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Oliga [24]
2 years ago
13

The brief wave of positive electrical charge that sweeps down the axon is _____.

Physics
1 answer:
kobusy [5.1K]2 years ago
5 0
I believe Action potential is the brief wave of positive charge that sweeps down the axon. Axon is part of the neuron that conducts impulses from the dendrites towards the cell body along the neuron. The action potential is brief since the sodium channels can only stay open for a very brief amount of time. As it travels along the neuron there is a change in polarity across the membrane of the axon .
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In a local diner, a customer slides an empty coffee cup down the counter for a refill. The cup slides off the counter and strike
zysi [14]

a) t=\sqrt{\frac{2h}{g}}

b) v=\frac{d}{\sqrt{\frac{2h}{g}}}

c) v=\sqrt{d^2(\frac{g}{2h})+(2gh)}

d) \theta=tan^{-1}(\frac{2h}{d}) (radians)

Explanation:

a)

The motion of the cup sliding off the counter is the motion of a projectile, consisting of two independent motions:

- A uniform motion along the horizontal direction

- A uniformly accelerated motion (free fall) along the vertical direction

The time of flight of the cup is entirely determined by the vertical motion, therefore we can use the suvat equation:

s=ut+\frac{1}{2}at^2

where here:

s=h (the vertical displacement is the height of the counter)

u=0 (the initial vertical velocity of the cup is zero)

a=g (the vertical acceleration is the acceleration of gravity)

Solving for t, we find the time of flight of the cup:

h=0+\frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}

b)

To solve this part, we just analyze the horizontal motion of the cup.

Here we know that the horizontal motion of the cup is uniform: this means that is horizontal speed is constant during the whole motion, and it is actually equal to the speed at which the mug leaves the counter.

For a uniform motion, the speed is given by

v=\frac{d}{t}

where

d is the distance covered

t is the time taken

Here, the distance covered is d, the distance from the base of the counter, while the time taken is the time of flight:

t=\sqrt{\frac{2h}{g}}

Substituting into the previous equation, we find the speed of the mug as it leaves the counter:

v=\frac{d}{\sqrt{\frac{2h}{g}}}

c)

Here we want to find the speed of the cup immediately before it hits the floor.

Here we have to consider that while the mug falls, its vertical speed increases, while the horizontal speed remains constant.

Therefore, the horizontal speed of the cup before it hits the ground is:

v_x=\frac{d}{\sqrt{\frac{2h}{g}}}=d\sqrt{\frac{g}{2h}}

The vertical speed instead is given by the suvat equation:

v_y=u_y + at

where:

u_y=0 is the initial vertical velocity

a=g is the acceleration

t=\sqrt{\frac{2h}{g}} is the time of flight

Substituting,

v_y = 0 +g(\sqrt{\frac{2h}{g}})=\sqrt{2gh}

The actual speed of the cup just before it hits the floor is the resultant of the horizontal and vertical speeds, so it is:

v=\sqrt{v_x^2+v_y^2}=\sqrt{d^2(\frac{g}{2h})+(2gh)}

d)

Just before hitting the floor, the velocity of the cup has two components:

v_x=d\sqrt{\frac{g}{2h}} is the horizontal component (in the forward direction)

v_y=\sqrt{2gh} is the vertical component (in the downward direction)

Since the two components are perpendicular to each other, the angle of the direction is given by the equation

tan \theta = \frac{v_y}{v_x}

where here \theta is measured as below the horizontal direction.

Substituting the expressions for v_x,v_y, we find:

tan \theta = \frac{\sqrt{2gh}}{d\sqrt{\frac{g}{2h}}}=\frac{2h}{d}

So

\theta=tan^{-1}(\frac{2h}{d}) (radians)

4 0
3 years ago
The strength of the gravitational pull between two object's is determined by?
iogann1982 [59]
Mass and distance are the two factors
4 0
3 years ago
Read 2 more answers
What happens in a tug of war if the net forces are balanced and why?
FinnZ [79.3K]

Answer:

Balanced forces are responsible for unchanging motion. Balanced forces are forces where the effect of one force is cancelled out by another. A tug of war, where each team is pulling equally on the rope, is an example of balanced forces. The forces exerted on the rope are equal in size and opposite in direction.

Explanation:

6 0
3 years ago
In order to sail through the frozen Arctic Ocean, the most powerful icebreaker ever built was constructed in the former Soviet U
professor190 [17]

Answer:

955.36 seconds ≈ 16 minutes

Explanation:

Power(P) is the rate of doing work(W)

That is, P = W/t, where t is the time.

multipying both sides with 't' and dividing with 'P', we get: t=W/P

Here, W = 5.35 x 10^10 J and P = 5.6 x 10^7 W ( 1 W = 1 J/s).

Therefore , on dividing W with P, we get 955.36 seconds.

6 0
3 years ago
The luxury liner Queen Elizabeth 2 has a diesel-electric powerplant with a maximum power of 90 MW at a cruising speed of 31.5 kn
AlexFokin [52]

Answer:

5558643.69 N

Explanation:

F = Force

v = Velocity = 31.5 knots

Converting to m/s

1\ knot=0.514\ m/s

31.5\ knot=31.5\times 0.514\ m/s=16.191\ m/s

Power is given by

P=Fv\\\Rightarrow F=\frac{P}{v}\\\Rightarrow F=\frac{90\times 10^6}{16.191}\\\Rightarrow F=5558643.69\ N

The forward force is exerted on the ship at this highest attainable speed is 5558643.69 N

5 0
3 years ago
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