Ch 31 hw exercise 31.10 7 of 15 constants you want the current amplitude through a inductor with an inductance of 4.90 mh (part of the circuitry for a radio receiver) to be 3.00 ma when a sinusoidal voltage with an amplitude of 12.0 v is applied across the inductor. part a what frequency is required? ff = nothing hz request answer provide feedback

f = 130 Khz

Explanation:

In a circuit driven by a sinusoidal voltage source, there exists a fixed relationship between the amplitudes of the current and the voltage through any circuit element, at any time.

For an inductor, this relationship can be expressed as follows:

VL = IL * XL (1) , which is a generalized form of Ohm's Law.

XL is called the inductive reactance, and is defined as follows:

XL = ω*L = 2*π*f*L, where f is the frequency of the sinusoidal source (in Hz) and  L is the value of the inductance, in H.

Replacing in (1), by the values given of VL, IL, and L, we can solve for f, as follows:

f = VL / 2*π*IL*L = 12 V / 2*π*(3.00*10⁻³) A* (4.9*10⁻³) H = 130 Khz

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft