Answer:
I = 0.002593 A = 2.593 mA
Explanation:
Current density = J = (3.00 × 10⁸)r² = Br²
B = (3.00 × 10⁸) (for ease of calculations)
The current through outer section is given by
I = ∫ J dA
The elemental Area for the wire,
dA = 2πr dr
I = ∫ Br² (2πr dr)
I = ∫ 2Bπ r³ dr
I = 2Bπ ∫ r³ dr
I = 2Bπ [r⁴/4] (evaluating this integral from r = 0.900R to r = R]
I = (Bπ/2) [R⁴ - (0.9R)⁴]
I = (Bπ/2) [R⁴ - 0.6561R⁴]
I = (Bπ/2) (0.3439R⁴)
I = (Bπ) (0.17195R⁴)
Recall B = (3.00 × 10⁸)
R = 2.00 mm = 0.002 m
I = (3.00 × 10⁸ × π) [0.17195 × (0.002⁴)]
I = 0.0025929449 A = 0.002593 A = 2.593 mA
Hope this Helps!!!
Answer:
2.76×10⁻¹⁰ C
Explanation:
Applying,
V = W/q................... Equation 1
Where V = Electric Potential, E = Electric potential energy, q = charge.
make q the subject of the equation
q = W/V................ Equation 2
From the question,
Given: W = 4.26×10⁻⁸ J, V = 154.5 V
Substitute these values into equation 2
q = 4.26×10⁻⁸/154.5
q = 2.76×10⁻¹⁰ C
Answer:
12 cm/s
Explanation:
Quite simply, you are looking for cm/s
so 60 cm / 5s = 12 cm/s
The Answer is A, the iris dilates the pupil.
Answer:
Frequency of sound wave = 198.83 hertz (Approx.)
Explanation:
Given:
Velocity of sound wave in air = 340 m/s
Wavelength = 1.71 meter
Find:
Frequency of sound wave
Computation:
Frequency = Velocity / Wavelength
Frequency of sound wave = Velocity of sound wave in air / Wavelength
Frequency of sound wave = 340 / 1.71
Frequency of sound wave = 198.8304
Frequency of sound wave = 198.83 hertz (Approx.)
