A 200-kg steel part is resting on a thin film of oil on a concrete floor. The oil is a Newtonian fluid and has a viscosity of μ = 6.52×10−2 N⋅s/m2 . The film thickness is d = 0.5 mm . What horizontal force is required to push the part across the floor at a constant velocity of A = 0.65 m/s?

2.097×10^-5 N

Explanation:

Viscosity = mass/distance×time

time (t) = mass/viscosity×distance

Mass of steel part = 200 kg

Viscosity of oil = 0.065 N.s/m^2

Distance = 0.5 mm = 0.5/1000 = 5×10^-4 m

t = 200/0.065×5×10^-4 = 6.2×10^6 s

Force (F) = mass×velocity/time = 200×0.65/6.2×10^6 = 2.097×10^-5 N

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: