Answer:
Explanation:
Use the equation:
(where T is the period)
(9 comes from the period of a full wavelength)
1200 times 2.5 is 3000. So it flew 3000 kilometers.
The answer is <span>nuclear fission. T</span>he source of the radioactive nuclei present in spent fuel rods is nuclear fission. Nuclear fission<span> is the </span>process<span> in which a large nucleus splits into two smaller nuclei with the release of energy. </span>
I would believe that this is false.
Answer:
A) r = 0.03 m
B) r = 0.0533 m
C) B_max = 0.00003 T
Explanation:
Formula for magnetic field inside the capacitor when it is parallel to the length element is;
B_in = (μ_o•I•r/(2πR²)
Formula for maximum magnetic field is;
B_max = (μ_o•I/(2πR)
Formula for magnetic field outside the capacitor is; B_out = (μ_o•I/(2πr)
A) Magnetic field inside the capacitor is gotten from our first equation above;
B_in = (μ_o•I•r/R²)
Since we want to find the radius at which the magnitude of the induced magnetic field equal to 75% or 0.75 of its maximum value.
Thus;
B_in = 0.75B_max
(μ_o•I•r/(2πR²) = 0.75((μ_o•I/(2πR))
μ_o•I and 2πR will cancel out to give;
r/R = 0.75
r = 0.75R
We are given R = 40 mm = 0.04 m
r = 0.75 × 0.04
r = 0.03 m
B) magnetic field outside the capacitor is; B_out = (μ_o•I/(2πr)
Thus for the magnitude of the induced magnetic field equal to 75% or 0.75 of its maximum value:
B_out = 0.75B_max
(μ_o•I/(2πr) = 0.75((μ_o•I/(2πR))
μ_o•I and 2π will cancel out to give;
1/r = 0.75/R
r = R/0.75
r = 0.04/0.75
r = 0.0533 m
C) B_max = μ_o•I/(2πR)
μ_o is a constant known as vacuum of permeability with a value of 4π × 10^(-7) T.m/A
Thus;
B_max = (4π × 10^(-7) × 6)/(2π × 0.04)
B_max = 0.00003 T
