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kramer
3 years ago
11

What is the wavelength of violet light that has a frequency of 7.5x10^14

Physics
2 answers:
Crank3 years ago
6 0
Λ= V/f 
<span>but change it to represent the speed of light, c </span>
<span>λ= c/f </span>
<span>c = 3.00 x 10^8 m/s </span>
<span>Plug in your given info and solve for λ(wavelength) </span>
<span>λ= 3.00 x 10^8 m/s / 7.5 x 10^14 Hz
(3.00 x 10^8) / (7.5 x 10^14) = 300,000,000 / 750,000,000,000,000 = 0.0000004
Hope this helps :)
</span>
levacccp [35]3 years ago
6 0

Wavelength of any wave = (speed of the wave) / (frequency)

Speed of light in vacuum = 3 x 10^8 m/s

Wavelength =  (3 x 10^8) / (7.5 x 10^14)  =<em>  4 x 10^-7 m</em>  (400 nanometers)


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The de Broglie wavelength of an electron traveling at 7.0 x 107m/s is 1.0 x 10-11m. (Remember that me = 9.1 x 10-31kg and h = 6.
ivolga24 [154]

The De Broglie's wavelength of a particle is given by:

\lambda=\frac{h}{p}

where

h=6.6 \cdot 10^{-34} Js is the Planck constant

p is the momentum of the particle


In this problem, the momentum of the electron is equal to the product between its mass and its speed:

p=m_e v=(9.1 \cdot 10^{-31} kg)(7.0 \cdot 10^7 m/s)=6.4 \cdot 10^{-23} kg m/s

and if we substitute this into the previous equation, we find the De Broglie wavelength of the electron:

\lambda=\frac{h}{p}=\frac{6.6 \cdot 10^{-34} Js}{6.4 \cdot 10^{-23} kg m/s}=1.0 \cdot 10^{-11} m


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7 0
3 years ago
A thundercloud has an electric charge of 48.8 C near the top of the cloud and –41.7 C near the bottom of the cloud. The magnitud
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Answer: 1.51 km

Explanation:

<u>Coulomb's Law:</u> The electrostatic force between two charge particles Q: and Q2 is directly proportional to product of magnitude of charges and inversely proportional to square of separation distance between them.

Or,   \vec{F}=k \frac{Q_{1} Q_{2}}{r^{2}}

Where Q1 and Q2 are magnitude of two charges and r is distance between them:

<u>Given:</u>

Q1 = Charge near top of cloud = 48.8 C

Q2 = Charge near the bottom of cloud = -41.7 C

Force between charge at top and bottom of cloud (i.e. between Q: and Q2) (F) = 7.98 x 10^6N

k = 8.99 x 109Nm^2/C^2

<u>So,</u>

\begin{aligned}&7.98 \times 10^{6}=\left(8.99 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right) \frac{48.8 \mathrm{C} \times 41.7 \mathrm{C}}{\mathrm{r}^{2}} \\&r=\sqrt{\frac{1.8294 \times 10^{13}}{7.98 \times 10^{6}}}=1.514  \times 10^{3} \mathrm{~m}=1.51 \mathrm{~km}\end{aligned}

Therefore, the separation between the two charges (r) = 1.51 km

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2 years ago
Lance Armstrong bikes at a constant speed up the Col d’Izoard, a famous mountain pass. Assume his teammates do such a good job r
svetoff [14.1K]

Work done against gravity to climb upwards is always stored in the form of gravitational potential energy

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W = mgh

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