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olasank [31]
2 years ago
7

If a hyperpolarizing graded potential and a depolarizing graded potential of similar magnitudes arrive at the trigger zone at th

e same time, what is most likely to occur
Physics
1 answer:
Flura [38]2 years ago
8 0

The likely outcome when a hyperpolarizing and a depolarizing potential of similar magnitudes arrive simultaneously at the trigger zone is the cancellation of charges.

<h3>What is membrane action potential?</h3>

Membrane action potential refers to differential electric charges within and outside of a given cell.

Hyperpolarization refers to a differential membrane potential of a cell associated with a negative charge, whereas depolarization is associated with a positive charge.

In conclusion, the likely outcome when a hyperpolarizing and a depolarizing potential of similar magnitudes arrive simultaneously at the trigger zone is the cancellation of charges.

Learn more about membrane action potential here:

brainly.com/question/6705448

#SPJ1

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PLS HELP I HAVE EXAM ILL GIVE U BRAINLIST
stiv31 [10]

Explanation:

False

electron and proton attract each other

3 0
3 years ago
If 34.7 g of O2 reacts with iron to form 79.34 g of iron oxide, how much iron was used in the reaction?
zhuklara [117]

Answer: B. 44.64 g

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

Mass of reactants = mass of iron + mass of oxygen = mass of iron + 34.7 g

Mass of product = mass of iron oxide = 79.34 g

As Mass of reactants = Mass of product

mass of iron + 34.7 g = 79.34 g

mass of iron = 44.64 g

Thus 44.64 g of iron was used in the reaction

6 0
3 years ago
An electric dipole is formed from ±1.00nC charges spaced 3.00 mm apart. The dipole is at the origin, oriented along the x-axis.
Ronch [10]

Answer:

Value of electric field along the axis and equitorial axis  E=31.25\ N/c and E = 15.625\ N/c respectively.

Explanation:

Given :

Distance between charges , d = 3 \ mm =\dfrac{3}{1000}\ m=3\times 10^{-3}\ m.

Magnitude of charges , q=1\ nC = 10^{-9}\ C.

Dipole moment , p=qL=10^{-9}\times 3\times 10^{-3}=3\times 10^{-12} \ C\ m.

Case A) (x,y) = (12.0 cm, 0 cm) :

Electric field of dipole in its axis ,

E=\dfrac{2kp}{r^3}

Putting all values and r=12\times 10^{-2}\ m.

We get , E=31.25\ N/c.

Case B) (x,y) = (0 cm, 12.0 cm) :

Electric field of dipole on equitorial axis ,

E = \dfrac{kp}{r^3}

Putting all values and r=12\times 10^{-2}\ m.

We get , E = 15.625\ N/c.

Hence , this is the required solution.

7 0
3 years ago
A ball of moist clay falls 17.3 m to the ground. It is in contact with the ground for 24.0 ms before stopping. (a) What is the a
gizmo_the_mogwai [7]

Answer:

Acceleration,  767.08\ m/s^2

Explanation:

Given that,

Height from a ball falls the ground, h = 17.3 m

It is in contact with the ground for 24.0 ms before stopping.

We need to find the average acceleration the ball during the time it is in contact with the ground.

Firstly, find the velocity when it reached the ground. So,

v^2=u^2+2ah

u = initial velocity=0 m/s

a = acceleration=g

v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 17.3} \\\\v=18.41\ m/s

It is in negative direction, u = -18.41 m/s

Let a is average acceleration of the ball. Consider, v = and u = -18.41 m/s.

a=\dfrac{v-u}{t}\\\\a=\dfrac{0-18.41}{24\times 10^{-3}}\\\\a=767.08\ m/s^2

So, the average acceleration of the ball during the time it is in contact is 767.08\ m/s^2.

4 0
3 years ago
Explain briefly why metals are good conductors of electric current
Mice21 [21]
The free electrons in metals can move through the metal, all while receiving and losing electrons, allowing metals to conduct electricity. Example: copper is a great conductor of electric current.
7 0
3 years ago
Read 2 more answers
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