Answer:
a = αR
Then We apply the Newton's second law of motion:
∑ F = ma
F - mg sinθ - f = ma.
F - mg sinθ - μmg cosθ = ma
Using given data we have:
F - 3217 = 470a ........................... (1)
Apply the Newton's law for rotation:
∑τ = I α
FR - (mg sinθ)R =
then: F - mg sinθ = 3ma / 2.
F - 2774 = 705a .......................... (2)
solving 1 and 2 we get:
F = 4103 N
a = 1.885 m/s²
b) In this case the time is given by:
t = it comes from: d =
starting from rest vo = 0
t = = 3.09 s
Answer:
a) The speed of the electron as it reaches the inner conductor is closest to:
v = 1.45 × 10⁷m/s
b) The electric field magnitude between the cylinders is
E = 10,000V/m
Explanation:
given
inner radius of the cylinder r₁ = 20mm = 0.02m
outter radius of the cylinder r₂ = 80mm = 0.08m
potential difference V= 600V
mass of electron = 9.1×10⁻³¹kg
charge on electron = 1.6×10⁻¹⁹C
calculating the work done in bringing electron at inner conductor is
note:
∴W = (ΔV)q
(ΔV)q =
(600)1.6×10⁻¹⁹ = ¹/₂ × 9.1×10⁻³¹ × v²
v² ≈ 2.11 × 10¹⁴
v = 1.45 × 10⁷m/s
According to the energy conservation law, the total energy of an isolated system is always constant.
The energy of an isolated system can neither be created nor be destroyed, it can only convert one form to another form.
∴ the maximum electric field
E = ΔV/d
E = 600/d
where d is the distance between the two points
where d = 0.06m
E = 600/0.06
E = 10,000V/m
Note: the electric field due to the potential difference between to points depends upon the potential difference V and the distance between both points d.