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schepotkina [342]
3 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]3 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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I’m not sure how to solve this
spayn [35]

Answer:

Option 10. 169.118 J/KgºC

Explanation:

From the question given above, the following data were obtained:

Change in temperature (ΔT) = 20 °C

Heat (Q) absorbed = 1.61 KJ

Mass of metal bar = 476 g

Specific heat capacity (C) of metal bar =?

Next, we shall convert 1.61 KJ to joule (J). This can be obtained as follow:

1 kJ = 1000 J

Therefore,

1.61 KJ = 1.61 KJ × 1000 J / 1 kJ

1.61 KJ = 1610 J

Next, we shall convert 476 g to Kg. This can be obtained as follow:

1000 g = 1 Kg

Therefore,

476 g = 476 g × 1 Kg / 1000 g

476 g = 0.476 Kg

Finally, we shall determine the specific heat capacity of the metal bar. This can be obtained as follow:

Change in temperature (ΔT) = 20 °C

Heat (Q) absorbed = 1610 J

Mass of metal bar = 0.476 Kg

Specific heat capacity (C) of metal bar =?

Q = MCΔT

1610 = 0.476 × C × 20

1610 = 9.52 × C

Divide both side by 9.52

C = 1610 / 9.52

C = 169.118 J/KgºC

Thus, the specific heat capacity of the metal bar is 169.118 J/KgºC

6 0
2 years ago
Find its moment of inertia about an axis perpendicular to its plane and passing through the midpoint of the line connecting its
antoniya [11.8K]

A) Moment of inertia about an axis passing through the point where the two segments meet : $I_A=\frac{1}{12} M L^2$

B) Moment of inertia passing through the point where the midpoint of the line connects to its two ends: $I x=\frac{1}{3} M L^2$

What is Moment of inertia?

The term "moment of inertia" refers to a physical quantity that quantifies a body's resistance to having its speed of rotation along an axis changed by the application of a torque (turning force). The axis might be internal or exterior, fixed or not.

A) The moment of inertia about an axis passing through the point where the two segments meet is $I_A=\frac{1}{12} M L^2$given that the rod is bent at the center and distance from all the points to the axis remains the same, the moment of inertia about the center will remain the same.

B) Determine the moment of inertia about an axis passing through the point midpoint of the line which connects the two ends

First step: determine the distance between the ends ( d )

After applying Pythagoras theorem$\mathrm{d}=\frac{\sqrt{2}}{2} L$

Next step : determine distance between the two axis $(\mathrm{x})$

After applying Pythagoras theorem

\mathrm{x}=\frac{\sqrt{2}}{4} L$$

Final step : Calculate the value of $\mathrm{I}_{\mathrm{x}}$

applying Parallel Axis Theorem

$$I_x=I_8+M x^2$$

$$\begin{aligned}& =\frac{1}{12} M L^2+\frac{1}{4} M L^2 \\& \therefore \quad I x=\frac{1}{3} M L^2 \\&\end{aligned}$$

Hence we can conclude that Moment of inertia about an axis passing through the point where the two segments meet: $I_A=\frac{1}{12} M L^2$, Moment of inertia passing through the point where the midpoint of the line connects its two ends: $I x=\frac{1}{3} M L^2$

To learn more about moment of inertia visit:brainly.com/question/15246709

#SPJ4

5 0
1 year ago
When the air resistance can be ignored the velocity of an object dropped initially from rest is given by the following equation
ad-work [718]

Answer:

I am confused of your question. Do you want final velocity? To get final velocity, use (initial V)+(Gravity*Time)

Explanation:

8 0
3 years ago
Read 2 more answers
What does the term Hubble time mean in cosmology, and what is the current best calculation for the Hubble time?
balu736 [363]

Hubble time in cosmology means the estimated age of the universe and the best calculation for it is T=1/H, where H is the Hubble constant
4 0
3 years ago
WILL MARK BRAINLIEST JUST PLEASE HELP
mixer [17]

Answer:

40N in either direction is the answer

6 0
3 years ago
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