All categories

3 years ago

How are the helium atoms in this model different from real helium atoms?

Physics

2 answers:

Taya2010 [7]3 years ago

5 0

Answer;

Real helium atoms are too small to see

Explanation;

-A helium atom is an atom of the chemical element helium.

-Helium is composed of two electrons bound by the electromagnetic force to a nucleus containing two protons along with either one or two neutrons, depending on the isotope, held together by the strong force.

KonstantinChe [14]3 years ago

3 0

With the information given in the question, we can only observe
that they are quite SIMILAR in at least one respect ... both the real
ones and those in 'this model' are quite invisible and intangible, and
we must take your word for it that they exist at all.

Like so many others in math and science, this question raises another
question: "Which model are you talking about ?"

You might be interested in

With what speed must you approach a source of sound to observe a 25% change in frequency?

insens350 [35]

Sound source is at rest, you are moving with velocity v, f = frequency, c = speed of sound:

f = f0(1 + v/c)

115 = 100(1 + v/343)
115 = 100 + 100v/343
15 = 100v/343
v = 15*343/100
<span> v = 51,45 m/s </span>

5 0

2 years ago

PLEASE HELP ME WITH THIS ONE QUESTION

Marizza181 [45]

Answer:

c) 2.02 x 10^16 nuclei

Explanation:

The isotope decay of an atom follows the equation:

ln[A] = -kt + ln[A]₀

<em>Where [A] is the amount of the isotope after time t, k is decay constant, [A]₀ is the initial amount of the isotope</em>

[A] = Our incognite

k is constant decay:

k = ln 2 / Half-life

k = ln 2 / 4.96 x 10^3 s

k = 1.40x10⁻⁴s⁻¹

t is time = 1.98 x 10^4 s

[A]₀ = 3.21 x 10^17 nuclei

ln[A] = -1.40x10⁻⁴s⁻¹*1.98 x 10^4 s + ln[3.21 x 10^17 nuclei]

ln[A] = 37.538

[A] = 2.01x10¹⁶ nuclei remain ≈

<h3>c) 2.02 x 10^16 nuclei</h3>

7 0

2 years ago

A ball of moist clay falls 17.3 m to the ground. It is in contact with the ground for 24.0 ms before stopping. (a) What is the a

gizmo_the_mogwai [7]

Answer:

Acceleration, $767.08\ m/s^2$

Explanation:

Given that,

Height from a ball falls the ground, h = 17.3 m

It is in contact with the ground for 24.0 ms before stopping.

We need to find the average acceleration the ball during the time it is in contact with the ground.

Firstly, find the velocity when it reached the ground. So,

$v^2=u^2+2ah$

u = initial velocity=0 m/s

a = acceleration=g

$v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 17.3} \\\\v=18.41\ m/s$

It is in negative direction, u = -18.41 m/s

Let a is average acceleration of the ball. Consider, v = and u = -18.41 m/s.

$a=\dfrac{v-u}{t}\\\\a=\dfrac{0-18.41}{24\times 10^{-3}}\\\\a=767.08\ m/s^2$

So, the average acceleration of the ball during the time it is in contact is $767.08\ m/s^2$ .

4 0

3 years ago

Cevab 2 anlatarak çözer misiniz

Assoli18 [71]

Translate please, i’d be able to help better:)

6 0

3 years ago

A motorcyclist changes the velocity of his bike from 20.0 meters/second to 35.0 meters/second under a constant acceleration of 4

Fiesta28 [93]

35-20 = 15m/s difference
15/4 = 3.75 seconds

4 0

3 years ago