Answer:
C) 24.4°
Explanation:
let nd = 2.419 be the index of refraction of diamond and na = 1.0 be the index of refraction of air and ∅c be the critical angle.
according to Snell's Law:
sin(∅c) = na/nd
sin(∅c) = (1.0)/(2.419)
∅c = 24.4°
Both a molten metallic core and reasonably fast rotation.
Answer:
A. 16.9 m
Explanation:
I think this is the answer i am not sure
but hope it helps
Answer:

Explanation:
GIVEN
diameter = 15 fm =
m
we use here energy conservation

there will be some initial kinetic energy but after collision kinetic energy will zero

on solving these equations we get kinetic energy initial
J ..............(i)
That is, the alpha particle must be fired with 35.33 MeV of kinetic energy. An alpha particle with charge q = 2 e
and gains kinetic energy K =e∆V ..........(ii)
by accelerating through a potential difference ∆V
Thus the alpha particle will
just reach the
nucleus after being accelerated through a potential difference ∆V
equating (i) and second equation we get
e∆V = 35.33 Me V

now you can yourself know to which part of electromagnetic spectrum the photon belongs....
not fitting sharply in green