E = hf
E = 6.63×10^-34 × 3.55×10 eV
1 eV = 1.60×10^-19 J
E = 6.63×10^-34 × 3.55×10 × 1.60×10^-19
E = 3.77×10^-51 J
Hope it helped!
Answer:
it's C (this is not a guess)
Explanation:
It is given that,
A planet were discovered between the sun and Mercury, with a circular orbit of radius equal to 2/3 of the average orbit radius of Mercury.
Mass of the Sun, 
Radius of Mercury's orbit, 
Radius of discovered planet, 

Let T is the orbital period of such a planet. Using Kepler's third law of planetary motion as :




T = 4135214.625 s
or
T = 47.86 days
So, the orbital period of such a planet is 47.86 days. Hence, this is the required solution.
Answer:
hyoid bone
Explanation:
Woodpeckers have a special bone that acts like a seat-belt for its skull. It's called the hyoid bone, and it wraps all the way around a woodpecker's skull. Every time the bird pecks, the hyoid acts like a seat-belt for the bird's skull and the delicate brain it protects.