Answer:
F = F₀ 0.2
Explanation:
For this exercise we apply Coulomb's law with the initial data
F₀ = k q_A q_B / d²
indicate several changes
q_A ’= ½ q_A
q_B ’= 1/10 q_B
d ’= ½ d
let's substitute these new values in the Coulomb equation
F = k q_A ’q_B’ / d’²
F = k ½ q_A 1/10 q_B / (1/2 d)²
F = (k q_A q_B / d2) ½ 1/10 2²
F = F₀ 0.2
Answer:
the answer is c
Explanation:
because if the force of gravity increases its not gonna change its weight it will just float
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
<u>The moon is not seen at all on the new moon because:</u>
The moon itself doesn't make a light, it reflects sunlight. At the point when the moon lies between the Earth and the Sun, it is just the posterior of the moon that is light by the Sun. In this circumstance, the side of the moon confronting the Earth can't mirror any light whatsoever and seems dull.
The distinction between another moon and a sunlight based over-shadowing is during a sun oriented obscuration, the moon is actually in the position to completely obstruct the sun during the day.
