Manganese has 2 (two) electron that would free floating and able to form a metallic bond.
The electronic configuration of manganese is (Ar) 3d5 4s2. The two electron in 4s orbital are the valence electron which can freely move from one place to another.
K = 1/2 m x v^2
m = mass on the cart
V = velocity imparted to the cart
KA = 1/2 mA x vA^2.......................(1)
KB = 1/2 mB x vB^2........................(2)
Diving equation 1 by equation 2, we get -
KA/KB = mA/mB
= 2
KA = 2 x KB
Option A is correct
Answer: A: 70/2=35
B: 35/2=17
C: 17.5/2=8.75
D: 8.75 of C-14 will be left
E: 5,730 years
F: 5,631
Explanation:
that’s all I got, hope I helped kinda
This shifts the star’s spectral lines toward the blue end of the spectrum. If the star is moving away from us, it’s waves are effectively stretched out when they reach earth, increasing their wavelength. This shifts the star’s spectral lines toward the red end of the spectrum.
Answer:
Explanation:
We need 2 different equations for this problem: first the velocity of sound equation, then the frequency of the sound equation.
The velocity of sound is found in:
v = 331.5 + .606T
We need to find that first in order to fill it into the frequency equation which is
where v is the velocity we will find the part a, f is frequency and lambda is the wavelength. Starting with the velocity of the sound:
v = 331.5 + .606(25) and
v = 331.5 + 15 and rounding correctly using the rules for sig fig when adding:
v = 347 m/s
Filling that into the frequency equation:
and
so
