An imaginary cubical surface of side L has its edges parallel to the x-, y- and z-axes, one corner at the point x = 0, y = 0, z
= 0 and the opposite corner at the point x=L, y=L, z=L. The cube is in a region of uniform electric field E⃗ =E1i^+E2j^, where E1 and E2 are positive constants. Calculate the electric flux through the cube face in the plane x = 0 and the cube face in the plane x=L. For each face the normal points out of the cube.
1 answer:
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Answer:
54%
Explanation:
We are given that
S.D=4.2 lb
Mean=
We have to find the percentage of household throw out at least 9 lb of paper a week.
Normal distribution formula :
We have a=9
%
Hence, the percentage of household throw out at least of paper a week=54%
Answer:
61.85 ohm
Explanation:
L = 12 m H = 12 x 10^-3 H, C = 15 x 10^-6 F, Vrms = 110 V, R = 45 ohm
Let ω0 be the resonant frequency.
ω0 = 2357 rad/s
ω = 2 x 2357 = 4714 rad/s
XL = ω L = 4714 x 12 x 10^-3 = 56.57 ohm
Xc = 1 / ω C = 1 / (4714 x 15 x 10^-6) = 14.14 ohm
Impedance, Z =
Z = \sqrt{45^{2}+\left ( 56.57-14.14 )^{2}} = 61.85 ohm
Thus, the impedance at double the resonant frequency is 61.85 ohm.