Answer:
Explanation:
Given:
dI/dt = 6.21 A/s
n = N/l
= 100 turns/cm
= 100 turns/cm × 100 cm/1 m
Radius, r = 14.7 cm
= 0.147 m
Inductance, L = uo × n^2 × A × l
L/l = 4pi × 10^-7 × (100 × 100)^2 × pi × 0.147^2
= 8.53 H
Emf, E = L × dI/dt
E/l = L/l × dI/dt
= 8.53 × 6.21
= 52.98 V/m
=
Answer:
E = 1/2 M V^2 + 1/2 I ω^2 = 1/2 M V^2 + 1/2 I V^2 / R^2
E = 1/2 M V^2 (1 + I / (M R^2))
For a cylinder I = M R^2
For a sphere I = 2/3 M R^2
E(cylinder) = 1 + 1 = 2 omitting the 1/2 M V^2
E(sphere) = 1 + 2/3 = 1.67
E(cylinder) / E(sphere) = 2 / 1.67 = 1.2
The cylinder initially has 1.20 the energy of the sphere
The PE attained is proportional to the initial KE
H(sphere) = 2.87 / 1/2 = 2.40 m since it has less initial KE
Answer:
For the distance range 50 to 500 km, the S-waves travel about 3.45 km/s and the P-waves around 8 km/s.
hope it helps.
Answer:
in a microscope the place you keep your eyes to observe
Answer:
(a) x=ASin(ωt+Ф₀)=±(√3)A/2
(b) x=±(√2)A/2
Explanation:
For part (a)
V=AωCos(ωt+Ф₀)⇒±0.5Aω=AωCos(ωt+Ф₀)
Cos(ωt+Ф₀)=±0.5⇒ωt+Ф₀=π/3,2π/3,4π/3,5π/3
x=ASin(ωt+Ф₀)=±(√3)A/2
For part(b)
U=0.5E and U+K=E→K=0.5E
E=K(Max)
(1/2)mv²=(0.5)(1/2)m(Vmax)²
V=±(√2)Vmax/2→ωt+Ф₀=π/4,3π/4,7π/4
x=±(√2)A/2
