The answer is 2.3 hope this helps texted me and tell me if it’s right
Answer:
a) The plasma membrane is called a selectively permeable membrane as it permits the movement of only certain molecules in and out of the cells. ... It allows hydrophobic molecules and small polar molecules diffuse through the lipid layer, but does not allow ions and large polar molecules cannot diffuse through the membrane
b) Plastids are present in the cells of plants. They are characterised by the presence of pigments. ... Chloroplasts contain chlorophyll and carotenoid pigments responsible for capturing the light energy that is necessary for photosynthesis. The chloroplasts are therefore known as the kitchen of the cell.
c) Lysosomes are known as the suicidal bag of the cell because it is capable of destroying its own cell in which it is present. It contains many hydrolytic enzymes which are responsible for the destruction process. This happens when either the cell is aged or gets infected by foreign agents like any bacteria or virus.
d) Mitochondria are often called the “powerhouses” or “energy factories” of a cell because they are responsible for making adenosine triphosphate (ATP), the cell's main energy-carrying molecule. ... In mitochondria, this process uses oxygen and produces carbon dioxide as a waste product.
e) In Hydra, the cells are arranged in two germinal layers—outer ectoderm and inner endoderm. Between these two layers is a layer of undifferentiated cells called mesoglea. Such kind of pattern of embryonic layers is seen in diploblastic animals. Hence, Hydra is a diploblastic animal.
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Answers A and C can be automatically eliminated because evaporation has to do with the water cycle and magnetism has to do with electric current and such. Next we can eliminate B or distillation because it also has to do with water/liquids. Therefore, the answer is D or filtration because it is the only answer left.
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soil i think is this from apex?
Velocity of submarine A is vs = 11.0m/s
frequency emitted by submarine A. F = 55.273 × 10∧3HZ
Velocity of submarine B = vO = 3.00m/s
The given equation is
f' = ((V + vO) ((v - vS)) × f
The observer on submarine detects the frequency f'.
The sign of vO should be positive as the observer of submarine B is moving away from the source of submarine A.
The speed of the sound used in seawater is 1533m/s
The frequency which is detected by submarine B is
fo = fs (V -vO/ v +vs)
= 53.273 × 10∧3hz) ((1533 m/s - 4.5 m/s)/ (1533 m/s +11 m/s)
fo = 5408 HZ
