Consider a neutron star of radius 10 km that spins with a period of 0.8 seconds. imagine a person is standing at the equator of this neutron star. calculate the centripetal acceleration of this person
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Ответ:
a = 616850.28 m/s²
Explanation:
Given that,
The radius of the neutron star, r = 10 Km
= 10,000 m
The time period of the neutron star, T = 0.8 s
The centripetal acceleration is given by the formula,
a = v²/r
The linear velocity is given by the relation,
v = rω
The time taken to complete one complete rotation is given by the relation
T = 2π /ω
Where,
ω = 2π / T
Substituting v and ω into the equation for centripetal acceleration. It becomes
a = 4π²r/T²
Substituting the given values in the above equation
a = 4π² x 10000 / 0.8²
= 616850.28 m/s²
Hence, the centripetal acceleration of this person is, a = 616850.28 m/s²
Ответ:
Repulsion.
Explanation:
Both magnets might be facing each other the same pole, if they are, then they repelled each other.