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faltersainse [42]
3 years ago
12

A quantum of electromagnetic radiation has an energy of 2.0 kev. what is its frequency? planck's constant is 6.63 × 10−34 j · s.

answer in units of hz.
Physics
1 answer:
xxMikexx [17]3 years ago
8 0
A quantum of electromagnetic radiation is called photon, and the energy of a photon is given by
E=hf (1)
where h is the Planck constant and f is the frequency of the photon.

In this problem, the photon energy is
E=2.0 keV=2.0 \cdot 10^3 eV=3.2 \cdot 10^{-16} J
If we re-arrange equation (1), we can find the energy of the photon:
f= \frac{E}{h}= \frac{3.2 \cdot 10^{-16} J}{6.6 \cdot 10^{-34} Js}=4.8\cdot 10^{17} Hz
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If block D weighs 300 lb and block B weighs 275 lb determine the required weight of block C and the angle theta for equilibrium?
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<span>and by reason, Tcd = Tbd </span>

<span>Tbd y = 275 - 300*sinθ </span>
<span>Tcd y = Tc - 300*sin30 </span>

<span>Tbd x = 300*cosθ </span>
<span>Tcdx = 300 * cos30 </span>

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A 2.0-kg object is attached to a spring (k = 55.6 N/m) that hangs vertically from the ceiling. The object is displaced 0.045 m v
rusak2 [61]

Answer:

Maximum acceleration will be 1.251m/sec^2

Explanation:

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We know that \omega ^2=\frac{k}{m}=\frac{55.6}{2}=27.8

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8 0
3 years ago
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pogonyaev
As we know that time period of simple pendulum is given as

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here we know that

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