Question

An AC source has a maximum voltage of 170 V and a frequency of 60 Hz. A capacitor circuit using this AC source and a capacitor of 3×10−6 F has a maximum current of 0. 192 A 0. 128 A. 0. 320 A 0. 256 A A smoke particle has a mass of about 10−19 kg and a de Broglie wavelength of 10− 18 m, what is the velocity of this particle (in order of magnitude)? 100 m/s 104 m/s 106 m/s 103 m/s

Answer 1

To determine the maximum current in the capacitor circuit, we can use the formula:

Imax = ω * C * Vmax

where:

Imax is the maximum current,

ω is the angular frequency (2πf),

C is the capacitance, and

Vmax is the maximum voltage.

Given:

Vmax = 170 V

f = 60 Hz

C = 3×10^(-6) F

First, we need to find the angular frequency ω:

ω = 2πf = 2π * 60 Hz = 120π rad/s

Now, we can calculate the maximum current Imax:

Imax = ω * C * Vmax

= (120π rad/s) * (3×10^(-6) F) * (170 V)

≈ 0.769 A

Rounded to three decimal places, the maximum current in the capacitor circuit is approximately 0.769 A.

Therefore, none of the options provided (0.192 A, 0.128 A, 0.320 A, or 0.256 A) match the calculated value.

Moving on to the second part of the question, we are given the de Broglie wavelength (λ) of a smoke particle:

λ = 10^(-18) m

The de Broglie wavelength is related to the velocity (v) of a particle using the formula:

λ = h / (mv)

where:

h is the Planck's constant (approximately 6.626 × 10^(-34) J·s)

m is the mass of the particle, and

v is the velocity of the particle.

Rearranging the formula, we can solve for the velocity v:

v = h / (mλ)

Given:

m = 10^(-19) kg

λ = 10^(-18) m

Substituting the values into the formula:

v = (6.626 × 10^(-34) J·s) / ((10^(-19) kg) * (10^(-18) m))

= 6.626 × 10^(15) m/s

Rounded to the nearest order of magnitude, the velocity of the smoke particle is approximately 10^(16) m/s.

Therefore, none of the options provided (100 m/s, 10^(4) m/s, 10^(6) m/s, or 10^(3) m/s) match the calculated value.