Air has pressure because

1. At the synaptic terminal, voltage-gated ______________ channels open, thereby stimulating the synaptic vesicles to release th

Sergeeva-Olga [200]

Ion

Explanation:

At the synaptic terminal, voltage-gated ion channels open, thereby stimulating the synaptic vesicles to release the neurotransmitters by exocytosis.

These ion channels are the signaling molecules in neurons. They are the transmembrane proteins that form ion channels. The membrane potential changes the conformation of the channel proteins that regulates their opening and closing. These channels play an important role in neurotransmitter release in presynaptic nerve endings.

For example - Ca²⁺ gated ion channel.

4 0

3 years ago

Two points charge of 4\mu C and 2\mu C are placed at theopposite corners of a rectangle. What is the potential difference Va- Vb

bulgar [2K]

Answer:

$Va-Vb=168KV$

Explanation:

From the question we are told that

Two points charge of 4\mu C and 2\mu C

Generally we find the Va and Vb individually to find there difference

Given a rectangle with two equal sides each,Assume lengths for bot sides

Length L=0.3

Breath B=0.4

Diagonal D= $\sqrt{0.3^2+0.4^2} =0.5$

at opposite sides

Mathematically Va can represented as

$Va =k(\frac{4*10^_-_6}{0.3} +\frac{-2*10^_-_6}{0.5} )$

$Va =9*10^9(\frac{4*10^_-_6}{0.3} +\frac{-2*10^_-_6}{0.5} )$

$Va =9*10^9(0.00001333333-0.000004} )$

$Va =84000V$

$Va =84KV$

Mathematically Vb is represented as

$Va =k(\frac{-4*10^_-_6}{0.3} +\frac{2*10^_-_6}{0.5} )$

$Va =9*10^9(\frac{-4*10^_-_6}{0.3} +\frac{+2*10^_-_6}{0.5} )$

$Va =9*10^9(-0.00001333333+0.000004} )$

$Va =-84000V$

$Va =-84KV$

Therefore

$Va-Vb=84-(-84)\\Va-Vb=84+84\\Va-Vb=168KV$

7 0

3 years ago

Why is more energy required to vaporise 1kg of water than to melt 1kg of ice?

KiRa [710]

Answer:

cause it takes a lotof energy to melt

Explanation:

7 0

3 years ago

Read 2 more answers

Calculate the electric field at the center of a square

pantera1 [17]

Answer:

$E_y=1175510.2\ N.C^{-1}$

The Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

Explanation:

Given:

side of a square, $a=52.5\ cm$

charge on one corner of the square, $q_1=+45\times 10^{-6}\ C$

charge on the remaining 3 corners of the square, $q_2=q_3=q_4=-27\times 10^{-6}\ C$

Distance of the center from each corners $=\frac{1}{2} \times diagonals$

$diagonal=\sqrt{52.5^2+52.5^2}$

$diagonal=74.25\cm=0.7425\ m$

∴Distance of center from corners, $b=0.3712\ m$

Now, electric field due to charges is given as:

$E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}$

For charge $q_1$ we have the field lines emerging out of the charge since it is positively charged:

$E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}$

$E_1=2938775.5\ N.C^{-1}$

Force by each of the charges at the remaining corners:

$E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}$

$E_2=E_3=E_4=1763265.3\ N.C^{-1}$

Now, net electric field in the vertical direction:

$E_y=E_1-E_4$

$E_y=1175510.2\ N.C^{-1}$

Now, net electric field in the horizontal direction:

$E_y=E_2-E_3$

$E_y=0\ N.C^{-1}$

So the Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

8 0

3 years ago

A 1.20 × 104 kg railroad car moving at 7.70 m/s to the north collides with and sticks to another railroad car of the same mass t

saul85 [17]

Answer: 4.77m/s

Explanation:

According to the law of conservation of momentum which states that the sum total of momentum of bodies before collision is equal to the sum of their momentum after collision. Note that the two bodies will move at a common velocity after colliding.

Let m1 and m2 be the mass of the first and second railroad cars

u1 and u2 be the velocities of the railroad cars

v be the common velocity

Using the formula

m1u1 + m2u2 = (m1 +m2)

m1 = 1.20×10⁴kg

m2 = 1.20×10⁴kg (body of same mass)

u1 = 7.70m/s

u2 = 1.84m/s

v = ?

(1.20×10⁴×7.7) + (1.20×10⁴×1.84) = (1.20×10⁴ + 1.20× 10⁴)v

9.24×10⁴ + 2.21×10⁴ = 2.4×10⁴v

11.45×10⁴ = 2.4×10⁴v

v = 11.45×10⁴/2.4×10⁴

v = 4.77m/s

The velocity of the cars after collision will be 4.77m/s

5 0

3 years ago