Aslender rod is 80.0 cm long and has mass 0.390 kg . a small 0.0200-kg sphere is welded to one end of the rod, and a small 0.0500-kg sphere is welded to the other end. the rod, pivoting about a stationary, frictionless axis at its center, is held horizontal and released from rest.
what is the linear speed of the 0.0500-kg sphere as its passes through its lowest point?

Answer

given,

length of slender rod =80 cm = 0.8 m

mass of rod = 0.39 Kg

mass of small sphere = 0.0200 kg

mass of another sphere weld = 0.0500 Kg

calculating the moment of inertia of the system

using conservation of energy

we know,

v = r ω

v = 1.084 m/s

v = 1.08 m/s

Explanation:

What is the linear speed of the 0.0500-kg sphere as its passes through its lowest point?

The decrease in PE is

d = 80.0cm * 1 / 1000m = 0.80m

h = 0.80 m /2 = 0.40 m

ΔPE = m*g*h

ΔPE = (0.0500 - 0.0200)kg * 9.8m/s² * 0.400 m

ΔPE = 0.1176 J

The moment of inertia of the assembly is

I = 1/12*m*L² + (m1 + m2)*(L/2)²

I = 1/12*0.390kg*(0.800m)² + 0.0700kg*(0.400m)²

I = 0.032 kg·m²

KE = ½Iω²

0.1176 J = ½ * 0.032kg·m² * ω²

ω = 2.71 rad/s

v = ωr = 2.71 rad/s * 0.400m

The linear velocity

v = 1.08 m/s

multiple

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: