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§Labor and management

§Productivity

§Wage levels

§Training needs

§Local infrastructure

Measuring site potential in step 3 involves determining whether a site can supply adequate resources needed to carry out the proposed business activity. Key issues include:•For many companies the most important resources will be labor and management.

•The productivity and wage levels of local labor and managers.

•The cost of training local managers, which can mean substantial investments of time and money.

•And the efficiency of local infrastructures, including roads, bridges, airports, seaports, and telecommunications systems.

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12y ago
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12y ago

Well, assuming you are talking about an action potential and the threshold potential of a nerves cellular membrane, it depends on whether the ion channel in question is a sodium, potassium, or chloride channel.

In the case of a sodium channel, when the threshold potential of the cell membrane is reached, sodium channels open, causing an influx of sodium, Na, and an outflux of potassium, K. Assuming that the influx of Na occurs at a much faster rate than the outflux of K, then the cell will build up a negative charge, relative to outside the cell. As the charge begins to build, the voltage in the cell reaches -45 and triggers the "all or none" rising phase of an action potential and positive feedback[that as the charge nears the threshold, voltage-sensitive sodium channels open, further increasing the magnitude of the charge and the cells permeability to sodium].

The positive feedback of the rising phase slows and comes to a halt as the sodium ion channels become maximally open. At the peak of the action potential, the sodium permeability is maximized and the membrane voltage Vm is nearly equal to the sodium equilibrium voltage ENa. However, the same raised voltage that opened the sodium channels initially also slowly shuts them off, by closing their pores; the sodium channels become inactivated. This lowers the membrane's permeability to sodium, driving the membrane voltage back down. At the same time, the raised voltage opens voltage-sensitive potassium channels; the increase in the membrane's potassium permeability drives Vm towards EK. Combined, these changes in sodium and potassium permeability cause Vm to drop quickly, repolarizing the membrane and producing the "falling phase" of the action potential.

The raised voltage opened many more potassium channels than usual, and some of these do not close right away when the membrane returns to its normal resting voltage. In addition, further potassium channels open in response to the influx of calcium ions during the action potential. The potassium permeability of the membrane is transiently unusually high, driving the membrane voltage Vm even closer to the potassium equilibrium voltage EK. Hence, there is an undershoot or hyperpolarization that persists until the membrane potassium permeability returns to its usual value.

Finally, each action potential is followed by a refractory period, which can be divided into an absolute refractory period, during which it is impossible to evoke another action potential, and then a relative refractory period, during which a stronger-than-usual stimulus is required. These two refractory periods are caused by changes in the state of sodium and potassium channel molecules. When closing after an action potential, sodium channels enter an inactivated state in which they cannot be made to open regardless of the membrane potential-this gives rise to the absolute refractory period. Even after a sufficient number of sodium channels have transitioned back to their resting state, it frequently happens that a fraction of potassium channels remains open, making it difficult for the membrane potential to depolarize, and thereby giving rise to the relative refractory period. Because the density and subtypes of potassium channels may differ greatly between different types of neurons, the duration of the relative refractory period is highly variable.

Hope this answers your question, it is important to note that the opposite process will occur with cal

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