<span>a)
Capacitance = k x ε° x area / separation
ε° = 8.854 10^-12 F/ m
k = 2.4max
average k = 0.78 / 1.27 * 2.4 +(1.27- 0.78) / 1.27 * 1 = 1.474 + 0.386 = 1.86
(61.4 % separation k = 2.4 --- 38.6 % k = 1 air --- average k = 0.614 * 2.34 + 0.386 * 1 = 1.86
area = 145 cm2 = 0.0145 m2
separation = 1.27 cm 0.0127 m
C = 1.86 * 8.854 10^-12 * 0.0145 / 0.0127 = 18.8 pF
b) Q = C * V --- 18.8 * 83 = 1560.4 pC = 1.5604 nC
c) E = V / d = 83 / 0.0127 = 6535.4 V/m </span>
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Answer:
Explained below:
Explanation:
Water quality is determined in terms of the physical, chemical and biological content of water. The quality of water of lakes and rivers variates with the seasons and also geographic areas, still in the absence of pollution. There are so many factors which affect water quality are as follows :
Pesticides
Temperature
Runoff
Sedimentation
Erosion
pH
Dissolved oxygen
Pesticides
Litter and rubbish
Decayed organic materials
Toxic and hazardous substances
Oils, grease, and other chemicals
Answer:
D
Explanation:
First we define our variables
V0=29.4
a=-9.8
V=0
We have to find the maximum displacement , which I will define as X
We use formula v^2=v0^2+2aX
All we do is substitute our values
0=29.4^2-19.6X
29.4^2=19.6X
X=29.4^2/19.6=44.1
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