It would slow down in the stage of growing or
.
Explanation:
1)The cell membrane functions as a semi-permeable barrier, allowing a very few molecules across it while fencing the majority of organically produced chemicals inside the cell. Electron microscopic examinations of cell membranes have led to the development of the lipid bilayer model (also referred to as the fluid-mosaic model). The most common molecule in the model is the phospholipid, which has a polar (hydrophilic) head and two nonpolar (hydrophobic) tails.
2) simple diffusion across the cell plasma membrane. The structure of the lipid bilayer allows small, uncharged substances such as oxygen and carbon dioxide, and hydrophobic molecules such as lipids, to pass through the cell membrane, down the concentration gradient is , by simple diffusion.
3) some molecules, such as carbon dioxide and oxygen, can diffuse across the plasma membrane directly, but others need help to cross its hydrophobic or however, because they are charged the polar, they can't cross the phospholipid part of the membrane without help .
4) during fission a copy of the DNA is made and attached to the cell membrane as well. As this cell elongate in preparation for fission, the two DNA copies are pulled apart two opposite ends of the cell. New membrane material is deposited between the two ends of the cell, and a new wall grows between them .
5) UMASS STEM-ED From Bubbles to Cell Membranes Workshop. Bubble ... dynamic nature which can't be properly appreciated in a static textbook. ... the small thread through one of the straws.
6) example of passive transport and active transport across a cell membrane so, cell membranes are semipermeable meaning they have control over what molecules can or cannot pass through. Some molecules can just drift Inn.
Answer:
yes
Explanation:
I think as people, we have to look at everything twice and consider the bad and the good to reqlly get the best result.
Answer:
It is given that a particle covers 10m in first 5s and 10m in next 3s. so using the equation of motion
Case I
s=ut+
2
1
at
2
10=5u+
2
1
a(5)
2
20=10u+25a
4=2u+5a..............(1)
Case 2
In next 3s the particle covers more 10m distance. So
20=8u+
2
1
a(8)
2
5=2u+8a.........(2)
On solving equation (1) and (2)
4=2u+5a
5=2u+8a
a=
3
1
m/s
2
Put the value of a in equation (1)
u=
6
7
m/s
Now to find distance in next 10 s. total time will be 10s
s=
6
7
×10+
2
1
×
3
1
×(10)
2
s=28.33m
Distance travelled in next 2 sec
s=28.33−20=8.33m
