Answer:
a. the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N
b. this force compare to the weight of the muon; the force is 1.38 × 10⁸ greater than muon
Explanation:
F= ma
v²=u² -2aS
(1.56 ✕ 10⁶)²=(2.40 ✕ 10⁶)²-2a(1220)
a=1.36×10⁹m/s²
recall
F=ma
F = 1.88 ✕ 10⁻²⁸ kg × 1.36×10⁹m/s²
F= 2.55 × 10⁻¹⁹N
the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N
b. this force compare to the weight of the muon
F/mg= 2.55 × 10⁻¹⁹/ (1.88 ✕ 10⁻²⁸ × 9.8)
= 1.38 × 10⁸
Magnetic domain structure is responsible for the magnetic behavior of ferromagnetic materials like iron, nickel, cobalt and their alloys, and ferrimagnetic materials like ferrite. ... Magnetic domains form in materials which have magnetic ordering; that is, their dipoles spontaneously align due to the exchange interaction.
We now that follow newton rules f=ma so net force equal to mass*acceleration=>f=50*1.5=75 N
A) 1 rev = 2π rad. Using this ratio, you can find the rad/s: 1160 rev/min x 2π rad/rev x 1 min/60 s = 121.5 rad/s
b) You can find linear speed from angular speed using this equation (note the radius is half the diameter given in the question): v = ωr = 121.5 rad/s x 1.175 m = 142.8 m/s
c) You can find centripetal acceleration using this equation: a = v^2/r = (142.8 m/s)^2 / 1.175 m = 17 355 m/s^2
