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Ksenya-84 [330]
3 years ago
15

The wave frequency is where 1 hertz equals 1 wave passed a fixed point in 1 second.

Physics
1 answer:
ahrayia [7]3 years ago
3 0

                     (10 waves) / (5 sec) 

                 =  (10/5) (wave/sec)

                 =      2      per sec

                 =      2        Hz .
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Answer:

A. The particle model, because only high-energy frequencies of light  can remove electrons .

Explanation:

Each photon of blue light has higher energy than each photon of red light has  . So when each photon strikes each electron , it gets ejected . But the photon of red light has not sufficient energy to eject electron . Once the photon of red light strikes the electron , the energy is wasted off . Energy of photon can not be accumulated . Thus photon behaves like particle .

4 0
3 years ago
Please choose the answer that describes the scientific notation 5098000
bonufazy [111]

So there is a decimal after the last zero and it looks like this 5098000. You have to move the decimal point six back to get in between the five and the zero which looks like this 5.098000 

<span>Scientific notation is the way that scientists easily handle very large numbers or very small numbers. For example, instead of writing 0.0000000056, we write 5.6 x 10^<span>9</span>.</span>

Being that we moved the decimal six places back the answer is 5.098 x 10^6

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4 years ago
When the medium is uniform, how do light waves pass through it?
LekaFEV [45]

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4 years ago
Q2) The position of an artillery, with a speed of projectile 650, which can fire in any direction above the horizontal plane and
Wittaler [7]

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3 years ago
Q 19.23: A proton is initially moving at 3.0 x 105 m/s. It moves 3.5 m in the direction of a uniform electric field of magnitude
DENIUS [597]

Answer:

The kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.

Explanation:

Given;

initial velocity of proton, v_p_i = 3 x 10⁵ m/s

distance moved by the proton, d = 3.5 m

electric field strength, E = 120 N/C

The kinetic energy of the proton at the end of the motion is calculated as follows.

Consider work-energy theorem;

W = ΔK.E

W =K.E_f - K.E_i

where;

K.Ef is the final kinetic energy

W is work done in moving the proton = F x d  = (EQ) x d = EQd

K.E_f =EQd + \frac{1}{2}m_pv_p_i^2

m_p \ is \ mass \ of \ proton = 1.673 \ \times \ 10^{-27} kg \\\\Q \ is \ charge \ of \ proton = 1.6 \times 10^{-19} C

K.E_f = 120\times 1.6 \times 10^{-19} \times 3.5   \ + \ \frac{1}{2}(1.673\times 10^{-27})(3\times 10^5)^2 \\\\

K.E_f = 6.72\times 10^{-17} \ + \ 7.53 \times 10^{-17} \\\\K.E_f = 14.25 \times 10^{-17} J\\\\K.E_f = 1.425\times 10^{-16} \ J

Therefore, the kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.

3 0
3 years ago
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