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3 years ago

An action potential arriving at the presynaptic terminal causes what to occur?

Physics

1 answer:

Ulleksa [173]3 years ago

5 0

Answer:

Voltage-gated calcium ion channels open, and calcium ions diffuse into the cell

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Why does diamond sparkles stars twinkles?

Helga [31]

Answer:

Because they want attention

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3 years ago

Describe the movement of particles at the melting point of a substance with temperature and energy.

Gnesinka [82]

Answer:

As a substance reaches the melting point, the particles begin to move faster, causing the substance to become a liquid.

Explanation:

lmk if you need a different or more detailed answer :)

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3 years ago

How long would it take for a car to travel 200 km if it has an average speed of 55 km hr?

Rainbow [258]

Answer:

3.63 hours or 3 and 37.5 minutes

Explanation:

200/55

Hope this helps :)

6 0

3 years ago

1. A step-up transformer increases 15.7V to 110V. What is the current in the secondary as compared to the primary? Assume 100 pe

Serjik [45]

1. $I_2 = 0.14 I_1$

Explanation:

We have:

$V_1 = 15.7 V$ voltage in the primary coil

$V_2 = 110 V$ voltage in the secondary coil

The efficiency of the transformer is 100%: this means that the power in the primary coil and in the secondary coil are equal

$P_1 = P_2\\V_1 I_1 = V_2 I_2$

where I1 and I2 are the currents in the two coils. Re-arranging the equation, we find

$\frac{I_2}{I_1}=\frac{V_1}{V_2}=\frac{15.7 V}{110 V}=0.14$

which means that the current in the secondary coil is 14% of the value of the current in the primary coil.

2. 5.7 V

We can solve the problem by using the transformer equation:

$\frac{N_p}{N_s}=\frac{V_p}{V_s}$

where:

Np = 400 is the number of turns in the primary coil

Ns = 19 is the number of turns in the secondary coil

Vp = 120 V is the voltage in the primary coil

Vs = ? is the voltage in the secondary coil

Re-arranging the formula and substituting the numbers, we find:

$V_s = V_p \frac{N_s}{N_p}=(120 V)\frac{19}{400}=5.7 V$

5 0

3 years ago

An electron is released from rest in a weak electric field given by E =-2.30 x 10-10 N/Cj. After the electron has traveled a ver

Goshia [24]

Explanation:

It is given that,

An electron is released from rest in a weak electric field of, $E=2.3\times 10^{-10}\ N/C$

Vertical distance covered, $s=1\ \mu m=10^{-6}\ m$

We need to find the speed of the electron. Let its speed is v. Using third equation of motion as :

$v^2-u^2=2as$

$v^2=2as$ .............(1)

Electric force is $F_e$ and force of gravity is $F_g$ . As both forces are acting in downward direction. So, total force is:

$F=mg+qE$

$F=9.1\times 10^{-31}\times 9.8+1.6\times 10^{-19}\times 2.3\times 10^{-10}$

$F=4.57\times 10^{-29}\ N$

Acceleration of the electron, $a=\dfrac{F}{m}$

$a=\dfrac{4.57\times 10^{-29}\ N}{9.1\times 10^{-31}\ kg}$

$a=50.21\ m/s^2$

Put the value of a in equation (1) as :

$v=\sqrt{2as}$

$v=\sqrt{2\times 50.21\times 10^{-6}}$

v = 0.010 m/s

So, the speed of the electron is 0.010 m/s. Hence, this is the required solution.

6 0

4 years ago