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lana66690 [7]
3 years ago
14

The difference in charge across the membrane is called the . It is measured in . At resting potential, the charge of the outside

of the neuron is and the inside is . Once a stimulus changes the membrane potential, the neuron conducts an impulse called a(n) . This causes the charge outside the cell to temporarily become and the inside to temporarily become . When the action potential arrives at the axon terminal it reaches a junction called the Because the nerve impulse cannot cross the synaptic cleft,
Biology
2 answers:
trasher [3.6K]3 years ago
5 0

Answer:

1. The difference in charge across the membrane is called the - membrane potential.

2. It is measured in mV

3. At resting potential, the charge of the outside of the neuron is 70 mV less negative than the inside

4. Once a stimulus changes the membrane potential, the neuron conducts an impulse called a(n) Action Potential

5. This causes the charge outside the cell to temporarily become positive and the inside to temporarily become negative

6. When the action potential arrives at the axon terminal it reaches a junction called the Chemical Synapse

Explanation:

1. The difference in charge across the membrane is called the - membrane potential.

2. It is measured in mV

3. At resting potential, the charge of the outside of the neuron is 70 mV less negative than the inside

4. Once a stimulus changes the membrane potential, the neuron conducts an impulse called a(n) Action Potential

5. This causes the charge outside the cell to temporarily become positive and the inside to temporarily become negative

6. When the action potential arrives at the axon terminal it reaches a junction called the Chemical Synapse

UkoKoshka [18]3 years ago
3 0

Answer:

1. The difference in charge across the membrane is called the - membrane potential.

2. It is measured in mV.

3. At resting potential, the charge of the outside of the neuron is 70 mV less negative than the inside.

4. Once a stimulus changes the membrane potential, the neuron conducts an impulse called a(n) Action Potential.

5. This causes the charge outside the cell to temporarily become positive and the inside to temporarily become negative.

6. When the action potential arrives at the axon terminal it reaches a junction called the Chemical Synapse.

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3 years ago
A fossil is discovered that has only 12.5% of the carbon-14 that it would have had originally (when the animal was alive). If th
mariarad [96]

Answer:

The fossil is 17,100 years old.

Explanation:

The decay equation:

\frac{dN}{dt}\propto -N

\Rightarrow \frac{dN}{dt}= -\lambda N

\Rightarrow \frac{dN}{N}= -\lambda dt

Integrating both sides

\Rightarrow \int\frac{dN}{N}= \int-\lambda dt

\Rightarrow ln |N|=-\lambda t+c

When t=0, N=N_0 = initial amount

ln |N_0|=-\lambda .0+c

\Rightarrow ln |N_0|=c

\therefore ln |N|=-\lambda t+ln|N_0|

\Rightarrow ln |N|-ln|N_0|=-\lambda t

\Rightarrow ln |\frac {N}{N_0}|=-\lambda t

\Rightarrow \frac {N}{N_0}=e^{-\lambda t}

\Rightarrow N=N_0e^{-\lambda t}

The decay equation is

N=N_0e^{-\lambda t}

Given that,

The half life of carbon - 14 is 5700 years.

For half life, N=\frac{1}{2} N_0

To find the value of  \lambda, we need to put the value of N and t in the decay equation.

\frac12N_0=N_0e^{-\lambda \times 5700}

\Rightarrow \frac12=e^{-\lambda \times 5700}             [ Divided N_0 both sides]

Taking ln both sides

\Rightarrow ln| \frac12|=ln|e^{-\lambda \times 5700}|

\Rightarrow ln| \frac12|={-\lambda \times 5700}

\Rightarrow \lambda= \frac{ln| \frac12|}{-5700}

\Rightarrow \lambda= \frac{ln|1|-ln|2|}{-5700}             [ ln|\frac mn|= ln |m|-ln |n|]

\Rightarrow \lambda= \frac{ln|2|}{5700}                      [ln 1= 0]

The fossil has only 12.5% of the carbon carbon-14 that it would have had originally.

So, N=\frac{12.5}{100} N_0

Then,

\frac{12.5}{100} N_0=N_0e^{-\frac{ln|2|}{5700}t

\Rightarrow \frac{12.5}{100} =e^{-\frac{ln|2|}{5700}t

Taking ln both sides

\Rightarrow ln|\frac{12.5}{100} |=ln|e^{-\frac{ln|2|}{5700}t}|

\Rightarrow ln|\frac{12.5}{100} |={-\frac{ln|2|}{5700}t}

\Rightarrow t=\frac{ ln|\frac{12.5}{100}|} {-\frac{ln|2|}{5700}}

\Rightarrow t=\frac{ ln|\frac{12.5}{100}|\times 5700} {-{ln|2|}}

\Rightarrow t=17,100

The fossil is 17,100 years old.

7 0
4 years ago
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