A 56.0-kg child takes a ride on a Ferris wheel that rotates four times each minute and has a diameter of 22.0 m. (a) What is the centripetal acceleration of the child? magnitude 1.76 m/s2 direction (b) What force (magnitude and direction) does the seat exert on the child at the lowest point of the ride? magnitude 647.4 N direction (c) What force does the seat exert on the child at the highest point of the ride? magnitude 450.2 N direction (d) What force does the seat exert on the child when the child is halfway between the top and bottom? (Assume the Ferris wheel is rotating clockwise and the child is moving upward.) magnitude N direction 79.8 ° counter-clockwise from the horizontal

a. 3.86 m/s²

b. 332.64 N directed upwards towards the center of the ferris wheel.

c. 764.96 N directed downwards towards the center of the ferris wheel.

d. 589.84 N at 21.5° counterclockwise from the horizontal direction.

Explanation:

a. Since the ferris wheel rotates 4 times per minute, its period, T = 60 s/4 = 15 s.

We now find it angular speed ω = 2π/T = 2π/15 = 0.418 rad/s

We then calculate its centripetal acceleration from a = rω² where r = radius of ferris wheel = 22.0 m.

So, a = 22 m × (0.418 rad/s)² = 3.86 m/s²

b. At the lowest point, the normal force, N and the centripetal force, F both act in opposite directions to the weight, mg of the object. So,

N + F = mg

N = mg - F      

N = mg - ma where a is the centripetal acceleration

N = m(g - a)

N = 56 kg(9.8 m/s² - 3.86 m/s²)

N = 56 kg × 5.94 m/s²

N = 332.64 m/s²

The normal force the seat exerts on the child is thus 332.64 N directed upwards towards the center of the ferris wheel.

c. At the highest point, the weight, mg of the object and the centripetal force, F both act in opposite directions to the normal force, N. So,

N = mg + F      

N = mg + ma where a is the centripetal acceleration

N = m(g + a)

N = 56 kg(9.8 m/s² + 3.86 m/s²)

N = 56 kg × 13.66 m/s²

N = 764.96 N

The normal force the seat exerts on the child is thus 764.96 N directed downwards towards the center of the ferris wheel.

d. Half way between the top and the bottom of the ferris wheel, the normal force must balance the weight and the centripetal force so the child doesn't fall off. For this to happen, the normal force is thus the resultanf of the centripetal force and the weight of the child. Since these two forces are perpendicular at this instance,

N = √[(mg)² + (ma)²] = m√(g² + a²) = 56 kg√[(9.8 m/s²)² + (3.86 m/s²)²] = 56 kg√[(97.196 (m/s²)² + 14.8996(m/s²)²] = 56kg√110.9396 (m/s²)² = 56 kg × 10.53 m/s² = 589.84 N.

Since the centripetal force acts towards the center of the ferris wheel in the horizontal direction, it is equal to the horizontal component of the normal force, Also, the weight acts downwards and is equal to the vertical component of the normal force.

So, the direction of the normal force is gotten from

tanθ = ma/mg = a/g

θ = tan⁻¹(a/g) = tan⁻¹(3.86 m/s² / -9.8 m/s²) = tan⁻¹(-0.3939) = -21.5°. Since the angle is it shows a counter clockwise direction.

So, the normal force is 589.84 N at 21.5° counterclockwise from the horizontal direction.

a solenoid is created by wrapping a l = 90 m long wire around a hollow tube of diameter d = 4.5 cm.   the wire diameter is d = 0.9 mm.   the solenoid wire is then connected to a power supply so that a current of i = 9 a flows through the wire.

randomized variablesl = 90 m

d = 4.5 cm

d = 0.9 mm

i = 9 a                                          

write an expression for the number of turns, n, in the solenoid. you do not need to take into account the diameter of the wire in this calculation.

calculate the number of turns, n, in the solenoid.

write an expression for the length of the solenoid (l2) in terms of the diameter of the hollow tube d.

calculate the length of the solenoid (l2)   in meters.

calculate the magnitude of the magnetic field at the center of the solenoid in teslas.

explanation:

no simpler subs form complex ones. synthesise = make