Answer:
have stars that might appear to wobble
often have one star that is brighter than the other
Explanation:
A binary star system is a star system made up of mostly two stars that moves round their common fixed center.
The two orbiting stars are gravitationally bonded to one another and they move round each other.
Most binary stars might appear wobble. One of the stars often appears brighter than the other.
Chromium , silver, zinc...
<u>Answer:</u> The original element is 
<u>Explanation:</u>
Alpha decay is defined as the process in which alpha particle is emitted. In this process, a heavier nuclei decays into a lighter nuclei. The alpha particle released carries a charge of +2 units.
The released alpha particle is also known as helium nucleus.

For the given alpha decay process of an isotope:

<u>To calculate A:</u>
Total mass on reactant side = total mass on product side
A = 208 + 4
A = 212
<u>To calculate Z:</u>
Total atomic number on reactant side = total atomic number on product side
Z = 82 + 2
Z = 84
The isotopic symbol of unknown element is 
Hence, the original element is 
Explanation:
(a) The given data is as follows.
Load applied (P) = 1000 kg
Indentation produced (d) = 2.50 mm
BHI diameter (D) = 10 mm
Expression for Brinell Hardness is as follows.
HB =
Now, putting the given values into the above formula as follows.
HB =
=
=
= 200
Therefore, the Brinell HArdness is 200.
(b) The given data is as follows.
Brinell Hardness = 300
Load (P) = 500 kg
BHI diameter (D) = 10 mm
Indentation produced (d) = ?
d = ![\sqrt{(D^{2} - [D - \frac{2P}{HB} \pi D]^{2})}](https://tex.z-dn.net/?f=%5Csqrt%7B%28D%5E%7B2%7D%20-%20%5BD%20-%20%5Cfrac%7B2P%7D%7BHB%7D%20%5Cpi%20D%5D%5E%7B2%7D%29%7D)
= ![\sqrt{(10 mm)^{2} - [10 mm - \frac{2 \times 500 kg}{300 \times 3.14 \times 10 mm}]^{2}}](https://tex.z-dn.net/?f=%5Csqrt%7B%2810%20mm%29%5E%7B2%7D%20-%20%5B10%20mm%20-%20%5Cfrac%7B2%20%5Ctimes%20500%20kg%7D%7B300%20%5Ctimes%203.14%20%5Ctimes%2010%20mm%7D%5D%5E%7B2%7D%7D)
= 4.46 mm
Hence, the diameter of an indentation to yield a hardness of 300 HB when a 500-kg load is used is 4.46 mm.
That question shoulda be true