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Answer:
displacement = 100 - 75 = 25 km east
The magnetic field at the center of the arc is 4 × 10^(-4) T.
To find the answer, we need to know about the magnetic field due to a circular arc.
<h3>What's the mathematical expression of magnetic field at the center of a circular arc?</h3>
- According to Biot savert's law, magnetic field at the center of a circular arc is
- B=(μ₀ I/4π)× (arc/radius²)
- As arc is given as angle × radius, so
B=( μ₀I/4π)×(angle/radius)
<h3>What will be the magnetic field at the center of a circular arc, if the arc has current 26.9 A, radius 0.6 cm and angle 0.9 radian?</h3>
B=(μ₀ I/4π)× (0.9/0.006)
= (10^(-7)× 26.9)× (0.9/0.006)
= 4 × 10^(-4) T
Thus, we can conclude that the magnitude of magnetic field at the center of the circular arc is 4 × 10^(-4) T.
Learn more about the magnetic field of a circular arc here:
brainly.com/question/15259752
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Well.....
Gravity from the sun pulls the planets torward it while inertia pulls it outward....but I guess that would be why it orbits sorry if this doesn't help but uh
Explanation:
Take north to be positive and south to be negative.
a = (v − v₀) / t
a = (-4.5 m/s − 4.5 m/s) / 8 s
a = -1.125 m/s²
The acceleration is 1.125 m/s² south.
