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Setler79 [48]
3 years ago
15

What is the closest distance the electrodes used in an NCV test can be placed on a nerve in order to measure the voltage change

as a response to the stimulus?
Physics
1 answer:
sammy [17]3 years ago
6 0

Answer:

0.1 m

Explanation:

The closest distance the electrodes used in an NCV test in oerder to measure

the voltage change as a response to the stimulus is 0.1 m.

This is because the shortest observable time period is not less than the action-potential time response of 1 mili second the length traveled by the sensation during this time is 1 m sec x 100 m / s =0.1 m, which is the shortest distance the electrodes could be positioned on the nerve.

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Select the correct answer.
r-ruslan [8.4K]

Answer:

That would be B. Hope this helps!

Explanation:

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3 0
3 years ago
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Reactions that release energy into the environment in the form of heat
meriva
DEFINITION:::::;;The type of reactions in which energy is releases to the environment are called Exothermic reactions.

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Hope it helps
6 0
2 years ago
The speed of a car during a trip is represented in the graph. Compare distance and displacement at four hours into the trip.
mezya [45]

Answer:

The answer to your question is: D.

Explanation:

Distance refers to the amount of space between two points, it is a scalar quantity.

Displacement refers to the space between two points, measure from the minimum path linking them, it is a vector quantity.

I'm not agree with these answers, because the total distance  is approximately 500km.

A) The distance traveled is 300 km.  This answer is not correct.

B) Distance is 300 km and displacement is 0 km.  This answer is not correct because the displacement is also 500 km.

C) Distance is 300 km/hour and displacement is 300 km. 300 km/h is a measure of speed not of distance, this option is wrong.

D) Both distance traveled and displacement are 300 km. I think this is the correct answer because distance and displacement measure the same. but I think both measure 500 km.

7 0
3 years ago
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A ball is thrown vertically upward, which is the positive direction. A little later, it returns to its point of release. The bal
Aleks [24]

Answer:

The initial velocity of the ball is <u>39.2 m/s in the upward direction.</u>

Explanation:

Given:

Upward direction is positive. So, downward direction is negative.

Tota time the ball remains in air (t) = 8.0 s

Net displacement of the ball (S) = Final position - Initial position = 0 m

Acceleration of the ball is due to gravity. So, a=g=-9.8\ m/s^2(Acting down)

Now, let the initial velocity be 'u' m/s.

From Newton's equation of motion, we have:

S=ut+\frac{1}{2}at^2

Plug in the given values and solve for 'u'. This gives,

0=8u-0.5\times 9.8\times 8^2\\\\8u=4.9\times 64\\\\u=\frac{4.9\times 64}{8}\\\\u=4.9\times 8=39.2\ m/s

Therefore, the initial velocity of the ball is 39.2 m/s in the upward direction.

3 0
3 years ago
The volume of an ideal gas is adiabatically reduced from 151 L to 80.6 L. The initial pressure and temperature are 1.50 atm and
Zolol [24]

Answer:

gas is dioatomic

T_f = 330.0 K

\eta = 7.07 mole

Explanation:

Part 1

below equation is used to determine the type Gas by determining \gamma value

\frac{V_{1}}{V_{F}}\gamma=\frac{P_{i}}{P_{f}}

where V_i and V_f is initial and final volume respectively

and P_i and P_f are initial and final pressure

\gamma = \frac{ln(P_f/P_i)}{ln(V_i/V_f)}

\gamma = \frac{ln(3.61/1.50)}{ln(151/80.6}

\gamma = 1.38

therefore gas is dioatomic

Part 2

final temperature in adiabatic process is given as

T_f = T_i*[\frac{v_i}{V_f}](^\gamma-1)

substituing value to get final temperature

T_f = 260*[\frac{151}{80.6}]^ {(1.38-1)}

T_f = 330.0 K

Part 3

determine number of moles by using following formula

\eta =\frac{PV}{RT}

\eta =\frac{1.013*10^{5}*0.151}{8.314*260}

\eta = 7.07 mole

4 0
3 years ago
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