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Answer:
When her hands extends, her momen of inertia is .
Explanation:
Given that,
Initial angular speed,
Initial moment of inertia,
Final angular speed,
Initially, a skater rotates with her arms crossed and finally she extends her arms. The momentum remains conserved. Using the conservation of momentum as :
is final moment of inertia
So, when her hands extends, her momen of inertia is . Hence, this is the required solution.
Answer:
r = 4.21 10⁷ m
Explanation:
Kepler's third law It is an application of Newton's second law where the forces of the gravitational force, obtaining
T² = ( ) r³ (1)
in this case the period of the season is
T₁ = 93 min (60 s / 1 min) = 5580 s
r₁ = 410 + 6370 = 6780 km
r₁ = 6.780 10⁶ m
for the satellite
T₂ = 24 h (3600 s / 1h) = 86 400 s
if we substitute in equation 1
T² = K r³
K = T₁²/r₁³
K =
K = 9.99 10⁻¹⁴ s² / m³
we can replace the satellite values
r³ = T² / K
r³ = 86400² / 9.99 10⁻¹⁴
r = ∛(7.4724 10²²)
r = 4.21 10⁷ m
this distance is from the center of the earth