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Rasek [7]
1 year 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]1 year 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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A system of two paint buckets connected by a lightweight rope is released from rest with the 12.0-kg bucket 2.00 m above the flo
NISA [10]

Explanation:

The given data is as follows.

    Mass of small bucket (m) = 4 kg

    Mass of big bucket (M) = 12 kg

    Initial velocity (v_{o}) = 0 m/s

    Final velocity (v_{f}) = ?

  Height H_{o} = h_{f} = 2 m

and,    H_{f} = h_{o} = 0 m

Now, according to the law of conservation of energy

         starting conditions = final conditions

  \frac{1}{2}MV^{2}_{o} + Mgh_{o} + \frac{1}{2}mv^{2}_{o} + mgh_{o} = \frac{1}{2}MV^{2}_{f} + Mgh_{f} + \frac{1}{2}mv^{2}_{f} + mgh_{f}

     \frac{1}{2}(12)(0)^{2} + (12)(9.81)(2) + \frac{1}{2}(4)(0)^{2} + (4)(9.81)(0) = \frac{1}{2}(12)V^{2}_{f} + (12)(9.81)(0) + \frac{1}{2}(4)V^{2}_{f} + (4)(9.81)(2)

                 235.44 = 8V^{2}_{f} + 78.48

                V_{f} = 4.43 m/s

Thus, we can conclude that the speed with which this bucket strikes the floor is 4.43 m/s.

3 0
1 year ago
Waves hitting at an angle and then bending around features of the coast is known as
Pavel [41]
<span>Waves hitting at an angle and then bending around features of the coast is known as Wave refraction
When waves hitting a specific angle, some part of the waves will be closer to the shallow part of the water and some part will be closer to the deeper part of the water, which makes the wave became somehow bent around the shore.</span>
6 0
2 years ago
The value of specific heat for copper is 390 J/kg⋅C∘, for aluminun is 900 J/kg⋅C∘, and for water is 4186 J/kg⋅C∘.
abruzzese [7]

Answer:

The equilibrium temperature is

21.97°c

Explanation:

This problem bothers on the heat capacity of materials

Given data

specific heat capacities

copper is Cc =390 J/kg⋅C∘,

aluminun Ca = 900 J/kg⋅C∘,

water Cw = 4186 J/kg⋅C∘.

Mass of substances

Copper Mc = 235g

Aluminum Ma = 135g

Water Mw = 825g

Temperatures

Copper θc = 255°c

Water and aluminum calorimeter θ1= 16°c

Equilibrium temperature θf =?

Applying the principle of conservation of heat energy, heat loss by copper equal heat gained by aluminum calorimeter and water

McCc(θc-θf) =(MaCa+MwCw)(θf-θ1)

Substituting our data into the expression we have

235*390(255-θf)=

(135*900+825*4186)(θf-16)

91650(255-θf)=(3574950)(θf-16)

23.37*10^6-91650*θf=3.57*10^6θf- +57.2*10^6

Collecting like terms and rearranging

23.37*10^6+57.2*10^6=3.57*10^6θf+91650θf

8.2*10^6=3.66*10^6θf

θf=80.5*10^6/3.6*10^6

θf =21.97°c

5 0
1 year ago
A 0.0140 kg bullet traveling at 205 m/s east hits a motionless 1.80 kg block and bounces off it, retracing its original path wit
makvit [3.9K]

Answer:

Final velocity of the block = 2.40 m/s east.

Explanation:

Here momentum is conserved.

Initial momentum = Final momentum

Mass of bullet = 0.0140 kg

Consider east as positive.

Initial velocity of bullet = 205 m/s

Mass of Block = 1.8 kg

Initial velocity of block = 0 m/s

Initial momentum = 0.014 x 205 + 1.8 x 0 = 2.87 kg m/s

Final velocity of bullet = -103 m/s

We need to find final velocity of the block( u )

Final momentum = 0.014 x -103+ 1.8 x u = -1.442 + 1.8 u

We have

            2.87 = -1.442 + 1.8 u

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Final velocity of the block = 2.40 m/s east.

7 0
2 years ago
A container of volume 0.6 m^3 contains 5.3 mol of argon gas at 24°C. Assuming argon behaves as an ideal gas, find the total inte
Vitek1552 [10]

Answer:

the internal energy of the gas is 433089.52 J

Explanation:

let n be the number of moles, R be the gas constant and T be the temperature in Kelvins.

the internal energy of an ideal gas is given by:

Ein = 3/2×n×R×T

     = 3/2×(5.3)×(8.31451)×(24 + 273)

     = 433089.52 J

Therefore, the internal energy of this gas is 433089.52 J.

5 0
1 year ago
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