The magnitude of the magnetic field on the axis of the ring 5 cm from its center is 143 pT.
The radius of the nonconducting ring is R = 10 cm.
The ring is uniformly charged q = 10 μC.
The angular speed of the ring, ω = 20 rad/s
The ring is x = 5 cm from the center of the ring.
Now,
R = 10 cm = 0.1 m
q = 10.0 μC = 10 × 10⁻⁶ C
x = 5 cm = 0.05 m
The magnetic field on the axis of a current loop is given as:
B = [ μ₀ IR² ] / [4π(x² + R²)^{3/2} ]
Now, I = q / [2π/ω]
So, the magnitude of the magnetic field which is directed away from the center is:
B = [ μ₀ ωqR² ] / [4π(x² + R²)^{3/2} ]
B = [ μ₀ (200) (10 × 10⁻⁶) (0.1)² ] / [4π((0.05)² + (0.1)²)^{3/2} ]
B = 1.43 × 10⁻¹⁰ T
B = 143 pT
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Answer:
D. When a chemical reaction releases energy, it is an example of an endothermic reaction.
Explanation:
"Endo" means within. Therefore, an endothermic reaction releasing energy makes no sense.
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Answer:
U = 9.1 m/s
Explanation:
from the question we are given the following
time (t) = 1.8 s
angle = 23 degrees
acceleration due to gravity (g) = 9.8 m/s^{2}
let us first calculate the initial velocity (u) which too the first ball to its maximum height from the equation below
v = u + 0.5at
- The final velocity (v) is zero since the ball comes to rest
- The time (t) it takes to get to the maximum height would be half the time it is in the air, t = 0.5 x 1.8 = 0.9
therefore
0 = u - (0.5 x 9.8 x 0.9)
u = 7.9 m/s
for the second ball to get to the maximum height of the first ball, the vertical component of its initial velocity (U) must be the same as the initial velocity of the first ball. therefore
U sin 60 = 7.9
U = 7.9 ÷ sin 60
U = 9.1 m/s