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3 years ago

What is the energy of a photon with a frequency of 1.7 × 1017 Hz? Planck’s constant is 6.63 × 10–34 J•s.

Physics

2 answers:

ANTONII [103]3 years ago

6 0

Answer:

The correct answer is B.

Explanation:

Got a 100% on the Exam

Karo-lina-s [1.5K]3 years ago

3 0

The energy carried by a single photon is given by
$E=hf$
where h is the Planck's constant and f is the frequency of the photon.

The photon of our exercise has a frequency of $f=1.7 \cdot 10^{17} Hz$ , therefore its energy is
$E=hf=(6.63 \cdot 10^{-34}Js)(1.7 \cdot 10^{17} Hz)=1.1 \cdot 10^{-16} J$

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Answer:

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Explanation:

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Answer:

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2 years ago

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Answer:

Sam's and Abigail speeds before colliding were a. 12.34 m/s and b. 2.86 m/s, respectively. Their total kinetic energy was diminished by c. 1484.42 J, approximately

Explanation:

By conservation of momentum, we have

$\Delta \vec{p} = 0\\\\m\vec{v}_i = m\vec{v}_f\\m_S v_S\hat{i} + m_A v_A\hat{j} = m_S \times 7 \cos{(40)} \hat{i} + m_S \times 7 \sin{(40)} \hat{j} + m_A \times 9.4 \cos{(18)} \hat{i} - m_A \times 9.4 \sin{(18)} \hat{j}.$

Writing for each direction at a time,

$m_S v_S = 7m_S \cos{(40)} +9.4 m_A \cos{(18)}\\v_S = 7 \cos{(40)} + 9.4 \frac{m_A}{m_S} \cos{(18)} = 7 \cos{(40)} + 9.4\times\frac{57}{73} \cos{(18)} \approx \mathbf{12.34}\\m_A v_A = 7m_S\sin{(40)} - 9.4m_A\sin{(18)}\\v_A = 7\frac{m_S}{m_A}\sin{(40)} - 9.4\sin{(18)} = 7\times\frac{73}{57}\sin{(40)} - 9.4\sin{(18)} \approx \mathbf{2.86}.$

Their kinetic energy changed by

$K_f - K_i = \left( \frac{1}{2}m_s v_{fs}^2 + \frac{1}{2}m_a v_{fa}^2 \right) - \left( \frac{1}{2}m_s v_{is}^2 + \frac{1}{2}m_a v_{ia}^2 \right) = \left( \frac{1}{2}\times 73 \times 7^2 + \frac{1}{2}\times 57 \times 9.4^2 \right) - \left( \frac{1}{2}\times 73 \times 12.34^2 + \frac{1}{2}\times 57 \times 2.86^2 \right) \approx \mathbf{-1484.418 J}.$

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3 years ago