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4 years ago

13

A gamma ray with an energy of 3.40 × 10-14 joules strikes a photographic plate. We know that Planck's constant is 6.63 × 10-34 j

oule·seconds. What is the frequency of the photon?
5.13 × 1019 hertz

5.60 × 1017 hertz

2.45 × 1017 hertz

2.45 × 10-12 hertz

7.80 × 10-19 hertz

2 answers:

suter [353]4 years ago

4 0

The correct answer is: "A. 5.13 x 10^19 hertz"

Wittaler [7]4 years ago

4 0

The answer is A, 5.13 x 10^19 hertz

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From the question we are told that

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Generally the radius is mathematically represented as $r_i = \frac{d_i}{2} = \frac{7}{2} = 3.5 \ m$

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Generally the radius is mathematically represented as $r_f = \frac{d_f}{2} = \frac{4}{2} = 2\ m$

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Here $I_{k_1 }$ is the initial moment of inertia of the first astronauts which is equal to $I_{p_1}$ the initial moment of inertia of the second astronauts So

$I_{k_1} = I_{p_1 } = m * r_i^2$

Also $w_{k_1 }$ is the initial angular velocity of the first astronauts which is equal to $w_{p_1}$ the initial angular velocity of the second astronauts So

$w_{k_1} =w_{p_1 } = w_1$

Here $I_{k_2 }$ is the final moment of inertia of the first astronauts which is equal to $I_{p_2}$ the final moment of inertia of the second astronauts So

$I_{k_2} = I_{p_2} = m * r_f^2$

Also $w_{k_2 }$ is the final angular velocity of the first astronauts which is equal to $w_{p_2}$ the final angular velocity of the second astronauts So

$w_{k_2} =w_{p_2 } = w_2$

So

$mr_i^2 w_1 + mr_i^2 w_1 = mr_f^2 w_2 + mr_f^2 w_2$

=> $2 mr_i^2 w_1 = 2 mr_f^2 w_2$

=> $w_f = \frac{2 * m * r_i^2 w_1}{2 * m * r_f^2 }$

=> $w_f = \frac{3.5^2 * 0.5}{ 2^2 }$

=> $w_f = 1.531 \ rad/ s$

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