As a spherical ammonia vapor bubble rises in liquid ammonia, its diameter changes from 1 cm to 3 cm. Calculate the amount of work produced by this bubble, in kJ, if the surface tension of ammonia is 0.07 N/m.

8.7976×10^-8 kJ

Explanation:

Change in diameter = 3 cm - 1 cm = 2 cm = 2/100 = 0.02 m

Area of spherical ammonia = πD^2 = 3.142×0.02^2 = 1.2568×10^-3 m^2

Work produced = surface tension × area = 0.07 × 1.2568×10^-3 = 8.7976×10^-5 Nm = 8.7976×10^-5 J = 8.7976×10^-5/1000 = 8.7976×10^-8 kJ

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:

0.863333

86%

hope this