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3 years ago

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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]3 years ago

4 0

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

Wittaler [7]3 years ago

4 0

The answer is A, 5.13 x 10^19 hertz

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A point charge of 4.0 µC is placed at a distance of 0.10 m from a hard rubber rod with an electric field of 1.0 × 103 . What is

yuradex [85]

Since

Electric potential energy = qV

Where V = Ed

Hence
Electric potential energy = q(Ed) --- (1)

Since E = 1.0 * 10^3 N/C
d = 0.10 m
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Plug in the values in (1)
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Electric potential energy = 400 μJ

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3 years ago

Which word means "to get away from something"? <br> a become b dream c escape d practice

klemol [59]

The correct answer is - c. escape.

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3 years ago

(a) Write an equation describing a sinusoidal transverse wave traveling on a cord in the positive of a y axis with an angular wa

Vedmedyk [2.9K]

Missing data: the wave number

$k=60 cm^{-1}$

(a) $z = 0.003 sin (6000y-31.4 t)$

For a transverse wave travelling in the positive y-direction and with vibration along the z-direction, the equation of the wave is

$z = A sin (ky-\omega t)$

where

A is the amplitude of the wave

k is the wave number

$\omega$ is the angular frequency

t is the time

In this situation:

A = 3.0 mm = 0.003 m is the amplitude

$k = 60 cm^{-1} = 6000 m^{-1}$ is the wave number

$T = 0.20 s$ is the period, so the angular frequency is

$\omega=\frac{2\pi}{T}=\frac{2\pi}{0.20}=31.4 rad/s$

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$z = 0.003 sin (6000y-31.4 t)$

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$v_t = z' = -A \omega cos (ky-\omega t)$

So the maximum transverse wave occurs when the cosine term is equal to 1, therefore the maximum transverse speed must be

$v_{t}_{max} =\omega A$

where

$\omega = 31.4 rad/s\\A = 0.003 m$

Substituting,

$v_{t}_{max}=(31.4)(0.003)=0.094 m/s$

(c) 5.24 mm/s

The wave speed is given by

$v=f \lambda$

where

f is the frequency of the wave

$\lambda$ is the wavelength

The frequency can be found from the angular frequency:

$f=\frac{\omega}{2\pi}=\frac{31.4}{2\pi}=5 Hz$

While the wavelength can be found from the wave number:

$\lambda = \frac{2\pi}{k}=\frac{2\pi}{6000}=1.05\cdot 10^{-3} m$

Therefore, the wave speed is

$v=(5)(1.05\cdot 10^{-3} )=5.24 \cdot 10^{-3} m/s = 5.24 mm/s$

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Bezzdna [24]

Let’s do this together!

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3 years ago