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Softa [21]
3 years ago
9

PSYCHOLOGY propelling neural signals is a function of ______ neurotransmitters.

Physics
1 answer:
Nesterboy [21]3 years ago
5 0
The answer is b. excitatory
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A constant electric field with magnitude 1.50 ✕ 103 N/C is pointing in the positive x-direction. An electron is fired from x = −
romanna [79]

Answer:

The speed of electron is 1.5\times10^{7}\ m/s

Explanation:

Given that,

Electric field E=1.50\times10^{3}\ N/C

Distance = -0.0200

The electron's speed has fallen by half when it reaches x = 0.190 m.

Potential energy P.E=5.04\times10^{-17}\ J

Change in potential energy \Delta P.E=-9.60\times10^{-17}\ J as it goes x = 0.190 m to x = -0.210 m

We need to calculate the work done by the electric field

Using formula of work done

W=-eE\Delta x

Put the value into the formula

W=-1.6\times10^{-19}\times1.50\times10^{3}\times(0.190-(-0.0200))

W=-5.04\times10^{-17}\ J

We need to calculate the initial velocity

Using change in kinetic energy,

\Delta K.E = \dfrac{1}{2}m(\dfrac{v}{2})^2+\dfrac{1}{2}mv^2

\Delta K.E=\dfrac{-3mv^2}{8}

Now, using work energy theorem

\Delta K.E=W

\Delta K.E=\Delta U

So, \Delta U=W

Put the value in the equation

\dfrac{-3mv^2}{8}=-5.04\times10^{-17}

v^2=\dfrac{8\times(5.04\times10^{-17})}{3m}

Put the value of m

v=\sqrt{\dfrac{8\times(5.04\times10^{-17})}{3\times9.1\times10^{-31}}}

v=1.21\times10^{7}\ m/s

We need to calculate the change in potential energy

Using given potential energy

\Delta U=-9.60\times10^{-17}-(-5.04\times10^{-17})

\Delta U=-4.56\times10^{-17}\ J

We need to calculate the speed of electron

Using change in energy

\Delta U=-W=-\Delta K.E

\Delta K.E=\Delta U

\dfrac{1}{2}m(v_{f}^2-v_{i}^2)=4.56\times10^{-17}

Put the value into the formula

v_{f}=\sqrt{\dfrac{2\times4.56\times10^{-17}}{9.1\times10^{-31}}+(1.21\times10^{7})^2}

v_{f}=1.5\times10^{7}\ m/s

Hence, The speed of electron is 1.5\times10^{7}\ m/s

4 0
3 years ago
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