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schepotkina [342]
2 years ago
6

When a nerve cell depolarizes, charge is transferred across the cell membrane, changing the potential difference. For a typical

nerve cell, 9.0 pC of charge flows in a time of 0.50 ms. You may want to review (Page 706) . Part A What is the average current
Physics
1 answer:
laila [671]2 years ago
6 0

Answer:

I = 18 x 10⁻⁹ A = 18 nA

Explanation:

The current is defined as the flow of charge per unit time. Therefore,

I = q/t

where,

I = Average Current passing through nerve cell

q = Total flow of charges through nerve cell

t = time period of flow of charges

Here, in our case:

I = ?

q = (9 pC)(1 x 10⁻¹² C/1 pC) = 9 x 10⁻¹² C

t = (0.5 ms)(1 x 10⁻³ s/1 ms) = 5 x 10⁻⁴ s

Therefore,

I = (9 x 10⁻¹² C)/(5 x 10⁻⁴ s)

<u>I = 18 x 10⁻⁹ A = 18 nA</u>

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A student drops a ball from the top of a 10-meter tall building. The ball leaves the thrower's hand with a zero speed. What is t
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Answer:

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u = 0, h = 10 m, g = 9.8 m/s^2

Use third equation of motion

v^2 = u^2 + 2 g h

Here, v be the velocity of ball as it just strikes with the ground

v^2 = 0 + 2 x 9.8 x 10

v^2 = 196

v = 14 m/s

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What's the difference between Uniform magnetic field and Non-uniform magnetic field?
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3 years ago
the frequency of the middle b note on a piano is 493.88 hz. what is the wavelength of this note in centimeters? the speed of sou
vekshin1

Answer:

69.5

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V=FX

343.06=493.88×X

X=343.06/493.88

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2 years ago
A rubber ducky is placed 20 cm from a thin convex lens with a focal length of 15 cm. Which statement correctly describes the nat
blondinia [14]
I believe the answer is B, a real and inverted image is formed on the side of the lens opposite the rubber ducky. The focal length is 15 cm and therefore the center of curvature (2F) will be 30 cm. When the object is placed between F and 2F (in this case 20 cm) in front of a convex lens, an inverted, real image is formed on the other side of the lens.
6 0
3 years ago
Read 2 more answers
If 2.0×10^−4 C of charge passes a point in 2.0×10^−6 s , what is the rate of current flow?1.0×10^−10 A1.0×10^2 A4.0×10^−1 A4.0×1
Ipatiy [6.2K]

This problem is about the rate of the current. It's important to know that refers to the quotient between the electric charge and the time, that's the current rate.

I=\frac{Q}{t}

Where Q = 2.0×10^−4 C and t = 2.0×10^−6 s. Let's use these values to find I.

\begin{gathered} I=\frac{2.0\times10^{-4}C}{2.0\times10^{-6}\sec } \\ I=1.0\times10^{-4-(-6)}A \\ I=1.0\times10^{-4+6}A \\ I=1.0\times10^2A \end{gathered}

<em>As you can observe above, the division of the powers was solved by just subtracting their exponents.</em>

<em />

<h2>Therefore, the rate of the current flow is 1.0×10^2 A.</h2>
3 0
11 months ago
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