An estuary. <span>Estuaries are partially enclosed bodies of water along coastlines where fresh water and salt water meet and mix. They act as a transition zone between oceans and continents.</span>
Taste bud receptors if not stimulated are normally polarized. With the presence of the food, chemicals from the food bind with the cell protein which depolarized the cell membrane receptor. Depolarization will lead into the propagation of electric impulse. This will lead into the creation of sensory impulse. The larger number of the chemicals in the food will lead into a stronger impulse.
<h3>Three
tools Oceanographers use</h3>
<h3>
Functions of the
tools.</h3>
- Thermometer helps to measure the temperature of the ocean. Temperature usually varies according to the season.
- Radar helps to measure ocean currents and and it works under any type of weather condition.
- Sonar is an instrument used to accurately measure the ocean depths through the use of sound waves.
Read more about Oceanographers here brainly.com/question/977380
Answer:
Órgano que forma parte del aparato digestivo. El estómago ayuda a digerir los alimentos al mezclarlos con jugos digestivos convirtiéndolos en líquido diluido.
Given question is incomplete. Complete question has been attached.
Answer:
C. Excessive potassium has diffused out causing hyperpolarization.
Explanation:
The nerve action potential can be divided into following stages:
- Stimulus is detected by the cell in resting stage.
- Sodium channels in the membrane open from where influx of sodium ions occur which is called depolarization
- After a while, sodium channels close and potassium channels open from where efflux of potassium ions occur which is called repolarization.
- The membrane potential further lowers due to continous efflux of potassium ions which is called hyperpolarization.
- After a while potassium channels close and membrane returns to its resting stage.
In the given figure, stage 4 depicts hyperpolarization because the membrane potential has dropped to the lowest point below -70mV. Hence, option C is correct.
