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Yuki888 [10]
2 years ago
9

During action potential, the electrical charge inside the neuron is __________ the electrical charge outside the neuron.

Physics
2 answers:
GaryK [48]2 years ago
7 0
During action potential, the electrical charge inside the neuron is positive compared to the electrical charge outside the neuron. 
FromTheMoon [43]2 years ago
5 0
<span>The correct answer is: Positive compared to

Explanation:
An action potential is the process that happens when a neuron fires a signal. During the action potential, part of the neural membrane opens to allow positively charged ions inside the cell and negatively charged ions out; consequently, the positive charge inside the neuron increases during that process. In a nutshell, during an action potential, the electrical charge inside the neuron is positive compared to the electrical charge outside the neuron.</span>
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An 800-N billboard worker stands on a 4.0-m scaffold weighing 500 N and supported by vertical ropes at each end. How far would t
Lynna [10]

Answer:

2.5 m

Explanation:

Weight of billboard worker = 800 N

Number of ropes = 2

Length of scaffold = 4 m

Weight of scaffold = 500 N

Tension in rope = 550 N

The sum of the torques will be

-800(4-x)-500\times 2+550\times 4=0\\\Rightarrow -800(4-x)=500\times 2-550\times 4\\\Rightarrow -800(4-x)=-1200\\\Rightarrow -x=\dfrac{1200}{800}-4\\\Rightarrow -x=-2.5\\\Rightarrow x=2.5\ m

The position of the person will be 2.5 m

7 0
2 years ago
"if a stream flow measures 12 meters in 60 seconds, what is the stream's average rate of flow?"
Vlad [161]
<span>Discharge is the volume of water moving down a stream or river per unit of time, commonly expressed in cubic feet per second or gallons per day. In general, river discharge is computed by multiplying the area of water in a channel cross section by the average velocity of the water in that cross section: discharge = area * velocity. In this case, the answer is 0.2 m/s.</span>
7 0
2 years ago
3. In 1989, Michel Menin of France walked on a tightrope suspended under a
Tamiku [17]

Answer: 80m

Explanation:

Distance of balloon to the ground is 3150m

Let the distance of Menin's pocket to the ground be x

Let the distance between Menin's pocket to the balloon be y

Hence, x=3150-y------1

Using the equation of motion,

V^2= U^s + 2gs--------2

U= initial speed is 0m/s

g is replaced with a since the acceleration is under gravity (g) and not straight line (a), hence g is taken as 10m/s

40m/s is contant since U (the coin is at rest is 0) hence V =40m/s

Slotting our values into equation 2

40^2= 0^2 + 2 * 10* (3150-y)

1600 = 0 + 63000 - 20y

1600 - 63000 = - 20y

-61400 = - 20y minus cancel out minus on both sides of the equation

61400 = 20y

Hence y = 61400/20

3070m

Hence, recall equation 1

x = 3150 - 3070

80m

I hope this solve the problem.

6 0
2 years ago
Consider three drinking glasses. All three have the same area base, and all three are filled to the same depth with water. Glass
Kay [80]

The glass which has the greatest liquid pressure at the bottom is all 3 have equal non-zero pressure at the bottom. The correct answer between all the choices given is the first choice or letter A. I am hoping that this answer has satisfied your query about and it will be able to help you.

4 0
1 year ago
Read 2 more answers
A radar used to detect the presence of aircraft receives a pulse that has reflected off an object 5 ✕ 10−5 s after it was transm
Sunny_sXe [5.5K]

Answer:

7500 m

Explanation:

The radar emits an electromagnetic wave that travels towards the object and then it is reflected back to the radar.

We can call L the distance between the radar and the object; this means that the electromagnetic wave travels twice this distance, so

d = 2L

In a time of

t=5\cdot 10^{-5}s

Electromagnetic waves travel in a vacuum at the speed of light, which is equal to

c=3.0\cdot 10^8 m/s

Since the electromagnetic wave travels with constant speed, we can use the equation for uniform motion ,so:

d=vt (1)

where

v=c=3.0\cdot 10^8 m/s

t=5\cdot 10^{-5}s

d=2L, where L is the distance between the radar and the object

Re-arranging eq(1) and substituting, we find L:

L=\frac{vt}{2}=\frac{(3.0\cdot 10^8)(5\cdot 10^{-5})}{2}=7500 m

7 0
2 years ago
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