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Furkat [3]
4 years ago
11

The minimum energy needed to eject an electron from a sodium atom is 4.41 x 10-19 j. what is the maximum wavelength of light, in

nanometers, that will show a photoelectric effect with sodium?
Physics
1 answer:
Minchanka [31]4 years ago
5 0
The energy of an electron as it is ejected from the atom can be calculated from the product of the Planck's constant and the frequency of the light energy. We can calculate the wavelength from the frequency we can calculate. We do as follows:

E = hv
 4.41 x 10-19  = 6.62607004 × 10<span>-34 (v)
v = 6.66x10^14 /s

wavelength = speed of light / frequency
</span>
wavelength = 3x10^8 / 6.66x10^14
wavelength = 4.51x10^-7 m = 450.75 nm
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The main morphological types of galaxies are elliptical, spiral, and irregular.

Based on their morphology , galaxies have been classified into 3 types namely elliptical, spiral, and irregular.

These galaxies have various sizes and shapes ranging from dwarf galaxies to giant galaxies.

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2 years ago
I need help with one - seven
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Answer:

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A motorcycle is travelling at a constant velocity of 30ms. The motor is in high gear and emits a hum of 700Hz. The speed of soun
timurjin [86]

Answer:

a) T=1.43\times 10^{-3}\ s

b) d=0.0429\ m

c) \lambda=0.4857\ m

d) f_o=767.7\ Hz

Explanation:

Given:

  • velocity of the sound from the source, S=340\ m.s^{-1}
  • original frequency of sound from the source, f_s=700\ Hz
  • speed of the source, v_s=30\ m.s^{-1}

(a)

We know time period is inverse of frequency:

Mathematically:

T=\frac{1}{f}

T=\frac{1}{700}

T=1.43\times 10^{-3}\ s

(b)

Distance travelled by the motorcycle during one period of sound oscillation:

d=v_s.T

d=30\times 1.43\times 10^{-3}

d=0.0429\ m

(c)

The distance travelled by the sound during the period of one oscillation is its wavelength.

\lambda=\frac{S}{f}

\lambda=\frac{340}{700}

\lambda=0.4857\ m

(d)

observer frequency with respect to a stationary observer:

<u>According to the Doppler's effect:</u>

\frac{f_o}{f_s}= \frac{S+v_o}{S-v_s} ...........................(1)

where:

f_o\ \&\ v_o are the observed frequency and the velocity of observer respectively.

Here, observer is stationary.

\therefore v_o=0\ m.s^{-1}

Now, putting values in eq. (1)

\frac{f_o}{700}= \frac{340+0}{340-30}

f_o=767.7\ Hz

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4 years ago
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Shakina and Juliette set the car's initial velocity to zero and set the acceleration to +1.2 m/s2, then clicked "start." Answer
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Let's start by differentiating the terms distance and displacement. They both refer to the length of paths. Distance only accounts for the total length regardless of the path taken. Displacement measures the linear path from the starting point to the end point. So, it does not necessarily follow the actual path. However, for this problem, assuming that the path is just in one direction, displacement and distance would just be equal. The equation would be:

Distance = Displacement =  v₀t + 0.5at² = 0(10 s) + 0.5(+1.2 m/s²)(10 s)²
Distance = Displacement = 60 meters
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