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3 years ago

Explain how neurons convey information using both electrical and chemical signals.

Physics

1 answer:

8090 [49]3 years ago

6 0

Answer:

Explanation:

Neurons communicate via both electrical signals and chemical signals. The electrical signals are action potentials, which transmit the information from one of a neuron to the other; the chemical signals are neurotransmitters, which transmit the information from one neuron to the next.

The electrical signal travels down the axon to the axon terminals where it tells the vesicles to release the neurotransmitters into the synaptic cleft which travel to the receptors of the receiving cell which releases the second messengers

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When a 20-V emf is placed across two resistors in series, a current of 2.0 A is present in each of the resistors. When the same

ehidna [41]

Answer:

7.24 ohm

Explanation:

Let R1 and R2 are resistance of two resistors.

Emf=E=20 V

Current,I=2 A

Current,I'=10 A

We have to find the magnitude of the greater of the two resistances.

In series

$R=R_1+R_2$

$V=IR$

By using the formula

$20=2(R_1+R_2)$

$R_1+R_2=\frac{20}{2}=10$ ...(1)

In parallel

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}$

$\frac{1}{R}=\frac{R_2+R_1}{R_1R_2}$

$R=\frac{R_1R_2}{R_1+R_2}$

$20=10(\frac{R_1R_2}{R_1+R_2}$

$2=\frac{R_1R_2}{10}$

$R_1R_2=20$

$R_2=\frac{20}{R_1}$

Substitute the value

$\frac{20}{R_1}+R_1=10$

$R^2_1+20=10R_1$

$R^2_1-10R_1+20=0$

$R_1=\frac{10\pm\sqrt{(-10)^2-4(20)}}{2}$

By using quadratic formula

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$R_1=\frac{10\pm 2\sqrt 5}{2}$

$R_1=\frac{10+2\sqrt 5}{2}=5+\sqrt 5=7.24 ohm$

$R_1=\frac{10-2\sqrt 5}{2}=2.76 ohm$

Substitute the value

$R_2=\frac{20}{7.24}=2.76 ohm$

$R_2=\frac{20}{2.76}=7.24 ohm$

Hence, the magnitude of the greater of the two resistance=7.24 ohm

8 0

3 years ago

Read 2 more answers

When a fixed amount of ideal gas goes through an isobaric expansion A) its internal (thermal) energy does not change.B) the gas

Bingel [31]

<h2>Answer: its temperature must increase.</h2>

Explanation:

In an isobaric process the pressure remains constant, which means the initial pressure and the final pressure will be the same.

In addition, during this thermodynamic process, the volume of the ideal gas expands or contracts in such a way that the variation of pressure $\Delta P$ is neutralized.

Now, according to the First law of Thermodynamics that establishes the conservation of energy:

$\Delta U=\Delta Q-\Delta W$ (1)

Where:

$\Delta U$ is the internal energy

$\Delta Q$ is the heat transferred

$\Delta W$ is the work

Now, for an isobaric process:

$\Delta W=P\Delta V$ (2)

Where:

$P$ is the pressure (always positive)

$\Delta V$ is the volume variation of the gas

Here we have two possible results:

-If the gas expands (positive $\Delta V$ ), the work is positive.

-If the gas compresses (negative $\Delta V$ ), the work is negative.

In this case we are talking about the first result (work is positive).

Then, according to the above, equation (1) can be written as follows:

$\Delta U=\Delta Q - P\Delta V$ (3)

Clearing $\Delta Q$ :

$\Delta Q=\Delta U+P \Delta V$ (4)

Then, for an ideal gas in an isobaric process, part of the heat ( $Q$ ) added to the system will be used to do work (positive in this case) and the other part will increase the internal energy, hence the temperature will increase as well.

7 0

3 years ago

The least common fossils are those that have been petrified frozen buried distilled

Alchen [17]

Answer: Frozen fossils

6 0

3 years ago

What is true about valence electrons?

luda_lava [24]

The valence electrons are the one furthest from the nucleus

8 0

3 years ago

Read 2 more answers

A copper telephone wire has essentially

Lunna [17]

Answer:

128.21 m

Explanation:

The following data were obtained from the question:

Initial temperature (θ₁) = 4 °C

Final temperature (θ₂) = 43 °C

Change in length (ΔL) = 8.5 cm

Coefficient of linear expansion (α) = 17×10¯⁶ K¯¹)

Original length (L₁) =.?

The original length can be obtained as follow:

α = ΔL / L₁(θ₂ – θ₁)

17×10¯⁶ = 8.5 / L₁(43 – 4)

17×10¯⁶ = 8.5 / L₁(39)

17×10¯⁶ = 8.5 / 39L₁

Cross multiply

17×10¯⁶ × 39L₁ = 8.5

6.63×10¯⁴ L₁ = 8.5

Divide both side by 6.63×10¯⁴

L₁ = 8.5 / 6.63×10¯⁴

L₁ = 12820.51 cm

Finally, we shall convert 12820.51 cm to metre (m). This can be obtained as follow:

100 cm = 1 m

Therefore,

12820.51 cm = 12820.51 cm × 1 m / 100 cm

12820.51 cm = 128.21 m

Thus, the original length of the wire is 128.21 m

5 0

3 years ago