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Rasek [7]
3 years ago
10

A nerve signal is transmitted through a neuron when an excess of Na+ ions suddenly enters the axon, a long cylindrical part of t

he neuron. Axons are approximately 10.0 μm in diameter, and measurements show that about 5.60×1011 Na+ions per meter (each of charge +e) enter during this process. Although the axon is a long cylinder, the charge does not all enter everywhere at the same time. A plausible model would be a series of nearly point charges moving along the axon. Let us look at a 0.100 mm length of the axon and model it as a point charge.1) What electric field (magnitude and direction) does the sudden influx of charge produce at the surface of the body if the axon is 5.00 cm below the skin? Note that distance to the skin is much greater than the length of the charged axon so that the axon charge is effectiely point-like.,E = (INSERT ANSWER) N/C2) If the charge that enters each meter of the axon gets distributed uniformly along it, how many coulombs of charge enter a 0.100 mm length of the axon? Answer in C3) the field is directed away from the axon or the field is directed into the axon
Physics
1 answer:
ivolga24 [154]3 years ago
3 0

Answer:

1.32.225 N/C, direction is away from the point charge

2. 8.972*10^-12 C

3. the field is directed away from the axon

Explanation:

The electric field can be calculated as shown below:

E = k*|q|/r^2

Where:

E = electric field; k = 8.98755*10^9 N*m^2/C^2; r = distance between the measured field and point charge = 0.05 m; q = the point charge

For 0.100 m of the axon, the value of q is:

q = (5.6*10^11)*(+e)*(0.001)

+e = charge of an electron = 1.60217*10^-19 C

Thus:

q = (5.6*10^11)*(1.60217*10^-19)*(0.0001) = 8.972*10^-12 C

Therefore:

E = (8.98755*10^9)*(8.972*10^-12)/0.05^2 = 32.255 N/C

A positive point charge always produce an electric field that is directed away from the field while a negative point charge produces an electric field that is directed toward the field

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The water in a river flows uniformly at a constant speed of 2.50 m/s between parallel banks 80.0 m apart. You are to deliver a p
NISA [10]

Answer:

a)  The swimmer should travel perpendicular to the bank to minimize the spent in getting to the other side.

b) 133.33 m

c) 53.13°

d) 106.67 m

Explanation:

a) The swimmer should travel perpendicular to the bank to minimize the spent in getting to the other side.

b) velocity = distance * time

Let the velocity of the swimmer be v_{s} = 1.5 m/s

The separation of the two sides of the river, d = 80 m

The time taken by the swimmer to get to the other end of the river bank,

t = \frac{d}{v_{s} }

t = 80/1.5

t = 53.33 s

The swimmer will be carried downstream by the river through a distance, s

Let the velocity of the river be v_{r} = 2.5 m/s

S = v_{r} t

S = 53.33 * 2.5

S = 133.33 m

c) To minimize the distance traveled by the swimmer, his resultant velocity must be perpendicular to the velocity of the swimmer relative to water

That is ,

cos \theta = \frac{v_{s} }{v_{r} } \\cos \theta = 1.5/2.5\\cos \theta = 0.6\\\theta = cos^{-1} 0.6\\\theta = 53.13^{0}

d) Downstream velocity of the swimmer, v_{y} = v_{s} sin \theta\\

v_{y} = 1.5 sin 53.13\\v_{y} = 1.2 m/s

The vertical displacement is given by, y = v_{y} t

80 = 1.2 t

t = 80/1.2

t = 66.67 s

the horizontal speed,

v_{x} = 2.5 - 1.5cos53.13\\v_{x} = 1.6 m/s

The downstream horizontal distance of the swimmer, x = v_{x} t

x = 1.6 * 66.67

x = 106.67 m

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3 years ago
A person on a merry go round is constantly changing direction
Salsk061 [2.6K]

Answer: yurp

Explanation:

because its spinning

7 0
3 years ago
Read 2 more answers
Two steel balls, of masses m1=1.00 kg and m2=2.00 kg, respectively, are hung from the ceiling with light strings next to each ot
Zolol [24]

