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Alborosie
3 years ago
12

Why is it important for car designers to keep accurate records of their experimental procedures

Physics
1 answer:
storchak [24]3 years ago
5 0
Like all experiments, it's important to keep records of the their experiment procedures so future car designers and experimenters can compare and contrast their results from their experiment and improve their experiment accordingly. I'm pretty sure this is it. Stay cool my man.
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What is a lack of physical, social, or economic access to safe and nutritious foods called? A) famine B) food security C) food s
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Food insecurity, because you are only lacking safe and nutritious food. It would be famine if you had no food whatsoever, and it would be a food shortage if you were lacking a small amount of ALL food
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A glass of root beer with a scoop of ice cream floating on top and a straw sticking out.
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These forces are all equal and cancel each other out. Gravity pushes downward on the ice cream. This can also be called the weight of the ice cream. Buoyant force pushes the ice cream upward

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Does a falling rock have potential or kinetic energy
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depends t what stage in the fall it is. If it is at the peak, it is fully potential. If it is in the middle, it has both. If it is at the bottom of the fall, it is completely kinetic

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20% of injury crashes in 2009 involve the reports of distracted driving. True or false
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Calculate the energy, wavelength, and frequency of the emitted photon when an electron moves from an energy level of -3.40 eV to
jasenka [17]

Answer:

(a) The energy of the photon is 1.632 x 10^{-8} J.

(b) The wavelength of the photon is 1.2 x 10^{-17} m.

(c) The frequency of the photon is 2.47 x 10^{25} Hz.

Explanation:

Let;

E_{1} = -13.60 ev

E_{2} = -3.40 ev

(a) Energy of the emitted photon can be determined as;

E_{2} - E_{1} = -3.40 - (-13.60)

           = -3.40 + 13.60

           = 10.20 eV

           = 10.20(1.6 x 10^{-9})

E_{2} - E_{1} = 1.632 x 10^{-8} Joules

The energy of the emitted photon is 10.20 eV (or 1.632 x 10^{-8} Joules).

(b) The wavelength, λ, can be determined as;

E = (hc)/ λ

where: E is the energy of the photon, h is the Planck's constant (6.6 x 10^{-34} Js), c is the speed of light (3 x 10^{8} m/s) and λ is the wavelength.

10.20(1.6 x 10^{-9}) = (6.6 x 10^{-34} * 3 x 10^{8})/ λ

λ = \frac{1.98*10^{-25} }{1.632*10^{-8} }

  = 1.213 x 10^{-17}

Wavelength of the photon is 1.2 x 10^{-17} m.

(c) The frequency can be determined by;

E = hf

where f is the frequency of the photon.

1.632 x 10^{-8}  = 6.6 x 10^{-34} x f

f = \frac{1.632*10^{-8} }{6.6*10^{-34} }

 = 2.47 x 10^{25} Hz

Frequency of the emitted photon is 2.47 x 10^{25} Hz.

6 0
2 years ago
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