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ElenaW [278]
3 years ago
14

Describe how personal fitness contributes to physical,mental/emotional,social health

Physics
1 answer:
Semenov [28]3 years ago
3 0
When you work out it helps clear your thoughts and think of nothing. It makes you feel good.
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A coin is located 35.0 mm from the center of a turntable which accelerates from rest to a speed of 7.95 rad/s in 0.270 s. What i
beks73 [17]

The minimum coefficient of friction to avoid slippage of the coin is 0.23.

<h3>Minimum coefficient of friction</h3>

The minimum coefficient of friction can determined from the maximum speed of the coin.

ma = μmg

mv²/r = μmg

v²/r = μg

v² = μgr

μ = v²/(gr)

μ = (ωr)²/(gr)

μ = ω²r/g

μ = (7.95² x 0.035)/(9.8)

μ = 0.23

Learn more about coefficient of friction here: brainly.com/question/20241845

#SPJ11

5 0
2 years ago
The minimum energy required for an effective collision is called
kifflom [539]

Answer:

Activation Energy

Explanation:

Effective collisions are those that result in a chemical reaction. In order to produce an effective collision, reactant particles must possess some minimum amount of energy. This energy, used to initiate the reaction, is called the activation energy.

6 0
4 years ago
What wavelength of light contains enough energy in a single photon to ionize a hydrogen atom?
BaLLatris [955]

There's probably a much quicker, easier way to do it, but I don't work with this stuff every day so this is the way I have to do it:

First, I searched the "ionization energy" of Hydrogen on Floogle.  That's how much work it takes to rip the one electron away from its Hydrogen atom, and it's 13.6 eV (electron-volts).

In order to find the frequency/wavelength of a photon with that energy, I need the energy in units of Joules.

1 eV = 1.602 x 10⁻¹⁹ Joule  (also from Floogle)

13.6 eV = 2.179 x 10⁻¹⁸ Joule

OK.  Now we can use the popular well-known formula for the energy of a photon:

Energy = h · (frequency)  

or  Energy = h · (light speed/wavelength)

' h ' is Max Planck's konstant ... 6.626 × 10⁻³⁴ m²-kg / s

Wow !  The only thing we don't know in this equation is the wavelength, which is what we need to find.  That's gonna be a piece-o'-cake now, because we know the energy, we know ' h ', and we know the speed of light.

Wavelength = h · c / energy

Wavelength =

(6.626 x 10⁻³⁴ m²-kg/sec) · (3 x 10⁸ m/s) / (2.179 x 10⁻¹⁸ joule)

<em>Wavelength = 9.117 x 10⁻⁸ meter </em>

That's  91.1 nanometers .

It's not visible light (visible is between about 390 to 780 nm), but it's not as short as I was expecting.  I thought it was going to be an X-ray, but it's not that short.  X-rays are defined as 0.1 to 10 nanometers.  This result is in the short end of Ultra-violet.

(You have no idea how happy I am with this result.  I figured it out exactly the way I showed you, and I never peeked.  Then, AFTER I had my solution, I went to Floogle and searched to see what it really is, and whether I came out anywhere close.  I found it in the article on the "Lyman Series".  It says the wavelength of the energy released by an electron that falls in from infinity and settles in the n=1 energy level of Hydrogen is  91.175 nm !  This gives me a big hoo-hah for the day, and I'm going to bed now.)

6 0
3 years ago
Read 2 more answers
the electric potential of a hydrogen atom is modeled by the equation where a0 is the bohr radius of the atom and q is the charge
zepelin [54]

In light of this, V=V 0 loge (r/r 0 ) Field E= dr dV =V 0(r0r) eE= r mV2 alternatively, reV0r0=rmV2. V=(m eV 0 r 0 ) \ s1 / 2mV=(m e V 0 r 0 ) 1/2 = constant mvr= 2 nh, also known as Bohr's quantum condition or Hermitian matrix.

Show that the eigenfunctions for the Hermitian matrix in review exercise 3a can be normalized and that they are orthogonal.  

Demonstrate how the pair of degenerate eigenvalues for the Hermitian matrix in review exercise 3b can be made to have orthonormal eigenfunctions.

Under the given Hermitian matrix, "border conditions," solve the following second order linear differential equation: d2x/ dt2 + k2x(t) = 0 where x(t=0) = L and dx(t=0)/ dt = 0.

To know more about Hermitian click on the link:

brainly.com/question/14671266

#SPJ4

3 0
1 year ago
A flare is dropped from an airplane flying horizontally at uniform velocity (constant speed in a straight line). Neglecting air
jolli1 [7]

Answer:

Option B (remain vertically under the plane) is the correct option.

Explanation:

  • A flare would follow a particle trajectory with horizontal direction somewhat like airplane velocity as well as initial maximum motion as null but instead, gravity will induce acceleration. It would be lowered vertically underneath the plane before flare had already sunk to something like the surface.
  • There is no different movement in the airplane nor even the flash. And none of them can change its horizontal level.  

Some other alternatives are given really aren't linked to the specified scenario. So choice B is the perfect solution to that.

7 0
3 years ago
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