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Olenka [21]
3 years ago
10

Earth's axis is tilted. Tell what would happen if it was vertical instead.

Chemistry
2 answers:
slamgirl [31]3 years ago
7 0
I think the like term is 6 hope this is right
pshichka [43]3 years ago
6 0

Answer:

If earths axis is tilted, we would practically be in a night lock day and night. And the seasons would be different too

Explanation:

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when hydrogen atom absorbs a photon, and an electron moves level 1 to level 2, what happens to the energy of the atom?
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When an electron is hit by a photon of light, it absorbs the quanta of energy the photon was carrying and moves to a higher energy state. One way of thinking about this higher energy state is to imagine that the electron is now moving faster, (it has just been "hit" by a rapidly moving photon).

Explanation: pls mark brainliest :))

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1.Bubbling or fizzing is an indicator of
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D is the answer since it is changing the element.
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3 years ago
2.3.12mL =<br> L*<br> O<br> a. .312<br> O b.31.2<br> O<br> C..00312<br> O d..0312
Anna11 [10]

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a..312 I guesss kkkkkkkkkkkkkkkkkk

6 0
3 years ago
A 10 mm Brinell hardness indenter is used for some hardness testing measurements of a steel alloy. a) Compute the HB of this mat
Lady_Fox [76]

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 = \frac{2P}{\pi D [D - \sqrt{(D^{2} - d^{2})}]}    

Now, putting the given values into the above formula as follows.

                HB = \frac{2P}{\pi D [D - \sqrt{(D^{2} - d^{2})}]}                            = \frac{2 \times 1000 kg}{3.14 \times 10 mm [D - \sqrt{((10 mm)^{2} - (2.50)^{2})}]}  

                       = \frac{2000}{9.98}                          

                       = 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})}

                         = \sqrt{(10 mm)^{2} - [10 mm - \frac{2 \times 500 kg}{300 \times 3.14 \times 10 mm}]^{2}}

                          = 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.

5 0
3 years ago
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