Answer:
Yes to answer is 3/5 of what
Answer:
A force is a push or pull that acts upon an object as a result of that objects interactions
Explanation:
Answer:
E = 9.4 10⁶ N / C
, The field goes from the inner cylinder to the outside
Explanation:
The best way to work this problem is with Gauss's law
Ф = E. dA = qint / ε₀
We must define a Gaussian surface, which takes advantage of the symmetry of the problem. We select a cylinder with the faces perpendicular to the coaxial.
The flow on the faces is zero, since the field goes in the radial direction of the cylinders.
The area of the cylinder is the length of the circle along the length of the cable
dA = 2π dr L
A = 2π r L
They indicate that the distance at which we must calculate the field is
r = 5 R₁
r = 5 1.3
r = 6.5 mm
The radius of the outer shell is
r₂ = 10 R₁
r₂ = 10 1.3
r₂ = 13 mm
r₂ > r
When comparing these two values we see that the field must be calculated between the two housings.
Gauss's law states that the charge is on the outside of the Gaussian surface does not contribute to the field, the charged on the inside of the surface is
λ = q / L
Qint = λ L
Let's replace
E 2π r L = λ L /ε₀
E = 1 / 2piε₀ λ / r
Let's calculate
E = 1 / 2pi 8.85 10⁻¹² 3.4 10-12 / 6.5 10-3
E = 9.4 10⁶ N / C
The field goes from the inner cylinder to the outside
Answer:
1.25 kgm²/sec
Explanation:
Disk inertia, Jd =
Jd = 1/2 * 3.7 * 0.40² = 0.2960 kgm²
Disk angular speed =
ωd = 0.1047 * 30 = 3.1416 rad/sec
Hollow cylinder inertia =
Jc = 3.7 * 0.40² = 0.592 kgm²
Initial Kinetic Energy of the disk
Ekd = 1/2 * Jd * ωd²
Ekd = 0.148 * 9.87
Ekd = 1.4607 joule
Ekd = (Jc + 1/2*Jd) * ω²
Final angular speed =
ω² = Ekd/(Jc+1/2*Jd)
ω² = 1.4607/(0.592+0.148)
ω² = 1.4607/0.74
ω² = 1.974
ω = √1.974
ω = 1.405 rad/sec
Final angular momentum =
L = (Jd+Jc) * ω
L = 0.888 * 1.405
L = 1.25 kgm²/sec
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