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zlopas [31]
4 years ago
9

Which of the following is not a property of electromagnetic waves?

Physics
1 answer:
lawyer [7]4 years ago
4 0

Answer: It does not include alpha rays

Explanation: It does not include alpha rays

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It indicates that you should proceed with caution. You should slow down and be prepared to stop to safely navigate this intersection.  
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4 years ago
What is the work Sarah,60kg does against gravity is she goes up a 4m high stairwell in 3 seconds?
ladessa [460]
Work can be obtained from the following expression

W = F*d

Where F is the force (In this case the force it takes to lift Sarah, which is mass*gravity)

And d is the distance traveled.

W = 60kg*9.8m/s^2 * 4m = 2352 Joules 
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3 years ago
In which of the following cases would sound reach each ear out of phase?
notsponge [240]

Answer:

D

Explanation:

Because when you stand directly near the source of sound the distance is less and sound is directly proportional to area.

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3 years ago
A ray of light of vacuum wavelength 550 nm traveling in air enters a slab of transparent material. The incoming ray makes an ang
alisha [4.7K]

Answer:

sin 40 = n sin 26    snell's law for incoming ray in air

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5 0
3 years ago
Coherent light with wavelength 608 nm passes through two very narrow slits, and the interference pattern is observed on a screen
Kipish [7]

Answer:

1.22 \mu m

Explanation:

In the double-slit interference, light passes through a double slit and produce a pattern of alternating bright and dark fringes on a distant screen. This pattern is due to the combined effect of the diffraction of each slit + the interference of the light coming from the two slits.

The condition to observe a maximum (bright fringe), so costructive interference, in the distant screen, is:

y=\frac{m\lambda D}{d}

where:

y is the distance of the m-th maximum from the central fringe

\lambda is the wavelength of the light used

D is the distance of the screen from the slits

d is the separation between the slits

In this problem, we know that:

\lambda=608 nm=608\cdot 10^{-9}m is the wavelength of light used

D=3.00 m is the distance of the screen

y=4.84 mm = 4.84\cdot 10^{-3} m is the distance of the first maximum (first-order bright fringe) from the central pattern, so when

m = 1

Solving for d, we find the separation of the slits:

d=\frac{m\lambda D}{y}=\frac{(1)(608\cdot 10^{-9})(3.00)}{4.84\cdot 10^{-3}}=3.77\cdot 10^{-4} m

The first dark fringe on the screen instead is given by the formula

y'=\frac{(\frac{\lambda'}{2})D}{d}

where

\lambda' is the wavelength of the new light

Here we want the first dark fringe of the new light to be coincident to the first bright fringe of the previous light, so

y=4.84\cdot 10^{-3}m

Therefore, solving for \lambda',

\lambda'=\frac{2y'd}{D}=\frac{2(4.84\cdot 10^{-3})(3.77\cdot 10^{-4})}{3.00}=1.22\cdot 10^{-6} m = 1.22 \mu m

4 0
3 years ago
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