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Sladkaya [172]
2 years ago
7

A stone of mass 5g is projected with a rubber catapult. if the catapult is stretched through a distance of 7cm by an average for

ce of 70 North, calculate the instantaneous velocity of the stone when released ​
Physics
1 answer:
weqwewe [10]2 years ago
4 0

Explanation:

solution: mass m = 5g = 0.005kg; extension e = 7cm = 0.07m; force f = 70 N; velocity = ?; using: work done in elastic material w = 1/2 fe = 1/2 ke2 = 1/2 mv2 - the kinetic energy of the moving stone. 1/2 fe =...

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Nerve cells transmit electric signals through their long tubular axons. These signals propagate due to a sudden rush of Na+ ions
iren [92.7K]

Answer:

I = \frac{9.12x10^{-8} C}{0.007 s}= 1.30285 x10^{-5} \frac{C}{s}=1.30285 x10^{-5} A = 13.02 \mu A

Explanation:

For this case we have the following info given:

Number of Na+ ions 5.7 x10^{11} ions

Each ion have a charge of +e and the crage of the electron is 1.6 x10^{-19}C

The time is given t = 7 ms if we convert this into seconds we got:

t = 7ms * \frac{1s}{1000 ms}= 0.007s

Now we can use the following formula given from the current passing thourhg a meter of nerve axon given by:

Q = Ne

Where N represent the number of ions, e the charge of the electron and Q the total charge

If we replace on this case we have this:

Q= 5.7x10^{11} * (1.6 x10^{-19}C) = 9.12x10^{-8} C

And from the general definition of current we know that:

I =\frac{Q}{t}

And since we know the total charge Q and the time we can replace:

I = \frac{9.12x10^{-8} C}{0.007 s}= 1.30285 x10^{-5} \frac{C}{s}=1.30285 x10^{-5} A = 13.02 \mu A

The current during the inflow charge in the meter axon for this case is 13.02 \mu A

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