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3 years ago

Conduction of a nerve impulse would be the fastest in _________

Physics

1 answer:

Vinil7 [7]3 years ago

8 0

The options of the given are:

A. A large diameter myelinated fiber

B. A small diameter myelinated fiber

C. A large unmyelinated fiber

D. A small unmyelinated fiber

E. A small fiber with multiple Schwann cells

Answer: Option A, A large diameter myelinated fiber.

Explanation:

The conduction of the nerve impulse would be greatest in the myelinated fiber because the main function of the myelin sheath is to increase the speed of the impulse at which the electrical signals propagate.

In case of the unmyelinated sheath the nerve impulse travels slowly as the conduction waves but in case of the large diameter myelinated sheath the signals travel via saltatory conduction( hop)

In this type of propagation the signals are transferred from the node of Ranvier in one neuron to next node which increases the overall velocity of the action potentials.

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omeli [17]

Answer:

objects with greater mass should have greater inertia

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3 years ago

The loudness of a sound is determined by the __________, or height, of the sound wave.

NISA [10]

Answer: Aplitude

Explanation: Loudness depends on the amplitude,or height, of sound waves. The greater the amplitude, the louder the sound perceived. Amplitude is measured in decibels.

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3 years ago

In an electrical circuit, a voltage difference:

liberstina [14]

A. supplies energy to make charge flow

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4 years ago

At what altitude h above the north pole is the weight of an object reduced to 67% of its earth-surface value? Assume a spherical

lara31 [8.8K]

Answer:

The answer to the question is

At an altitude of 1413 km or 1.222·R above the north pole the weight of an object reduced to 67% of its earth-surface value

Explanation:

We make use of the gravitational formula as follows

F = G $\frac{m_{1} m_{2} }{R^{2} }$ where

m₁ = mass of the object

m₂ = mass of the earth

d = distance between the two objects and

G = gravitational constant

if at the altitude the weight is reduced to 67 % of its weight on earth then

with all other variables remaining constant, we have

67% F = G $\frac{m_{1} m_{2} }{R_{2} ^{2} }$ =0.67× G× $\frac{m_{1} m_{2} }{R_{1} ^{2} }$

cancelleing like ternss from both sides we have

1/R₂² =0.67/R₁² or r₁²/R₂² =0.67 and R₁/R₂ = 0.8185

or R₂ = R₁/0.8185 or 1.222·R = (6.37×10⁶ m)/0.8185 = 7.783×10⁶ m

Hence at an altitude = 1.413×10⁶ m the weight of the object will reduce to 67% its earth-surface value

6 0

3 years ago

Tom used 75 N of force to push a box 3 meters across the floor. How much work did he do?

yuradex [85]

Answer:

It must be 225 J.

Explanation:

According to

Work done = Force × distance travelled

= 75 × 3

= 225 J

This answer is taken so since no angles are stated in the question.

3 0

3 years ago