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Nataly_w [17]
2 years ago
6

Help me fill this out please, i give brainliest

Chemistry
2 answers:
stealth61 [152]2 years ago
7 0

Answer:

first one waining gibbious

secound one waining crecent

third one new moon

fourth one first quarter

fithe one third quarter

sixth one  full moon

sevnth one waxing crecent

eithe one waxing gibbious

Explanation:

xxTIMURxx [149]2 years ago
4 0

Answer:

Okay, Left top= waning gibbous

left second= new moon

left third= third quarter

left fourth= waxing crescent

right top= Waning crescent

right second= first quarter

right third= full moon

right fourth= Waxing gibbous

hope you do good :)

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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
2 years ago
Which of the following is a chemical change?I
Sergeu [11.5K]

Answer:

combustion is a chemical change

3 0
3 years ago
Determine the number of moles within a sample of 10g of propane(C3H8). Round your answer to 2 decimal places with proper units.
a_sh-v [17]

Given :

10 gram sample of propane( C₂H₈ ).

To Find :

The number of moles of propane in given sample.

Solution :

Molecular mass of propane, M = (2 × 12) + ( 1 × 8 ) gram/mol

M = 32 gram/mol

We know, number of moles is given by :

Number of moles, n = m/M

n = 10/32 mol

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Therefore, number of moles in given sample is 0.3125 mol.

7 0
2 years ago
Becky is preparing tea. When the water for the tea boils, she pours it in a cup of cold milk. Assuming that the system is closed
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Answer:B

Explanation:

4 0
3 years ago
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The reaction represented above goes essentially to completion. The reaction takes place in a rigid, insulated vessel that is ini
Leni [432]

Answer:

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45. (D) 3Pi

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Explanation:

Here we have

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