Answer:
0.01 H
Explanation:
V = 12 cos (1000t + 45)
C = 100 micro farad
Let the inductance be L .
When the current and the voltage are in the same phase so it is the condition of resonance.
So capacitive reactance = inductive reactance
Xc = XL
1/ωC = ωL
L = 1 / ω²C
By comparisonV = Vo Cos (ωt + Ф)
ω = 1000 rad/s
L = 1 / (1000 x 1000 x 100 x 10^-6)
L = 1 / 100
L = 0.01H
thus, the inductance of the inductor is 0.01 H.
Answer:
A) 1.4167 × 10^(-11) F
B) r_a = 0.031 m
C) E = 3.181 × 10⁴ N/C
Explanation:
We are given;
Charge;Q = 3.40 nC = 3.4 × 10^(-9) C
Potential difference;V = 240 V
Inner radius of outer sphere;r_b = 4.1 cm = 0.041 m
A) The formula for capacitance is given by;
C = Q/V
C = (3.4 × 10^(-9))/240
C = 1.4167 × 10^(-11) F
B) To find the radius of the inner sphere,we will make use of the formula for capacitance of spherical coordinates.
C = (4πε_o)/(1/r_a - 1/r_b)
Rearranging, we have;
(1/r_a - 1/r_b) = (4πε_o)/C
ε_o is a constant with a value of 8.85 × 10^(−12) C²/N.m
Plugging in the relevant values, we have;
(1/r_a - 1/0.041) = (4π × 8.85 × 10^(−12) )/(1.4167 × 10^(-11))
(1/r_a) - 24.3902 = 7.8501
1/r_a = 7.8501 + 24.3902
1/r_a = 32.2403
r_a = 1/32.2403
r_a = 0.031 m
C) Formula for Electric field just outside the surface of the inner sphere is given by;
E = kQ/r_a²
Where k is a constant value of 8.99 × 10^(9) Nm²/C²
Thus;
E = (8.99 × 10^(9) × 3.4 × 10^(-9))/0.031²
E = 3.181 × 10⁴ N/C
Producers get energy from the sun to make food from matter.
Answer:
83.72 K
Explanation:
= Polarizability of argon = 
I = First ionization = 1521 kJ/mol
r = Distance between atoms = 3.8 A
R = Gas constant = 8.314 J/mol K
T = Boiling point
Potential energy due to dispersion of gas is given by
Kinetic energy is given by
The potential and kinetic energy will balance each other
The boiling point of argon is 83.72 K
(4.2*10^5)*(4*10^-9) = 1.68* 10^-3
