You slide a chair across a rough, horizontal surface. The chair's mass is 18.8 kg. The force you exert on the chair is 156 N directed 26 degrees below the horizontal. While you slide the chair a distance of 5.00 m , the chair's speed changes from 1.30 m/s to 3.10 m/s . Find the work done by friction on the chair.

626.612 J

Explanation:

Work done by friction on the chair is given as

W-ΔEk = W' Equation 1

Where W' = Work done by friction on the chair, W = Work done on the chair by me, Ek = change in Kinetic energy of the chair as a result of the slide.

From the question,

W = FdcosФ Equation 2

ΔEk = 1/2m(v²-u²) Equation 3

Where F = Force applied on the chair, d = distance of slide, Ф = angle between the force and the horizontal, m = mass of the chair, v = final velocity of the chair, u = initial velocity of the chair

Substitute equation 2 and equation 3 into equation 1

W' = FdcosФ-1/2m(v²-u²) Equation 4

Given: F = 156, d = 5 m, Ф = 26°, m = 18.8 kg, v = 3.1 m/s, u = 1.3 m/s

Substitute into equation 4

W' = 156×5×cos26°-1/2×18(3.1²-1.3²)

W' = 701.06-74.448

W' = 626.612 J.

Hence the work done by friction = 626.612 J

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: