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notsponge [240]
3 years ago
10

If a poison is ingested that prevents the release of acetylcholine, which one of the following would occur at a myoneural juncti

on?
Physics
2 answers:
Marta_Voda [28]3 years ago
8 0
<h3><u>Answer;</u></h3>

The sodium and potassium gates on the motor end plate will not open

<h3><u>Explanation;</u></h3>
  • <em><u>Poisons are substances that are harmful or toxic to living organisms. They include biological compounds and also artificial compounds. </u></em>
  • Upon exposure, poisons may disrupts metabolic process or destroy tissues through chemical reactions with cells.
  • <em><u>Poisons such as Botulinum toxin, when ingested cause oxygen deprivation in the body.</u></em> <em><u>This poison prevents the release of acetylcholine by neurons at the neuromuscular junction since the potassium and sodium ions gates do not open.  Without this neurotransmitter, the respiratory muscles cannot contract. </u></em>

kirill115 [55]3 years ago
4 0
Had to look for the options and here is my answer.
Given the situation that a poison is consumed and this prevented the acetylcholine release, the one that would most likely occur at a myoneural junction is that "sodium and potassium gates on the motor end plate will not open". Hope this helps.
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Answer:

the equivalent mass : m_e = m_1+m_2+\frac{I}{R^2}

the equation of the motion of the block of mass m_1 in terms of its displacement is = (m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)

Explanation:

Let use m₁ to represent the mass of the block and m₂ to represent the mass of the cylinder

The radius of the cylinder  be = R

The distance between the center of the pulley to center of the block to be = x

Also, the angles of inclinations of the cylinder and the block with respect to the ground to be \phi and \beta respectively.

The velocity of the block to be = v

The equivalent mass of the system = m_e

In the terms of the equivalent mass, the kinetic energy of the system can be written as:

K.E = \frac{1}{2} m_ev^2       --------------- equation (1)

The angular velocity of the cylinder = \omega  :  &

The inertia of the cylinder about its center to be = I

The angular velocity of the cylinder can be written as:

v = \omega R

\omega =\frac{v}{R}

The kinetic energy of the system in terms of individual mass can be written as:

K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I\omega^2

By replacing \omega with \frac{v}{R} ; we have:

K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I(\frac{v}{R})^2

K.E = \frac{1}{2}(m_1+ m_2+ \frac{I}{R} )v^2   ------------------ equation (2)

Equating both equation (1) and (2); we have:

m_e = m_1+m_2+\frac{I}{R^2}

Therefore, the equivalent mass : m_e = m_1+m_2+\frac{I}{R^2}    which is read as;

The equivalent mass is equal to the mass of the block plus the mass of the cylinder plus the inertia by  the square of the radius.

The expression for the force acting on equivalent mass due to the block is as follows:

f_{block }=m_1gsin \beta

Also; The expression for the force acting on equivalent mass due to the cylinder is as follows:

f_{cylinder} = m_2gsin \phi

Equating the above both equations; we have the equation of motion of the  equivalent system to be

m_e \bar x = f_{cylinder}-f_{block}

which can be written as follows from the previous derivations

(m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)

Finally; the equation of the motion of the block of mass m_1 in terms of its displacement is = (m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)

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