Answer:

(a) The maximum height achieved by the first ball, m₁ is 0.11 m

(a) The maximum height achieved by the second ball, m₂ ball is 0.44 m

Explanation:

Given;

mass of the first ball, m₁ = 1 kg

mass of the second ball, m₂ = 2 kg

The velocity of the first when released from a height of 1 m before collision;

u₁² = u₀² + 2gh

u₀ = 0, since it was released from rest

u₁² =  2gh

u₁² = 2 x 9.8 x 1

u₁² = 19.6

u₁ = √19.6

u₁ = 4.427 m/s

The velocity of the second ball before collision, u₂ = 0

Apply the principle of conservation of linear momentum, to determine the velocity of the balls after an elastic collision.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

where;

v₁ is the final velocity of the first ball after an elastic collision

v₂ is the final velocity of the second ball after an elastic collision

m₁u₁ + m₂(0) = m₁v₁ + m₂v₂

m₁u₁ =  m₁v₁ + m₂v₂

1 x 4.427 = v₁ + 2v₂

v₁ + 2v₂ = 4.427

v₁  = 4.427 - 2v₂  ----- equation (1)

one directional velocity;

u₁ + v₁ = u₂ + v₂

u₂ = 0

u₁ + v₁ = v₂

v₁ = v₂ - u₁

v₁ = v₂ - 4.427 ------ equation (2)

Substitute v₁ into equation (1)

v₂ - 4.427 = 4.427 - 2v₂

3v₂ = 4.427 + 4.427

3v₂  = 8.854

v₂ = 8.854 / 3

v₂  = 2.95 m/s (→ forward direction)

v₁ = v₂ - 4.427

v₁ = 2.95 - 4.427

v₁  = - 1.477 m/s

v₁  = 1.477 m/s ( ← backward direction)

Apply the law of conservation of mechanical energy

mgh_{max} = \frac{1}{2}mv_{max}^2

(a) The maximum height achieved by the first ball (v₁  = 1.477 m/s)

mgh_{max} = \frac{1}{2}mv_{max}^2 \\\\gh_{max} = \frac{1}{2}v_{max}^2\\\\ h_{max}  =  \frac{1}{2g}v_{max}^2\\\\ h_{max}  = \frac{1}{2*9.8}(1.477^2)\\\\ h_{max}  = 0.11 \ m

(b) The maximum height achieved by the second ball (v₂  = 2.95 m/s)

mgh_{max} = \frac{1}{2}mv_{max}^2 \\\\gh_{max} = \frac{1}{2}v_{max}^2\\\\ h_{max}  =  \frac{1}{2g}v_{max}^2\\\\ h_{max}  = \frac{1}{2*9.8}(2.95^2)\\\\ h_{max}  = 0.44 \ m

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3 years ago
Para cargar un camión, se suele utilizar una tabla entre el contenedor y el suelo, con el fin de subir la carga, ya sea desplaza
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Answer:

A. No

B. si

Explanation:

A. El trabajo realizado en la carga es la energía potencial ganada por la carga al elevar la carga al nivel del camión y colocar la carga dentro del camión.

El trabajo realizado para elevar la carga W = m × g × h

Dónde;

m = masa de la carga

g = aceleración debido a la gravedad

h = Nivel de altura donde se coloca la carga en el camión

Por lo tanto, el trabajo realizado depende de la masa, m, de la carga y el nivel de altura, h, donde la carga se coloca en el camión y el trabajo realizado es el mismo para todos los métodos utilizados para colocar la carga en el camión

B. La ecuación para el trabajo realizado, W, también se puede escribir de la siguiente manera;

W = Fuerza, F × Distancia, D

De lo que tenemos;

F = W/D

Por lo tanto, cuando la mesa aumenta la distancia, como una rampa o un plano inclinado, la fuerza requerida disminuirá.

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