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3 years ago

What general roles does the cell membrane play in maintaining homeostasis

Physics

1 answer:

mr_godi [17]3 years ago

6 0

Answer:

Generally, cell membrane plays an important role in maintaining the homeostasis by controlling the substances that penetrate the cell, either entering the cell, or leaving the cell. Since the cell membrane is a phospholipid bilayer, it can control the transition of water and ions.

Explanation:

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How many moles of MgCl2 are there in 329 g of the compound? Your Answer:

Nookie1986 [14]

Answer:3.46

Explanation:

mass=329

molar mass=71

no. of moles=329/71

=3.46

6 0

3 years ago

Read 2 more answers

A fisherman is fishing from a bridge and is using a "44.0-N test line." In other words, the line will sustain a maximum force of

vitfil [10]

Answer:

a) 4.485 kg b) 3.94 kg

Explanation:

since the maximum tension the line can stand is 44 N and for question a the speed is constant (acceleration must be zero since the velocity or speed is not changing), F(tension) = mass * acceleration due to gravity (g) .

44 = m * 9.81m/s^2

m = 44/9.81 = 4.485kg

b) F(tension) = ma + mg ( where a is the acceleration of the body and g is the acceleration of the gravity)

44 = m (a +g)

44 = m (1.37 + 9.81)

44/11.18 = m

m = 3.94 kg

3 0

3 years ago

You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headligh

choli [55]

Answer:

They can be seen from a distance of 4.372 kilometers.

Explanation:

Using the Reyligh creterion for diffraction through a circular aperture we have

$\frac{x}{D}=\frac{1.22\lambda }{d}$

where symbol's have their usual meaning

thus applying values we get

$D=\frac{dx}{1.22\lambda }$

$\therefore D=\frac{0.633\times 4.61\times 10^{-3}}{1.22\times 547\times 10^{-9}}\\\\D=4372.77m\\=4.372km$

5 0

3 years ago

The resistance of a circuit is doubled. How is the value of the current affected?

tekilochka [14]

A because it makes more sense

3 0

3 years ago

suppose that you look into a photometer's eyepiece and the fluorescent disks appear to be equal in intensity. If the distance be

ehidna [41]

Use the Inverse square law, Intensity (I) of a light is inversely proportional to the square of the distance(d).

I=1/(d*d)

Let Intensity for lamp 1 is L1 distance be D1 so on, L2 D2 for Intensity for lamp 2 and its distance.

L1/L2=(D2*D2)/(D1*D1)

L1/15=(200*200)/(400*400)
L1=15*0.25
L1=3.75 candela

3 0

3 years ago