Install Steam
sign in
|
language
简体中文 (Simplified Chinese)
繁體中文 (Traditional Chinese)
日本語 (Japanese)
한국어 (Korean)
ไทย (Thai)
Български (Bulgarian)
Čeština (Czech)
Dansk (Danish)
Deutsch (German)
Español - España (Spanish - Spain)
Español - Latinoamérica (Spanish - Latin America)
Ελληνικά (Greek)
Français (French)
Italiano (Italian)
Bahasa Indonesia (Indonesian)
Magyar (Hungarian)
Nederlands (Dutch)
Norsk (Norwegian)
Polski (Polish)
Português (Portuguese - Portugal)
Português - Brasil (Portuguese - Brazil)
Română (Romanian)
Русский (Russian)
Suomi (Finnish)
Svenska (Swedish)
Türkçe (Turkish)
Tiếng Việt (Vietnamese)
Українська (Ukrainian)
Report a translation problem

Average atomic mass = f1M1 + f2M2 +… + fnMn where f is the fraction representing the natural abundance of the isotope and M is the mass number (weight) of the isotope.
The average atomic mass of an element can be found on the periodic table, typically under the elemental symbol. When data are available regarding the natural abundance of various isotopes of an element, it is simple to calculate the average atomic mass.
For helium, there is approximately one isotope of Helium-3 for every million isotopes of Helium-4; therefore, the average atomic mass is very close to 4 amu (4.002602 amu).
Chlorine consists of two major isotopes, one with 18 neutrons (75.77 percent of natural chlorine
|z−z1|2=c2|z−z2|2
|z−z1|2=c2|z−z2|2
You will end up with an equation of the form
(1−c2)x2+(1−c2)y2+2αx+2βy+γ=0
(1−c2)x2+(1−c2)y2+2αx+2βy+γ=0
for some constants α,β,γα,β,γ. Since c≠1c≠1, you can divide by (1−c2)(1−c2) (Note that cc is a real number and ≠−1≠−1 because of what you started with). This will give you the equation of a circle.
For a spherical object whose actual diameter equals {\displaystyle d_{\mathrm {act} },} d_{\mathrm {act} }, and where {\displaystyle D} D is the distance to the centre of the sphere, the angular diameter can be found by the formula
{\displaystyle \delta =2\arcsin \left({\frac {d_{\mathrm {act} }}{2D}}\right)} \delta =2\arcsin \left({\frac {d_{\mathrm {act} }}{2D}}\right)
The difference is due to the fact that the apparent edges of a sphere are its tangent points, which are closer to the observer than the centre of the sphere. For practical use, the distinction is only significant for spherical objects that are relatively close, since the small-angle appr