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AleksAgata [21]
3 years ago
11

Ice causes a crack to form in a rock. This is an example of

Physics
1 answer:
kakasveta [241]3 years ago
4 0
Molecular weathering
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What is one piece of evidence that does NOT support the Big Bang Theory?​
svetoff [14.1K]

Answer:

The redshift data does not support the Big Bang Theory and  without the redshift interpretation favored by the Big bang, there is  no Big Bang.  

Edwin Hubble did not believe that the redshift indicated  expansion. Hubble saw the redshift as an hitherto unrecognized  principle of nature.  The redshift is the effect of centripetal force mv2/r = mv2/ct =  v/c(mv/t) = v/c(p/t)= (v/c)m

The unrecognized principle of nature is the refraction effect of  gravity on light.  

n=c/v or v= c/n.

The Universe is older than 13.7 Billion years, likely around  16.5 billion years.  

The proper law of Gravity is W = -mGM/r + cmV = -mu/r + cP where  cP is the so-called "Dark Energy", or the actual vector Momentum  Energy cmV= cP. This is The Quaternion Universe with scalar and  vector energy!

Hope this helps!

3 0
3 years ago
An object is placed at 1.5 m from a convex lens with a focal length of 1.0 m.
AleksAgata [21]
A. Use the thin lens equation to determine the image distance from the lens. Is the correct answer because I have no idea why
8 0
3 years ago
Electrons are added into the outermost what in groups 1 & 2
aivan3 [116]

S orbital.

Group 1 elements have a general configuration ns^{1}, where n represents the highest occupied Principal Energy Level. For example, Lithium has the valence configuration 2s^{1} whereas Cesium has 6s^{1}. Both of them belong to Group 1 of Periodic Table.

Group 2 elements have a general configuration of ns^{2}. For example, Magnesium has 3s^{2} as its outer shell configuration while Strontium has the same as 5s^{2}.

We see that in both the cases, the outermost S orbital is being filled.

3 0
3 years ago
Is mercury a terrestrial or gaseous planet
liraira [26]

Mercury a terrestrial. It isn't made of gas.

8 0
3 years ago
Assume the radius of an atom, which can be represented as a hard sphere, is r = 1.95 Å. The atom is placed in a (a) simple cubic
Nuetrik [128]

Answer:

(a) A = 3.90 \AA

(b) A = 4.50 \AA

(c) A = 5.51 \AA

(d) A = 9.02 \AA

Solution:

As per the question:

Radius of atom, r = 1.95 \AA = 1.95\times 10^{- 10} m

Now,

(a) For a simple cubic lattice, lattice constant A:

A = 2r

A = 2\times 1.95 = 3.90 \AA

(b) For body centered cubic lattice:

A = \frac{4}{\sqrt{3}}r

A = \frac{4}{\sqrt{3}}\times 1.95 = 4.50 \AA

(c) For face centered cubic lattice:

A = 2{\sqrt{2}}r

A = 2{\sqrt{2}}\times 1.95 = 5.51 \AA

(d) For diamond lattice:

A = 2\times \frac{4}{\sqrt{3}}r

A = 2\times \frac{4}{\sqrt{3}}\times 1.95 = 9.02 \AA

6 0
2 years ago
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