So the answer is B. because the mass have Kg as a international unit and velocity is m/s, they are international units in physics.
Complete Question:
A hollow cylinder with an inner radius of 4.0mm and an outer radius of 30mm conducts a 3.0-A current flowing parallel to the axis of the cylinder. If the current density is uniform throughout the wire, what is the magnitude of the magnetic field at a point 12mm from its center?
Answer:
The magnitude of the magnetic field = 7.24 μT
Explanation:
Inner radius, a = 4.0 mm = 0.004 m
Outer radius, b = 30 mm = 0.03 m
Radius, r = 12 mm = 0.012 m
let h² = b² - a²
h² = 0.03² - 0.004²
h² = 0.000884
Let d² = r² - a²
d² = 0.012² - 0.004²
d² = 0.000128
Current I = 3A
μ = 4π * 10⁻⁷
The magnitude of the magnetic field is given by:
B = 7.24 * 10⁻⁶T
B = 7.24 μT
Answer:
This might be because since they reproduce without mating, the genes of their offspring have are not mixed with anyone elses, making them have very little variation in contrast with those that reproduce our way.
Answer:
vT = v0/3
Explanation:
The gravitational force on the satellite with speed v0 at distance R is F = GMm/R². This is also equal to the centripetal force on the satellite F' = m(v0)²/R
Since F = F0 = F'
GMm/R² = m(v0)²/R
GM = (v0)²R (1)
Also, he gravitational force on the satellite with speed vT at distance 3R is F1 = GMm/(3R)² = GMm/27R². This is also equal to the centripetal force on the satellite at 3R is F" = m(vT)²/3R
Since F1 = F'
GMm/27R² = m(vT)²/3R
GM = 27(vT)²R/3
GM = 9(vT)²R (2)
Equating (1) and (2),
(v0)²R = 9(vT)²R
dividing through by R, we have
9(vT)² = (v0)²
dividing through by 9, we have
(vT)² = (v0)²/9
taking square-root of both sides,
vT = v0/3
