Explanation:
We have,
Semimajor axis is 
It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :

G is universal gravitational constant
M is solar mass
Plugging all the values,

Since,

So, the orbital period of a dwarf planet is 138.52 years.
Answer:

Explanation:
= Cambio en la longitud de la cuerda = 0.25 cm
T = tensión en cuerda
A = Área de la cadena = 
d = Diámetro de la cuerda = 0.2 cm
L = Longitud original de la cuerda = 1.6 m
El cambio de longitud de una cuerda viene dado por

La tensión en la cuerda es
.
Complete question:
What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field 0.425 T. (This is 60 rev/s.)
Answer:
The peak emf generated by the coil is 15.721 kV
Explanation:
Given;
Radius of coil, r = 0.250 m
Number of turns, N = 500-turn
time of revolution, t = 4.17 ms = 4.17 x 10⁻³ s
magnetic field strength, B = 0.425 T
Induced peak emf = NABω
where;
A is the area of the coil
A = πr²
ω is angular velocity
ω = π/2t = (π) /(2 x 4.17 x 10⁻³) = 376.738 rad/s = 60 rev/s
Induced peak emf = NABω
= 500 x (π x 0.25²) x 0.425 x 376.738
= 15721.16 V
= 15.721 kV
Therefore, the peak emf generated by the coil is 15.721 kV
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