 Hz
Explanation:
Given;
length of the open pipe, L = 0.8 m
velocity of sound, v = 340 m/s
The wavelength of the second harmonic is calculated as follows;
L = A ---> N + N--->N + N--->A
where;
L is the length of the pipe in the second harmonic
A represents antinode of the wave
N represents the node of the wave
![L = \frac{\lambda}{4} + \frac{\lambda}{2} + \frac{\lambda}{4} \\\\L = \lambda](https://tex.z-dn.net/?f=L%20%3D%20%5Cfrac%7B%5Clambda%7D%7B4%7D%20%2B%20%5Cfrac%7B%5Clambda%7D%7B2%7D%20%2B%20%5Cfrac%7B%5Clambda%7D%7B4%7D%20%5C%5C%5C%5CL%20%3D%20%5Clambda)
The frequency is calculated as follows;
![F_1 = \frac{V}{\lambda} = \frac{340}{0.8} = 425 \ Hz](https://tex.z-dn.net/?f=F_1%20%3D%20%5Cfrac%7BV%7D%7B%5Clambda%7D%20%3D%20%5Cfrac%7B340%7D%7B0.8%7D%20%3D%20425%20%5C%20Hz)
Therefore, the frequency of the second harmonic of the pipe is 425 Hz.
C. Radiosonde is the answer
the above mentioned is not correct