Answer:
I = (1.80 × 10⁻¹⁰) A
Explanation:
From Biot Savart's law, the magnetic field formula is given as
B = (μ₀I)/(2πr)
B = magnetic field = (1.0 × 10⁻¹⁵) T
μ₀ = magnetic constant = (4π × 10⁻⁷) H/m
r = 3.6 cm = 0.036 m
(1.0 × 10⁻¹⁵) = (4π × 10⁻⁷ × I)/(2π × 0.036)
4π × 10⁻⁷ × I = 1.0 × 10⁻¹⁵ × 2π × 0.036
I = (1.80 × 10⁻¹⁰) A
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Answer:
Following are the responses to these question:
Explanation:
Since the 
is the current of ckt which depend on the reactance which inductor that also enables the ckt and inductor resistance 
for capacities
for
When 
then
therefore, 
remains at the same so, the maximum current remains the in same ckt.
Answer:
Explanation:
Given that,
Mass of the thin hoop
M = 2kg
Radius of the hoop
R = 0.6m
Moment of inertial of a hoop is
I = MR²
I = 2 × 0.6²
I = 0.72 kgm²
Period of a physical pendulum of small amplitude is given by
T = 2π √(I / Mgd)
Where,
T is the period in seconds
I is the moment of inertia in kgm²
I = 0.72 kgm²
M is the mass of the hoop
M = 2kg
g is the acceleration due to gravity
g = 9.8m/s²
d is the distance from rotational axis to center of of gravity
Therefore, d = r = 0.6m
Then, applying the formula
T = 2π √ (I / MgR)
T = 2π √ (0.72 / (2 × 9.8× 0.6)
T = 2π √ ( 0.72 / 11.76)
T = 2π √0.06122
T = 2π × 0.2474
T = 1.5547 seconds
T ≈ 1.55 seconds to 2d•p
Then, the period of oscillation is 1.55seconds
Answer:
The answer is B, velocity.
Explanation:
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