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3 years ago

Please help..

2 answers:

7 0

The correct answer for this question would be the first option. In a neuromuscular synapse, it is the enzyme acetylcholinesterase, which is located at the motor end plate, that breaks down the neurotransmitter released from the axon. This <span>is responsible for stimulating terminating the stimulation of the skeletal muscle cell to make it contract. Hope this answer helps. </span>

QveST [7]3 years ago

6 0

Hello!

Acetylcholine is the neurotransmitter released by cholinergic neurons responsible for contracting muscles (one of the major) and by learning and memory. In the case of muscle contraction, nerve transmission (in this case called neuromuscular transmission) occurs at a site called the neuromuscular junction, where the axon splits into several endings and binds to skeletal muscle fiber. It is the product of the reaction between choline (lecithin component) and acetyl-CoA in the presence of the enzyme choline acetyltransferase.

Answer:

A. The enzyme acetylcholinesterase

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If the voltage across the first capacitor (the one with capacitance

RoseWind [281]

The answer to this question is: it depends. It depends on the arrangement of the capacitors in a circuit: it can be either in series or in parallel. The difference is shown in the picture.

Capacitors are like batteries in a way that they store power from the source. It has some rules depending on the type of circuit. For parallel circuits, the voltage across each capacitor is equal. Therefore, V₁=V₂=V₃.

On the other hand, if the capacitors are arranged in series, the voltage across each capacitor should add up to the total voltage of the source. Therefore, V₁+V₂+V₃ = Total Voltage.

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3 years ago

Coulomb’s law and static point charge ensembles (15 points). A test charge of 2C is located at point (3, 3, 5) in Cartesian coor

fenix001 [56]

Answer:

a) $F_{r}= -583.72MN i + 183.47MN j + 6.05GN k$

b) $E=3.04 \frac{GN}{C}$

Step-by-step explanation.

In order to solve this problem, we mus start by plotting the given points and charges. That will help us visualize the problem better and determine the direction of the forces (see attached picture).

Once we drew the points, we can start calculating the forces:

$r_{AP}^{2}=(3-0)^{2}+(3-0)^{2}+(5+0)^{2}$

which yields:

$r_{AP}^{2}= 43 m^{2}$

(I will assume the positions are in meters)

Next, we can make use of the force formula:

$F=k_{e}\frac{q_{1}q_{2}}{r^{2}}$

so we substitute the values:

$F_{AP}=(8.99x10^{9})\frac{(1C)(2C)}{43m^{2}}$

which yields:

$F_{AP}=418.14 MN$

Now we can find its components:

$F_{APx}=418.14 MN*\frac{3}{\sqrt{43}}i$

$F_{APx}=191.30 MNi$

$F_{APy}=418.14 MN*\frac{3}{\sqrt{43}}j$

$F_{APy}=191.30MN j$

$F_{APz}=418.14 MN*\frac{5}{\sqrt{43}}k$

$F_{APz}=318.83 MN k$

And we can now write them together for the first force, so we get:

$F_{AP}=(191.30i+191.30j+318.83k)MN$

We continue with the next force. The procedure is the same so we get:

$r_{BP}^{2}=(3-1)^{2}+(3-1)^{2}+(5+0)^{2}$

which yields:

$r_{BP}^{2}= 33 m^{2}$

Next, we can make use of the force formula:

$F_{BP}=(8.99x10^{9})\frac{(4C)(2C)}{33m^{2}}$

which yields:

$F_{BP}=2.18 GN$

Now we can find its components:

$F_{BPx}=2.18 GN*\frac{2}{\sqrt{33}}i$

$F_{BPx}=758.98 MNi$

$F_{BPy}=2.18 GN*\frac{2}{\sqrt{33}}j$

$F_{BPy}=758.98MN j$

$F_{BPz}=2.18 GN*\frac{5}{\sqrt{33}}k$

$F_{BPz}=1.897 GN k$

And we can now write them together for the second, so we get:

$F_{BP}=(758.98i + 758.98j + 1897k)MN$

We continue with the next force. The procedure is the same so we get:

$r_{CP}^{2}=(3-5)^{2}+(3-4)^{2}+(5-0)^{2}$

which yields:

$r_{CP}^{2}= 30 m^{2}$

Next, we can make use of the force formula:

$F_{CP}=(8.99x10^{9})\frac{(7C)(2C)}{30m^{2}}$

which yields:

$F_{CP}=4.20 GN$

Now we can find its components:

$F_{CPx}=4.20 GN*\frac{-2}{\sqrt{30}}i$

$F_{CPx}=-1.534 GNi$

$F_{CPy}=4.20 GN*\frac{2}{\sqrt{30}}j$

$F_{CPy}=-766.81 MN j$

$F_{CPz}=4.20 GN*\frac{5}{\sqrt{30}}k$

$F_{CPz}=3.83 GN k$

And we can now write them together for the third force, so we get:

$F_{CP}=(-1.534i - 0.76681j +3.83k)GN$

So in order to find the resultant force, we need to add the forces together:

$F_{r}=F_{AP}+F_{BP}+F_{CP}$

so we get:

$F_{r}=(191.30i+191.30j+318.83k)MN + (758.98i + 758.98j + 1897k)MN + (-1.534i - 0.76681j +3.83k)GN$

So when adding the problem together we get that:

$F_{r}=(-0.583.72i + 0.18347j +6.05k)GN$

which is the answer to part a), now let's take a look at part b).

Basically, we need to find the magnitude of the force and divide it into the test charge, so we get:

$F_{r}=\sqrt{(-0.583.72)^{2} + (0.18347)^{2} +(6.05)^{2}}$

which yields:

$F_{r}=6.08 GN$

and now we take the formula for the electric field which is:

$E=\frac{F_{r}}{q}$

so we go ahead and substitute:

$E=\frac{6.08GN}{2C}$

$E=3.04\frac{GN}{C}$

7 0

3 years ago

I need guidance pls

seropon [69]

The slowest line is the solid line and the fastest is the dotted line that crosses the solid line
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3 years ago

If the temperature of the metal oxide and water solution is increased,this would most likely..

kiruha [24]

Chemical Reaction between metal oxide and water solution

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3 years ago

an object weighs 98 n on earth. How much does it weigh on planet x where the acceleration due to gravity in 6 m/s^2

Degger [83]

60 N because 98N=mg (here g= 9.8 on earth) thus mass can be calculated which is 98/9.8 = 10kg Now,new weight with g = 6m/s^2 =m×g' (here g' is new acceleration of the new planet) = 10×6=60N

7 0

3 years ago

Read 2 more answers