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enyata [817]
3 years ago
6

The student in charge of the experiment wishes to present his/her findings in support of the theory that motor imagery and actio

n observation together facilitate a greater increase in motor excitability than either protocol alone.
Which of the following would be the best reason to withhold this presentation?

A. The procedure used did not include MEP recordings prior to each task.
B. MEP amplitudes in an individual are typically highly consistent.
C. The motor tasks performed in the experiment were too simple.
D. The six different conditions were run in random order.
Physics
1 answer:
Katen [24]3 years ago
3 0

Answer:

C. The motor tasks performed in the experiment were too simple.

On this case is the best option since the student wants to explain the effect of motor imagery and action observation together into the excitability. And maybe is too simple, since we need to cover other possibilities in order to analyze the excitability.

Explanation:

A. The procedure used did not include MEP recordings prior to each task.

Not true, is not a requisite record MEP prior to the task to evaluate the variable of interest on this case.

B. MEP amplitudes in an individual are typically highly consistent.

The Motor evoked potentials (MEP) "are electrical signals recorded from neural tissue or muscle after activation of central motor pathways". But on this case that's a technical aspect related to the topic and this not would be the reason why we need to withhold the presentation

C. The motor tasks performed in the experiment were too simple.

On this case is the best option since the student wants to explain the effect of motor imagery and action observation together into the excitability. And maybe is too simple, since we need to cover other possibilities in order to analyze the excitability.

D. The six different conditions were run in random order.

That's not true the student are not analyzing 6 different conditions, just 2.

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Consider a force of 57.3 N, pulling 3 blocks of
andrezito [222]

Block 1 (the rightmost block) has

• net horizontal force

∑ <em>F</em> = <em>F</em> - <em>T₁</em> - <em>f₁</em> = <em>m₁a</em>

• net vertical force

∑ <em>F</em> = <em>N₁</em> - <em>m₁g</em> = 0

where <em>F</em> = 57.3 N, <em>T₁</em> is the tension in the string connecting blocks 1 and 2, <em>f₁</em> is the magnitude of kinetic friction felt by block 1, <em>m₁</em> = 0.8 kg is its mass, <em>a</em> is the acceleration you want to find, and <em>N₁</em> is the magnitude of the normal force exerted by the surface.

Block 2 (middle) has much the same information:

• net horiz. force

∑ <em>F</em> = <em>T₁</em> - <em>T₂</em> - <em>f₂</em> = <em>m₂a</em>

• net vert. force

∑ <em>F</em> = <em>N₂</em> - <em>m₂g</em> = 0

with similarly defined symbols.

The same goes for block 3 (leftmost):

• net horiz. force

∑ <em>F</em> = <em>T₂</em> - <em>f₃</em> = <em>m₃a</em>

• net vert. force

∑ <em>F</em> = <em>N₃</em> - <em>m₃g</em> = 0

We have <em>m₁</em> = <em>m₂</em> = <em>m₃</em> = 0.8 kg, so I'll just replace each with <em>m</em>. It follows that each normal force has the same magnitude, <em>N₁</em> = <em>N₂</em> = <em>N₃</em> = <em>mg</em>. And as a consequence of that, each frictional force has the same magnitude, <em>f₁</em> = <em>f₂</em> = <em>f₃</em> = 0.4<em>mg.</em>

In short, the relevant equations are

[1] … 57.3 N - <em>T₁</em> - 0.4<em>mg</em> = <em>ma</em>

[2] …<em>T₁</em> - <em>T₂</em> - 0.4<em>mg</em> = <em>ma</em>

[3] … <em>T₂</em> - 0.4<em>mg</em> = <em>ma</em>

<em />

Adding [1], [2] and [3] together eliminates the tension forces, and we get

57.3 N - 1.2<em>mg</em> = 3<em>ma</em>

<em />

Solve for <em>a</em> :

57.3 N - 1.2 (0.8 kg) (9.8 m/s²) = 3 (0.8 kg) <em>a</em>

57.3 N - 9.408 N = (2.4 kg) <em>a</em>

<em>a</em> = (47.892 N) / (2.4 kg)

<em>a</em> ≈ 20.0 m/s²

3 0
2 years ago
Assuming the ball's initial velocity was 51 ∘ above the horizontal and ignoring air resistance, what did the initial speed of th
Umnica [9.8K]

horizontal distance of home run is 400 ft = 122 m

height of the home run is 3 ft = 0.9 m

now the angle of the hit is 51 degree

now we have equation of trajectory of the motion

x = vcos\theta * t

y = v sin\theta * t - \frac{1}{2} gt^2

solving above two equations we have

y = xtan\theta - \frac{gx^2}{2v^2cos^2\theta}

now here we will plug in all data

0.9 = 122 tan51 - \frac{9.8 * 122^2}{2*v^2 * cos^251}

0.9 = 150.65 - \frac{184150.2}{v^2}

\frac{184150.2}{v^2} = 149.75

v = 35.1 m/s

<em>so the ball was hit with speed 35.1 m/s from the ground</em>

8 0
3 years ago
Determine the mechanical energy of this object: 1-kg ball rolls on the ground at 2 m/s
Fiesta28 [93]
The potential energy would be zero. Only kinetic energy is present in this case. To find out what the answer is we do the equation: mv^2/2 soo...

KE =mv^2/2 
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Julli [10]

Answer:

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Explanation:

It is a measure of how well a machine reduces friction.

7 0
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Harrizon [31]
There would be no mass or weight and he would float away
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3 years ago
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