Weight of the child m = 50 kg
Radius of the merry -go-around r = 1.50 m
Angular speed w = 3.00 rad/s
(a)Child's centripetal acceleration will be a = w^2 x r = 3^2 x 1.50 => a = 9 x
1.5
Centripetal Acceleration a = 13.5m/sec^2
(b)The minimum force between her feet and the floor in circular path
Circular Path length C = 2 x 3.14 x 1.50 => c = 3 x 3.14 => C = 9.424
Time taken t = 2 x 3.14 / w => t = 6.28 / 3 => t = 2.09
Calculating velocity v = distance / time = 9.424 / 2.09 m/s => v = 4.5 m/s
Calculating force, from equation F x r = mv^2 => F = mv^2 / r => 50 x (4.5)^2
/ 1.5
F = 50 x 3 x 4.5 => F = 150 x 4.5 => F = 675 N
(c)Minimum coefficient of static friction u
F = u x m x g => u = F / m x g => u = 675/ 50 x 9.81 => 1.376
u = 1.376
Hence with the force and the friction coefficient she is likely to stay on merry-go-around.
The horizontal component of the magnetic field is 12.6 μT.
The magnetic influence on moving electric currents, electric charges, and magnetic materials is described by a magnetic field, which is a vector field. When a charge moves through a magnetic field, a force that is perpendicular to both its own velocity and the magnetic field operates on it.
The horizontal component of the Earth's magnetic field is perpendicular to the axis of a circular coil with five turns and a diameter of D = 30.0 cm that is vertically orientated.
A coil current of I = 0.600 A causes a horizontal compass to deflect 45.0° from magnetic north when it is positioned in the coil's center.
Let B be the magnetic field and R be the radius of the circular coil.
Then the horizontal component of the Earth's magnetic field is given as:
B(h) = B(coil) = μ₀ NI / 2R
B(h) = (4π × 10⁻⁷ ) (5)(0.6) / 0.3
B(h) = 12.6 μT
Learn more about magnetic field here:
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Its d all the above your welcome
