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olganol [36]
3 years ago
14

Estimate the time it would take for such axons to carry a message from a foot stepping on a sharp object to the brain and then b

ack to the foot. This calculation gives a rough prediction of human reaction time.
Engineering
1 answer:
melisa1 [442]3 years ago
7 0

Answer:

<em>The Myelinated axons-fibers that are placed in an insulating myelin shealth.  </em>

<em>In myelinated axons, charges interchange through the membrane that happens to move between nodes By knowing how quickly the wall (membrane and sheath) of axon part between these bulge tins be electrically charged and discharged by changes in membrane potential at one node, we estimate the haste of tendon impulses along the axon. </em>

Explanation:

<em>The axons use an active direction 1 um (10 m) Myelinated courage axon Insulating myelin cover 10 um (10m) conducts  axoplasm 1 mm node differentiation Evenly spaced bulge of Ranvier of levy transfer across the membrana to create an action potential that continually amplifies and replenishes clectrical signals traveling along the axons</em>

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How does java achieve portable
sergejj [24]

Answer:

Java is called portable because you can compile a java code which will spew out a byte-code, and then you run that code with Java Virtual Machine. Java Virtual Machine is like an interpreter, which reads the compiled byte-code and runs it. So first of all, you need to install the JVM on the system you want.

Explanation:

5 0
3 years ago
What is the gear ratio of the given train
Olin [163]

Answer:

1/4

Explanation:

.......................

7 0
2 years ago
the tire restraining device or barrier shall be removed immediately from service for any of these defects except
lora16 [44]

Restraining devices and barriers shall be visually inspected on the rim wheel components or sudden release of contained air.

Restraining device means an apparatus such as a <em>cage, rack, assemblage of bars and other components</em> that will constrain all rim wheel components.

Restraining devices and barriers shall be visually inspected on the rim wheel components or sudden release of contained air. Any restraining device or barrier exhibiting damage such as the following defects shall be immediately removed from service.

Find out more on Restraining devices at: brainly.com/question/24647450

5 0
2 years ago
The Energy Losses Associated with Valves and Fittings: a)- are generally associated with a K factor b)- are generally associated
madam [21]

Answer:

a)Are generally associated with factor.

Explanation:

We know that losses are two types

1.Major loss  :Due to friction of pipe surface

2.Minor loss  :Due to change in the direction of flow

As we know that when any hindrance is produced during the flow of fluid then it leads to generate the energy losses.If flow is along uniform diameter pipe then there will not be any loss but if any valve and fitting placed is the path of fluid flow due to this direction of fluid flow changes and  it produce losses in the energy.

Lot' of experimental data tell us that loss in the energy due to valve and fitting are generally associated with K factor.These losses are given as

Losses=K\dfrac{V^2}{2g}

8 0
3 years ago
Steam at 20 bars is in the saturated vapor state (call this state 1) and contained in a pistoncylinderdevice with a volume of 0.
saul85 [17]

Answer:

Explanation:

Given that:

<u>At state 1:</u>

Pressure P₁ = 20 bar

Volume V₁ = 0.03 \mathbf{m^{3}}

From the tables at saturated vapour;

Temperature T₁ = 212.4⁰ C  ; v_1 = vg_1 = 0.0996 \mathbf{m^{3}} / kg

The mass inside the cylinder is m = 0.3 kg, which is constant.

The specific internal energy u₁ = ug₁ = 2599.2 kJ/kg

<u>At state 2:</u>

Temperature T₂ = 200⁰ C

Since the 1 - 2 occurs in an isochoric process v₂ = v₁ = 0.099 \mathbf{m^{3}} / kg

From temperature T₂ = 200⁰ C

v_f_2 = 0.0016 \ m^3/kg  

vg_2 = 0.127 \ m^3/kg  

Since  vf_2 < v_2 , the saturated pressure at state 2 i.e. P₂ = 15.5 bar

Mixture quality x_2 = \dfrac{v_2-vf_2}{vg_2 -vf_2}

x_2 = \dfrac{(0.099-0.0016)m^3/kg}{(0.127 -0.0016) m^3/kg}

x_2 = \dfrac{(0.0974)m^3/kg}{(0.1254) m^3/kg}

\mathsf{x_2 =0.78}

At temperature T₂, the specific internal energy u_f_2 = 850.6 \ kJ/kg , also ug_2 = 2594.3 \ kJ/kg

Thus,

u_2 = uf_2 + x_2 (ug_2 -uf_2)

u_2 =850.6  +0.78 (2594.3 -850.6)

u_2 =850.6  +1360.086

u_2 =2210.686 \ kJ/kg

<u>At state 3:</u>

Temperature T_3=T_2 = 200 ^0 C ,

V_3 = 2V_1 = 0.06 \ m^3

Specific volume v_3 = 0.2  \ m^3/kg

Thus; vg_3 =vg_2 = 0.127 \ m^3/kg ,

SInce v_3 > vg_3, therefore, the phase is in a superheated vapour state.

From the tables of superheated vapour tables; at v_3 = 0.2  \ m^3/kg and T₃ = 200⁰ C

The pressure = 10 bar and v =0.206 \ m^3/kg

The specific internal energy u_3 at the pressure of 10 bar = 2622.3 kJ/kg

The changes in the specific internal energy is:

u_2-u_1

= (2210.686 - 2599.2) kJ/kg

= -388.514 kJ/kg

≅ - 389 kJ/kg

u_3-u_2

= (2622.3 - 2210.686)  kJ/kg

= 411.614 kJ/kg

≅ 410 kJ/kg  

We can see the correct sketches of the T-v plot showing the diagrammatic expression in the image attached below.

3 0
3 years ago
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