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Basile [38]
3 years ago
9

The red light emitted by a helium–neon laser has a wavelength of 632.8 nm. What is the frequency of the light waves?

Physics
1 answer:
Monica [59]3 years ago
5 0

f = 4.74 × 10 ^14  Hz

hope this helps:)

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A gas is enclosed in a cylinder fitted with a light frictionless piston and maintained at atmospheric pressure. When 3300 kcal o
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Answer

given,

heat added to the gas,Q = 3300 kcal

initial volume, V₁ = 13.7 m³

final volume, V₂ = 19.7 m³

atmospheric pressure, P = 1.013 x 10⁵ Pa

a) Work done by the gas

    W = P Δ V

    W = 1.013 x 10⁵ x (19.7 - 13.7)

    W = 6.029 x 10⁵ J

b) internal energy of the gas = ?

  now,

 change in internal energy

  Δ U = Q - W

    Q = 3300 x 10³ cal

    Q = 3300 x 10³ x 4.186 J

    Q = 1.38 x 10⁷ J

now,

  Δ U = 1.38 x 10⁷  - 6.029 x 10⁵

  Δ U = 1.32 x 10⁷ J

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3 years ago
In an electrochemical cell, the anode is ____.
harkovskaia [24]
A) the electrode at which oxidation takes place
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A 58 g firecracker is at rest at the origin when it explodes into three pieces. The first, with mass 12 g , moves along the x ax
alexdok [17]

Answer:

Explanation:

We shall apply conservation of momentum law in vector form to solve the problem .

Initial momentum = 0

momentum of 12 g piece

= .012 x 37 i since it moves along x axis .

= .444 i

momentum of 22 g

= .022 x 34 j

= .748 j

Let momentum of third piece = p

total momentum

= p + .444 i + .748 j

so

applying conservation law of momentum

p + .444 i + .748 j  = 0

p = - .444 i -  .748 j  

magnitude of p

= √ ( .444² + .748² )

= .87 kg m /s

mass of third piece = 58 - ( 12 + 22 )

= 24 g = .024 kg

if v be its velocity

.024 v = .87

v = 36.25 m / s .

6 0
3 years ago
What role do decisions play in wellness?
Tcecarenko [31]

Answer: If you make certain decisions like changing your diet (for good or worse) increasing physical activity, or playing more video games, things like these can affect your wellness and health for the worse or better.

8 0
3 years ago
Read 2 more answers
SHOW WORK
Helga [31]

Answer:

Follows are the solution to the given question:

Explanation:

For point a:

T= 2\pi \sqrt{\frac{m}{k}}\\\\k = \frac{4 \pi^2 m}{T^2}\\\\= \frac{4 \times (3.14)^2 \times 3}{2^2}\\\\=29.578 \ \frac{N}{m}\\\\

For Point b:

E=\frac{1}{2} m a^2 w^2\\\\

   =\frac{1}{2} \times m \times a^2 \times \frac{4\pi^2}{T^2}\\\\=\frac{1}{2} \times 3 \times (0.15)^2 \times \frac{4\times 3.14^2}{2^2}\\\\=0.332 \ J

For Point C:

V_{max}= a w

        = (0.15) \times \frac{2\pi}{T}\\\\= (0.15) \times \frac{2\times 3.14}{2}\\\\=0.471 \frac{m}{s}

For point D:

X= a \sin (wt+ \phi)\\\\0.91=0.15 \sin(\frac{2\pi}{T} \times t+\phi)\\\\0.91=0.15 \sin(\frac{2\times 3.14}{2} \times 0.5+\phi)\\\\0.60 = \sin(3.14 \times 0.5+\phi)\\\\0.60 = \sin(1.57+\phi)\\\\1.57 +\phi =\sin^{-1} 60^{\circ}\\\\1.57 +\phi = 36.86^{\circ}\\\\=35.29^{\circ}\\\\So, X=15 \sin(3.14t+35.29^{\circ}) \ cm

5 0
3 years ago
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