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Readme [11.4K]
4 years ago
5

The inner and outer surfaces of a cell membrane carry a negative and positive charge, respectively. Because of these charges, a

potential difference of about 0.063 V exists across the membrane. The thickness of the membrane is 9.9 x 10-9 m. What is the magnitude of the electric field in the membrane
Physics
1 answer:
sertanlavr [38]4 years ago
7 0

Answer:

Electric field, E = 6.36\times 10^{6}\ V/m

Explanation:

Given that,

The potential difference across the membrane, \Delta V=0.063\ V

The thickness of the membrane is, \Delta x=9.9\times 10^{-9}\ m

We need to find the magnitude of the electric field in the membrane. We know that the relation between electric field and electric potential is given by :

E=\dfrac{-\Delta V}{\Delta x}\\\\E=\dfrac{0.063\ V}{9.9\times 10^{-9}\ m}\\\\E=6.36\times 10^{6}\ V/m

So, the magnitude of the electric field in the membrane is 6.36\times 10^{6}\ V/m.

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In a type of chemical reaction called thermal decomposition, heat is added to a substance, which then splits apart to form sever
exis [7]
The answer is a. endothermic.
In endothermic reaction, as the name suggests, the reactants require energy absorption to form and complete the chemical reaction to form products.
And even if you don't know the answer, you can exclude all other answers by scientific logic and you will be left with the correct answer.

Hope this helps.
5 0
3 years ago
A car starts from rest and accelerates uniformly at 6.6 m/s2 for 10.s. How far does the car travel?
quester [9]

Explanation:

s = ut + 1/2at²

= 0(10) + 1/2×6.6×(10)²

= 3.3 × 100

= 330 m

5 0
3 years ago
This problem has been solved!
Liono4ka [1.6K]

Answer:

Explanation:

In this problem we can use Bohr's atomic model, to deal with the electronic transition, so we can have transitions between two given states.

           ΔE = E_{f} -E₀

Lower state     final state                     energy

Fundamental      first excited state          ΔE₁ = -2.4 - (-4.7)  = 2.3 eV

Fundamental      second excited state   ΔE₂ = -1.0 - (-4.7) = 3.7 eV

Fundamental      third excited state        ΔE₃ = -0.3 - (-4.7) = 4.4 eV

As they indicate that there are no electrons in the excited states these are the only possible transitions.

When a wide range of light strikes, the frequencies of these energies are absorbed and observed as black bands (absence of radiation) in the spectrum.

3 0
3 years ago
SOMEBODY PLEEASEEE HELP A STRUGGLING HIGHSCHOOLERRRR +(
CaHeK987 [17]

Explanation:

What is the weight of a 2.00-kilogram object on the surface of Earth?

2.00 N

4.91 N

9.81 N

19.6 N

Given parameters:

Mass of the object = 2kg

Unknown:

Weight of the object  = ?

Solution:

The weight of an object is the force of gravity acting on the object;

 Weight  =  mass x acceleration due to gravity

Acceleration due to gravity  = 9.8m/s²

 Now insert the parameters and solve;

    Weight  = 2 x 9.8 = 19.6N

A person weighing 785 Newtons on the surface of the Earth would weigh 47 Newtons on the surface of Pluto. What is the magnitude of the gravitational acceleration on the surface of Pluto?

1.7 m/s²

0.59 m/s²

0.29 m/s²

9.8 m/s²

Given parameters:

Weight on Earth  = 785N

Weight on Pluto = 47N

Unknown:

Acceleration due to gravity on Pluto = ?

Solution

The mass of the body both on Earth and Pluto is the same.

  Weight = mass x acceleration due to gravity

Now find the mass on Earth;

  Acceleration due to gravity on Earth  = 9.8m/s²

    785   = mass x 9.8

         mass  = \frac{785}{9.8}   = 80.1kg

So;

  Acceleration due to gravity on Pluto = \frac{Weight on Pluto}{mass }  

  Acceleration due to gravity  = \frac{47}{80.1}   = 0.59m/s²

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3 years ago
some male coworkers of suzanne are bullying her because she is a woman . which type of hazard is this ?
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This is most definitely a physiological hazard
4 0
4 years ago
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