Answer:
2.36 μ H
Explanation:
Given,
Number of turns= 90
diameter = 1.3 cm = 0.013 m
unscratched length = 57 cm = 0.57 m
Area, A = π r²
= π x 0.0065² = 1.32 x 10⁻⁴ m²
we know,
L = 2.36 μ H
Hence, the inductance of the unstretched cord is equal to 2.36 μ H
Answer:
An asteroid that has an orbital period of 3 years will have an orbital with a semi-major axis of about 2 years.
Explanation:
Given;
orbital period of 3 years, P = 3 years
To calculate the years of an orbital with a semi-major axis, we apply Kepler's third law.
Kepler's third law;
P² = a³
where;
P is the orbital period
a is the orbital semi-major axis
(3)² = a³
9 = a³
a =
Therefore, An asteroid that has an orbital period of 3 years will have an orbital with a semi-major axis of about 2 years.