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Kazeer [188]
3 years ago
14

Hans observed properties of four different waves and recorded observations about each one in his chart.

Physics
2 answers:
bogdanovich [222]3 years ago
8 0
The answer to your question is (Wave W is being diffracted, Wave X is being reflected, and waves Y and Z are being refracted.)


HOPE THIS HELPS YOU CAUSE IM DOING TEST AND I HOPE IM RIGHT AS WELL

PLZ GIVE ME THE BRAINLIEST ANSWER HAVE A GOOD DAY




weeeeeb [17]3 years ago
6 0

The correct answer is Wave W is being diffracted, Wave X is being reflected, and waves Y and Z are being refracted.

Wave X is bouncing off the surface thats why it is being reflected. As the waves pass from one medium to another they change speed i.e. when passing through water into air. This causes waves to bend and change color hence waves X and Z. The scattering of waves through a small opening is called diffraction so wave W

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A. True

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2 years ago
Which statement is NOT true?
Nuetrik [128]

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4 0
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Suppose that instead of being inclined to Earth's orbit around the Sun, the Moonâs orbit was in the same plane as Earthâs orbit
ycow [4]

Answer:

c) 12

Explanation:

A Solar eclipse occurs when The Sun, The Earth and The Moon comes in a straight line with the Moon being in between the Earth and the Sun. At this point the Moon appears to block the Sun and Moon's shadow falls on Earth. This would occur only on the day of the New Moon.

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6 0
3 years ago
An electron emitted in the beta decay of bismuth-210 has a mean kinetic energy of 390 keV. (a) Find the de Broglie wavelength of
Sauron [17]

Explanation:

Given that,

The mean kinetic energy of the emitted electron, E=390\ keV=390\times 10^3\ eV

(a) The relation between the kinetic energy and the De Broglie wavelength is given by :

\lambda=\dfrac{h}{\sqrt{2meE}}

\lambda=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 1.6\times 10^{-19}\times 390\times 10^3}}

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n\lambda=2d\ sin\theta

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For nickel, d=0.092\times 10^{-9}\ m

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\theta=sin^{-1}(\dfrac{1.96\times 10^{-12}}{2\times 0.092\times 10^{-9}})

\theta=0.010^{\circ}

As the angle made is very small, so such an electron is not useful in a Davisson-Germer type scattering experiment.

4 0
3 years ago
To the nearest tenth, what is the area of the shaded segment when JA=8ft ??
Murrr4er [49]
A=\dfrac{1}{6}\pi 8^2-\dfrac{8^2\sqrt{3}}{4}\approx 5.8\;[ft^2]
Answer: <span>C. 5.8 ft squared</span>
7 0
3 years ago
Read 2 more answers
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