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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 ?"

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What is the answer?

Andrews [41]

Answer:

6.26 m/s

Explanation:

Pretty slow.... the PE (Potential Energy) at 2m will be converted to KE (Kinetic Energy) at the bottom of the track (neglecting friction)

PE = KE

mgh = 1/2 mv^2 divide both sides of the equation by 'm'

gh = 1/2 v^2 multiply both sides by 2

2 gh = v^2 take sqrt of both sides

v = sqrt ( 2gh) = sqrt ( 2*9.81*2) = 6.26 m/s

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2 years ago

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IRISSAK [1]

Answer:

Explanation:

Force I think

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3 years ago

19) A child on a sled starts from rest at the top of a 15.0° slope. If the trip to the bottom takes 15.2 s,

sveticcg [70]

Answer: 288.8 m

Explanation:

We have the following data:

$t=15.2 s$ is the time it takes to the child to reach the bottom of the slope

$V_{o}=0$ is the initial velocity (the child started from rest)

$\theta=15\°$ is the angle of the slope

$d$ is the length of the slope

Now, the Force exerted on the sled along the ramp is:

$F=ma$ (1)

Where $m$ is the mass of the sled and $a$ its acceleration

In addition, if we draw a free body diagram of this sled, the force along the ramp will be:

$F=mg sin \theta$ (2)

Where $g=9.8 m/s^{2}$ is the acceleration due gravity

Then:

$ma=mg sin \theta$ (3)

Finding $a$ :

$a=g sin \theta$ (4)

$a=9.8 m/s^{2} sin(15\°)$ (5)

$a=2.5 m/s^{2}$ (6)

Now, we will use the following kinematic equations to find $d$ :

$V=V_{o}+at$ (7)

$V^{2}=V_{o}^{2}+2ad$ (8)

Where $V$ is the final velocity

Finding $V$ from (7):

$V=at=(2.5 m/s^{2})(15.2 s)$ (9)

$V=38 m/s$ (10)

Substituting (10) in (8):

$(38 m/s)^{2}=2(2.5 m/s^{2})d$ (11)

Finding $d$ :

$d=288.8 m$

6 0

3 years ago

The Type K thermocouple has a sensitivity of about 41 uV /°C, i.e. for each degree difference in the junction temperature, the o

Novosadov [1.4K]

Answer:

ΔTmin = 3.72 °C

Explanation:

With a 16-bit ADC, you get a resolution of $2^{16}=65536$ steps. This means that the ADC will divide the maximum 10V input into 65536 steps:

ΔVmin = 10V / 65536 = 152.59μV

Using the thermocouple sensitiviy we can calculate the smallest temperature change that 152.59μV represents on the ADC:

$\Delta Tmin = \frac{\Delta Vmin}{41 \mu V/C}= 3.72 C$

3 0

3 years ago

A car starts from rest and accelerates to 14 m/s in 2 seconds. What was its acceleration

Romashka [77]

Answer:

7m/s^2

Explanation:

using v=u+at

since the car started from rest, u=0 , v=14m/s t=2s

a =acceleration.

14=0+a×2

14=0+2a

14=2a

a= 14/2 =7

a=7m/s^2

7 0

2 years ago