What is the radius of a communications satellite’s orbital path that is in a uniform circular orbit around Earth and has a perio
1 answer:
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Answer:
The electric current in the wire is 0.8 A
Explanation:
We solve this problem by applying the formula of the magnetic field generated at a distance by a long and straight conductor wire that carries electric current, as follows:
B= Magnetic field due to a straight and long wire that carries current
u= Free space permeability
I= Electrical current passing through the wire
a = Perpendicular distance from the wire to the point where the magnetic field is located
Magnetic Field Calculation
We cleared (I) of the formula (1):
Formula(2)
a =8cm=0.08m
We replace the known information in the formula (2)
I=0.8 A
Answer: The electric current in the wire is 0.8 A
Answer:
<em>0.85c </em>
Explanation:
Rest mass of Kaon = 494 MeV/c²
Rest mass of proton = 938 MeV/c²
The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²
for the kaon, rest energy = 494c² MeV
for the proton, rest energy = 938c² MeV
Recall that the rest energy, and the total energy are related by..
= γ
which can be written in this case as
= γ
...... equ 1
where = total energy of the kaon, and
= rest energy of the kaon
γ = relativistic factor =
where
But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...
=
......equ 2
where is the total energy of the kaon, and
is the rest energy of the proton.
From =
= 938c²
equ 1 becomes
938c² = γ494c²
γ = 938c²/494c² = 1.89
γ = = 1.89
1.89 = 1
squaring both sides, we get
3.57( 1 - ) = 1
3.57 - 3.57 = 1
2.57 = 3.57
= 2.57/3.57 = 0.72
= 0.85
but,
v/c = 0.85
v = <em>0.85c </em>