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nexus9112 [7]
3 years ago
6

is the outer electron in potassium more or less strongly attracted to the positive nucleus than the outer electron in sodium is?

Physics
1 answer:
DiKsa [7]3 years ago
8 0

Answer:

Less strongly attracted.

Explanation:

Potassium and sodium are both alkali metals. That means that they have the ability to be able to gather more valence electrons. They will both gain a positive particle. The potassium electron is less attracted because of the fact that now the sodium has a positive chargem and the electron has a negative charge.

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Motion is a change in
butalik [34]
I think it’s position measured by distance and time.
8 0
3 years ago
The ________ of a neuron contain ________ that house neurotransmitters
Nina [5.8K]

Answer:

The <u>terminal buttons</u> of a neuron contain <u>synaptic vesicles</u> that house neurotransmitters

Explanation:

<em><u>Some terms explained:</u></em>

Terminal buttons:

The terminal Buttons of a neuron contain vesicles holding the neurotransmitters, they are the small knobs at the end of the neuron that send out information to other neurons. The terminal buttons essentially convert the electrical signals that reaches it into chemical signals.

Neurotransmitters:

Neurotransmitters are produced by the neurons in the axon terminal buttons and stored in the synaptic vesicles. Neurotransmitters are chemicals contained in terminal buttons that relays signals across the synapses between neurons. Neurotransmitters are the chemical messengers of the nervous system, they have an excitatory or inhibitory effect on another neuron.

Synaptic vesicles:

Synaptic vesicles are sacs in terminal button, they store neurotransmitters and release them into synaptic space in response to electrical signaling within the cell.

4 0
3 years ago
The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 5.0 rev/s in 8.0 s. At thi
tatiyna

Answer:

50 revolutions

Explanation:

Data provided:

case I: From rest to top spin

The initial angular speed of the washer, ωi = 0 rev /s

Final angular speed of the washer ωf = 5 rev /s

Time taken, t₁ = 8 s

now,  

The angular displacement or the number of revolutions taken (θ₁) is calculated as:

  θ₁ = ωi t₁ + (1/2)α₁t₁²

where,

α is the angular acceleration

The angular acceleration can be calculated as:

  ωf - ωi = α₁t₁

on substituting the values, we get

8α₁ = 5 - 0

or

α₁ = 0.625 rev/s²

substituting the values in the equation for the number of revolutions, we get

θ₁ = 0 + (1/2) (0.625)(8)²

or

θ₁ = 20 revolutions

also,  

For the case II: From top spin to rest

we have

The initial angular speed, ωi = 5 rev /s

and the final angular speed, ωf = 0 rev /s

Total time taken, t₂ = 12 s

Now, angular acceleration for this case

  ωf - ωi = α₂t₂

on substituting the values, we have

  12α₂ = 0 - 5

α₂ = - 0.4166 rev/s²

Therefore, the number of revolutions ( i.e angular displacement  )

θ₂ = ωit₂ + (1/2)α₂t₂²

on substituting the values, we have

θ₂ = 5 × 12 + (1/2)(-0.4166)(12)²

or

θ₂ = 30 rev

Hence,

the total number of revolutions made by the washer during the 20s is  

θ = θ₁ + θ₂

or

θ = 20 rev + 30 rev

or

θ = 50 revolutions

7 0
3 years ago
A 1.7-m-long string is under 22 N of tension. A pulse travels the length of the string in 54 ms. what is the mass of the string?
Mariana [72]

Answer:

Mass of the string, m = 37.7 grams

Explanation:

It is given that,

Length of the string, l = 1.7 m

Tension in the string, T = 22 N

Time taken by the string, t=54\ ms=54\times 10^{-3}\ s

The speed in the wave in the string is given by the following formula as :

v=\sqrt{\dfrac{T}{\mu}}

\mu is the mass per unit length, \mu=\dfrac{m}{l}

v=\sqrt{\dfrac{Tl}{m}}

Also, v=\dfrac{d}{t}

\dfrac{l}{t}=\sqrt{\dfrac{Tl}{m}}

m=\dfrac{Tt^2}{l}

m=\dfrac{22\times (54\times 10^{-3})^2}{1.7}

m = 0.0377 kg

or

m = 37.7 grams

So, the mass of the strings is 37.7 grams. Hence, this is the required solution.

6 0
3 years ago
Samantha is checking the weather for her upcoming trip to Mexico City. The weather forecast predicts a high-pressure system for
Zanzabum

Calm, sunny days with wind moving away from the center.

5 0
3 years ago
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