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marissa [1.9K]
3 years ago
13

Describe 3 diffferences between animal and plant cell's.

Chemistry
2 answers:
valentinak56 [21]3 years ago
4 0

Answer:

he key difference between plant and animal cells lies in the structural differences. Plant cells are rectangular wheres animal cells are round and plant cells contain chloroplasts, a cell wall, and vacuoles while animal cells do not.

Another difference is that plant cells have chloroplasts while animal cells do not. Chloroplasts, rich in chlorophyll, absorb light and conduct photosynthesis, providing the cell with the energy it needs.

The third is the fact that the plant cell doesnt require much nutrients as an animal cell would because for a plant  it uses Glucose for food as though animal require protien several vitamins water,and iron and vitamin b3

Explanation:

IrinaK [193]3 years ago
3 0
Animal cells are round, plant cells are circle (most of them) Plant cells have a rigid cell wall that surrounds the cell membrane. Animal cells do not have a cell wall.
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A 10 mm Brinell hardness indenter is used for some hardness testing measurements of a steel alloy. a) Compute the HB of this mat
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Explanation:

(a)  The given data is as follows.  

         Load applied (P) = 1000 kg

         Indentation produced (d) = 2.50 mm

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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.

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