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3 years ago

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

1 answer:

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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A 3.53-g sample of aluminum completely reacts with oxygen to form 6.67 g of aluminum oxide. find percent composition

katrin2010 [14]

The mass percent composition of aluminum is 52.9% in aluminum oxide.

Mass of the aluminum = 3.53 g

Mass of the aluminum oxide = 6.67 g.

The mass percent of a substance is the mass of the substance divided by the mass of the compound into 100.

Aluminum reacts with oxygen to form aluminum oxide.

The overall balanced equation for the reaction is,

$Aluminum + Oxygen→Aluminum \: oxygen$

$4Al +3O _{2}→2Al _{2}O _{3}$

The mass percent composition of aluminum in the aluminum oxide is,

$Mass \: percent = \frac{ Mass \: of \: the \: substance}{ Mass \: of \: the \: compound} \times 100$

$Mass \: percent = \frac{Mass \: of \: the \: aluminum}{ Mass \: of \: the \: aluminum \: oxide } \times 100$

$Mass \: percent = \frac{3.53}{6.67 } \times 100$

= 52.9 %

Therefore, the mass percent composition of aluminum is 52.9% in aluminum oxide.

To know more about aluminum oxide, refer to the below link:

brainly.com/question/25869623

#SPJ4

7 0

2 years ago

<img src="https://tex.z-dn.net/?f=%7B%5Chuge%5Cfcolorbox%7Bblue%7D%7Bblack%7D%7B%5Cpink%7B%5C%3A%5C%3A%5C%3A%5C%3A%5C%3A%5C%3A%5

kumpel [21]

the branch of optics that studies interference, diffraction, polarization, and other phenomena for which the ray approximation of geometric optics is not valid.

3 0

2 years ago

1. A 25.35-g piece of iron absorbs 1562.75 Joules of heat energy, and its temperature

Lera25 [3.4K]

Answer:

Q = C M T where C is the specific, M the mass, T the temperature change

Note 1 cal = 4.19 Joules

1562.75 J / (4.19 J/cal) = 378 cal

C = Q / (M * T) = 378 cal / (25.35 g * 155 deg C)

C = .096 cal / g deg C

8 0

2 years ago

A ball of mass 0.120 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.820 m. W

Vikentia [17]

Answer:

1.0752 kgm/s

Explanation:

Considering when the drop was dropped from rest from a height,

mass of the ball, m = 0.120 kg

height, h = - 1.25 m

the initial velocity, u = 0 m/s

the acceleration due to gravity, g = - 9.8 m/s²

From equation of motion

$V^{2} = U^{2} + 2gh$

Substituting the values,

$V^{2} = 0^{2} + 2(-9.8 m/s^{2})(-1.25 m)$

$V^{2} = 24.5 m/s$

$V = \sqrt{24.5} \ m/s$

$V = 4.95 \ m/s$

V = ± 4.95 m/s

V = - 4.95 m/s

Since the ball is moving downward, the final velocity of the ball when it hits the floor is V = - 4.95 m/s

Considering when the ball rebounds from the floor,

assume the mass of the ball still remain, m = 0.120 kg

height, h = 0.820 m

the final velocity, v = 0 m/s

the acceleration due to gravity, g = - 9.8 m/s²

From equation of motion

$V^{2} = U^{2} + 2gh$

Substituting the values,

$0^{2} = U^{2} + 2(-9.8 m/s^{2})(0.820 m)$

$0 = U^{2} - 16.072 m/s$

$U^{2} = 16.072 m/s$

$U = \sqrt{16.072} \ m/s$

U = ± 4.01 m/s

U = + 4.01 m/s

Since the ball is moving upward, the initial velocity of the ball from the bounce from the floor is U = + 4.01 m/s

From Newton's second law of motion, applied force is directly proportional to the rate of change in momentum.

$F = \frac{mv - mu}{t}$

$F.t = m(v - u)$

⇒ Impulse = Change in momentum

To calculate the impulse, the moment before the ball hits the ground will be the initial momentum while the moment the ball rebounces will be the final velocity,

∴ F.t = 0.120 kg(4.01 m/s - (-4.95 m/s) )

F.t = 0.120 kg(4.01 m/s + 4.95 m/s) )

F.t = 0.120 kg × 8.96 m/s

Impulse = 1.0752 kgm/s

The impulse given to the ball by the floor is 1.0752 kgm/s

6 0

3 years ago

26. A 40 kg boy jumps from a height of 4m onto a plate-form mounted on springs. As the

denpristay [2]

Answer:

c. 1600J

Explanation:

The loss in potential energy of the boy is given by:

$U=mg \Delta h$

where

m = 40 kg is the mass of the boy

g = 9.8 m/s^2 is the acceleration of gravity

$\Delta h = 4 m + 0.02 m = 4.02 m$ is the total change in the height of the boy (4 metres + 2 cm due to the compression of the spring)

Substituting, we find

$\Delta U = (40 kg)(9.8 m/s^2)(4.02 m) = 1577 J \sim 1600 J$

4 0

3 years ago