Suppose that she pushes on the sphere tangent to its surface with a steady force of F = 60 N and that the pressured water provides a frictionless support. How long will it take her to rotate the sphere one time, starting from rest?

25.06s

Explanation:

Remaining part of the question.

(A large stone sphere has a mass of 8200 kg and a radius of 90 cm and floats with nearly zero friction on a thin layer of pressurized water.)

Solution:

F = 60N

r = 90cm = 0.9m

M = 8200kg

Moment of inertia for a sphere (I) = ⅖mr²

I = ⅖ * m * r²

I = ⅖ * 8200 * (0.9)²

I = 0.4 * 8200 * 0.81

I = 2656.8 kgm²

Torque (T) = Iα

but T = Fr

Equating both equations,

Iα = Fr

α = Fr / I

α = (60 * 0.9) / 2656.8

α = 0.020rad/s²

The time it will take her to rotate the sphere,

Θ = w₀t + ½αt²

Angular displacement for one revolution is 2Π rads..

θ = 2π rads

2π = 0 + ½ * 0.02 * t²

(w₀ is equal to zero since sphere is at rest)

2π = ½ * 0.02 * t²

6.284 = 0.01 t²

t² =6.284 / 0.01

t² = 628.4

t = √(628.4)

t = 25.06s

probably not the "magnets"in the magnet, so to speak, are thought to be electrons orbiting in shells. vurrent loops two sides to the loop

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft