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:
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Answer:
C. Waves transfer energy, but not matter.
Explanation: hope this helps :)
Answer:
T = 1.2 s
T = 15.1 m = 15 m
Explanation:
This is a case of projectile motion:
TOTAL TIME OF FLIGHT:
The formula for total time of flight in projectile motion is:
T = 2 V₀ Sinθ/g
where,
T = Total Time of Flight = ?
V₀ = Launch Speed = 13.9 m/s
θ = Launch Angle = 25°
g = 9.8 m/s²
Therefore,
T = (2)(13.9 m/s)(Sin 25°)/(9.8 m/s²)
<u>T = 1.2 s</u>
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RANGE OF BALL:
The formula for range in projectile motion is:
R = V₀² Sin2θ/g
where,
R = Horizontal Distance Covered by ball = ?
Therefore,
T = (13.9 m/s)²(Sin 2*25°)/(9.8 m/s²)
<u>T = 15.1 m = 15 m</u>
Explanation:
Polaris also called North star is the brightest star in the the constellation Ursa Minor. It remains stationary whereas entire north sky move around it. It the point of north celestial pole.
A) False. This is because All of the stars move in a counterclockwise path around Polaris.
B) False. Reason stated above.
C) True, As the stars move towards the Polaris there orbital radius decreases as they move smaller circles.
D) False, As the stars move towards the Polaris there orbital radius decreases as they move smaller circles.
E) True.
F) False.
Answer:
-v/2
Explanation:
Given that:
- Collides with the wall going through a sliding motion on on the plane smooth surface.
- Upon rebounding from the wall its kinetic energy becomes one-fourth of the initial kinetic energy before collision.
<u>We know, kinetic energy is given as:</u>
consider this to be the initial kinetic energy of the body.
<u>Now after collision:</u>
Considering that the mass of the body remains constant before and after collision.
Therefore the velocity of the body after collision will become half of the initial velocity but its direction is also reversed which can be denoted by a negative sign.
