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ira [324]
2 years ago
9

Conduction of a nerve impulse would be the fastest in _________

Physics
1 answer:
Vinil7 [7]2 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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The magnetic field around the head has been measured to be approximately 3.00×10−8 gauss . Although the currents that cause this
konstantin123 [22]

Answer:

3.81972\times 10^{-7}\ A

Explanation:

B = Magnetic field = 3\times 10^{-8}\ G

d = Diameter of loop = 16 cm

r = Radius = \frac{d}{2}=\frac{16}{2}=8\ cm

i = Current

\mu_0 = Vacuum permeability = 4\pi \times 10^{-7}\ H/m

The magnetic field of a loop is given by

B=\frac{\mu_0i}{2r}\\\Rightarrow i=\frac{B2r}{\mu_0}\\\Rightarrow i=\frac{3\times 10^{-8}\times 10^{-4}\times 2\times 0.08}{4\pi\times 10^{-7}}\\\Rightarrow i=3.81972\times 10^{-7}\ A

The current needed to produce such a field at the center of the loop is 3.81972\times 10^{-7}\ A

5 0
2 years ago
Convert the volume 8.06 in.3 to m3, recalling that1in. =2.54cmand100cm=1m. Answer in units of m3.
galina1969 [7]
1 in=2.54 cm=(2.54 cm)(1 m/100 cm)=0.0254 m
Therefore:
1 in=0.0254 m
1 in³=(0.0254 m)³=1.6387064 x 10⁻⁵ m³

Therefore:

8.06 in³=(8.06 in³)(1.6387064 x 10⁻⁵ m³ / 1 in³)≈1.321 x 10⁻⁴ m³.

Answer: 8.06 in³=1.321 x 10⁻⁴ m³
8 0
2 years ago
3. If you are playing seesaw with your younger sister who weighs
defon

Answer:

move the point of the fulcrum

Explanation:

you can do this by moving yourself closer to the pivot point, or moving the sea saw so that your sister has more of the board on her side

8 0
2 years ago
An organ pipe is made to play a low note at 27.5 Hz, the same as the lowest note on a piano. Assuming a sound speed of 343 m/s,
timama [110]

Answer:

The length of open-open pipe needed is 6.23 m

The length of open-close  pipe needed is 3.11 m

Explanation:

Fundamental frequency for standing wave mode of  an open- open pipe is given by

f=\frac{v}{2L}

where v is the velocity and L is the length

The length of open-open pipe needed is

L=\frac{v}{2f} \\L=\frac{343}{2\times 27.5} \\L=6.23 m

Fundamental frequency for standing wave mode of  an open- close pipe is given by

f=\frac{v}{2L}

The length of open-close pipe needed is

L=\frac{v}{2f} \\L=\frac{343}{2\times 27.5} \\L=6.23 m

7 0
2 years ago
Two objects (45.0 and 21.0 kg) are connected by a massless string that passes over a massless, frictionless pulley. The pulley h
Oksanka [162]

a) The acceleration of the objects is 3.56 m/s^2

b) The tension in the string is 280.8 N

Explanation:

a)

We start by writing the equations of motion for the two masses attached to the pulley.

For the heavier mass, we have:

m_1 g - T = m_1 a (1)

where

m_1 = 45.0 kg is the mass

g=9.8 m/s^2 is the acceleration of gravity

T is the tension in the string

a is the acceleration of the system (here we assumed that the heavier mass accelerates downward)

For the lighter mass, we have

T-m_2 g = m_2 a (2)

where

T is the tension in the string

m_2 = 21.0 kg is the mass

g=9.8 m/s^2 is the acceleration of gravity

a is the acceleration of the system (here we assumed that the lighter mass accelerates upward)

From (1) we get

T=m_1g - m_1 a

And substituting into (2),

(m_1 g - m_1 a)-m_2 g = m_2 a\\(m_1 -m_2)g  = (m_1+m_2)a\\a=\frac{m_1 - m_2}{m_1+m_2}g=\frac{45-21}{45+21}(9.8)=3.56 m/s^2

b)

From the previous part of the problem we got an expression for the tension in the string:

T=m_1g - m_1 a

Where we have

m_1 = 45.0 kg

g=9.8 m/s^2

a=3.56 m/s^2 is the acceleration, found in part a)

Susbtituting, we find

T=(45.0)(9.8)-(45.0)(3.56)=280.8 N

Learn more about forces and acceleration:

brainly.com/question/11411375

brainly.com/question/1971321

brainly.com/question/2286502

brainly.com/question/2562700

#LearnwithBrainly

4 0
2 years ago
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