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2 years ago

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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A friend throws a heavy ball toward you while you are standing on smooth ice. You can either catch the ball or deflect it back t

baherus [9]

Answer:

Explanation:

My speed after the interaction will depend upon the impulse the ball will make on me . Now impulse can be expressed as follows

Impulse = change in momentum

change in momentum in the ball will be maximum when the ball bounces back with the same velocity which can be shown as follows

change in momentum = mv - ( - mv ) = 2mv

So when ball is bounced back with same velocity , it suffers greatest impulse from my hand . In return , it reacts with the same impulse on my hand pushing me with greatest impulse according to third law of motion. this maximizes my speed after the interaction.

6 0

2 years ago

There are two different size spherical paintballs and the smaller one has a diameter of 5 cm and the larger one is 9 cm in diame

slavikrds [6]

Answer:

145.8 cm³ of paint

Explanation:

d₁ = Smaller diameter paintball = 5 cm

d₂ = Larger diameter paintball = 9 cm

V₂ = Volume of larger diameter paintball

Volume of smaller diameter paintball

$V_1=\frac{4}{3}\pi r_1^3\\\Rightarrow V_1=\frac{4}{3}\pi \left(\frac{d_1}{2}\right)^3\\\Rightarrow V_1=\frac{4}{24}\pi d_1^3$

Similarly

$V_2=\frac{4}{24}\pi d_2^3$

Dividing the above two equations, we get

$\frac{V_1}{V_2}=\frac{d_1^3}{d_2^3}\\\Rightarrow V_2=\frac{V_1}{\frac{d_1^3}{d_2^3}}\\\Rightarrow V_2=\frac{28}{\frac{125}{729}}\\\Rightarrow V_2=163.296\ cm^3$

∴ The larger one hold 163.296 cm³ of paint

5 0

2 years ago

A sample of an unknown volatile liquid was injected into a Dumas flask (mflask = 27.0928 g, Vflask = 0.1040 L) and heated until

NNADVOKAT [17]

Answer:

The gas was Hexane

Explanation:

taking the diference between the mass of the flask and the final mass qe can calculate the mass of liquid injected (assuming none escaped the flask):

$m_{l} = 27.4593g - 27.0928g = 0.3665g$

with the volume of the flask we can get the density of the gas at the indicated pressure and temperature:

$d_{g} = \frac{0.3665 g}{0.1040L} = 3.524 g/L$

From the ideal gases law we have that the density can be calculated as:

$d_{g} = \frac{P*M}{R*T}$

Where R is the ideal gases constant = , and M the molecular weight of the fluid. Solving for M:

$M=\frac{d_{g}*R*T}{P}=\frac{3.524g/L*0,082atmL/molK*291K}{0.976atm}$

$M=86.16 g/mol$

Note that the temperature is computed in Kelvin T= 18+273=291K

The gas with the closer molar mass is Hexane

4 0

2 years ago

An 80-g particle moving with an initial speed of 50 m/s in the positive x direction strikes and sticks to a 60-g particle moving

liubo4ka [24]

The collision is a form of inelastic collision because the it forms a single mass after is collides. So it can be solve by momentum balance

( 0.08 kg * 50 m/s ) + ( 0.06 kg * 50 m/s) = ( 0.08 + 0.06 kg ) v

V = 50 m/s

So the kinetic energy lost is

KE = 0.5 (50 m/s)^2) *( 0.14 – 0.08kg )

KE = 75 J

8 0

2 years ago

La luz pasa del medio A al medio B formando un ángulo de 35° con la frontera horizontal entre ambos. Si el ángulo de refracción

zaharov [31]

Answer:

Índice de refracción entre los dos medios = 1,43

Refractive index between the two media = 1.43

Explanation:

El índice de refracción entre dos medios se explica mejor entendiendo primero la refracción.

Cuando las olas se mueven de un medio a otro, a menudo experimentan un cambio de dirección con respecto al medio en el que viajan.

Por lo tanto, el índice de refracción se expresa como el seno del ángulo de incidencia dividido por el seno del ángulo de refracción.

El seno del ángulo de incidencia y la refracción utilizados en esta fórmula de índice de refracción se miden respectivamente con respecto a la vertical.

En esta pregunta Ángulo de incidencia = 35° a la horizontal = (90° - 35°) a la vertical = 55° a la vertical.

Ángulo de refracción = 35°

Índice de refracción entre los dos medios

= (Sin 55°) ÷ (Sin 35°)

= 0.8192 ÷ 0.5736

= 1.428 = 1.43 a 2 d.p.

¡¡¡Espero que esto ayude!!!

English Translation

The light passes from medium A to medium B at an angle of 35 ° with the horizontal border between the two. If the angle of refraction is also 35 °, what is the relative refractive index between the two media?

Solution

The refractive index between two media is best explained by first understanding refraction.

When waves move from one medium to another, they often experience a change in direction with respect to the medium in which they are travelling.

Hence, refractive index is expressed as the sine of angle of incidence dibided by the sine of angle of refraction.

The sine of angle of incidence and refraction used in this refractive index formula are both respectively measured with respect to the vertical.

In this question,

Angle of incidence = 35° to the horizontal = (90° - 35°) to the vertical = 55° to the vertical.

Angle of refraction = 35°

Refractive index between the two media

= (Sin 55°) ÷ (Sin 35°)

= 0.8192 ÷ 0.5736

= 1.428 = 1.43 to 2 d.p.

Hope this Helps!!!

3 0

2 years ago