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shutvik [7]
4 years ago
15

How does a cell use endocytosis to form a vesicle

Physics
1 answer:
DerKrebs [107]4 years ago
4 0
<em>Endocytosis</em> is an active transport system. Vesicles have a membrane similar to the phospholipid membrane of a cell. When a solid material enters a cell, the plasma membrane forms a U-shaped membrane and surrounds the solid particle. This is done because of Pseudopodium. Eventually, the pseudopodium completely surrounds the solid particle and forms a phagosome, which is a vesicle. This is similar for fluids, too. However, there is another type of endocytosis known as Receptor-mediated endocytosis; this is a bit more complicated and for middle school, it is not required to know this. But in case if you are curious, here is what happens: certain receptors surround the cell. The receptors cling to particles and transfer the particles inside with the help of coat protein. 

In brief, there is a process called Pinocytosis or endocytosis, which is how vesicles are formed. Solid and fluid particles are enclosed by the plasma membrane and are brought into the cell. Eventually, a small spherical shape is formed, similar to a bubble. Eventually, that piece of plasma membrane splits off from the cellular membrane and becomes the vesicular membrane. It also has phospholipids, where the lipids are hydrophobic and the phosphorus is hydrophilic. 

I hope this helped!
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A person who weighs 800N on the earth's surface will weigh 200N at what height above the earth
Marina86 [1]

Answer: 6,400 km

Explanation:

The weight of a person is given by:

W=mg

where m is the mass of the person and g is the acceleration due to gravity. While the mass does not depend on the height above the surface, the value of g does, following the formula:

g=\frac{GM}{r^2}

where

G is the gravitational constant

M is the Earth's mass

r is the distance of the person from the Earth's center


The problem says that the person weighs 800 N at the Earth's surface, so when r=R (Earth's radius):

800 N= W=mg=m \frac{GM}{R^2} (1)

Now we want to find the height h above the surface at which the weight of the man is 200 N:

200 N = W' = mg' = m \frac{GM}{(R+h)^2} (2)

If we divide eq.(1) by eq.(2), we get

\frac{800 N}{200 N}=\frac{W}{W'}=\frac{(R+h)^2}{R^2}

4=\frac{(R+h)^2}{R^2}

By solving the equation, we find:

4R^2 = (R+h)^2=R^2+2Rh+h^2\\h^2 +2Rh-3R^2 =0

which has two solutions:

h=-3R --> negative solution, we can ignore it

h=R --> this is our solution

Since the Earth's radius is R=6.4\cdot 10^6 m, the person should be at h=R=6.4\cdot 10^6 m=6400 km above Earth's surface.

5 0
4 years ago
A size-5 soccer ball of diameter 22.6 cm and mass 426 g rolls up a hill without slipping, reaching a maximum height of 5.00 m ab
Drupady [299]

Answer

According the conservation of energy

\dfrac{1}{2}mv_i^2+\dfrac{1}{2}I\omega_i^2+0 = mgh + 0

I for ball = \dfrac{2}{3}mr^2

\dfrac{1}{2}mv_i^2+\dfrac{1}{2} \dfrac{2}{3}mr^2\omega_i^2= mgh

\omega_i = \dfrac{v_i}{r}

v_i^2+\dfrac{2}{3}r^2(\dfrac{v_i}{r})^2 = 2gh

v_i^2+\dfrac{2}{3}v_i^2 = 2gh

v_i^2+[1+\dfrac{2}{3}]=2gh

v_i^2\dfrac{5}{3}=2gh

v_i=\sqrt{\dfrac{6gh}{5}}=\sqrt{\dfrac{6\times 9.8\times 5}{5}}

v_i = 7.67\ m/s

a) \omega_i = \dfrac{v_i}{R}

\omega_i = \dfrac{7.67}{0.113}

\omega_i =67.87\ rad/s

b) K_{rot} = \dfrac{1}{2}\dfrac{2}{3}mR^2\omega_i^2

   K_{rot} = \dfrac{1}{3}\times 0.426\times 0.112^2\times 67.87^2

   K_{rot} = 8.205\ J

4 0
3 years ago
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