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Zina [86]
3 years ago
9

Which element does not nomally form chemical bonds?

Physics
1 answer:
Kruka [31]3 years ago
7 0
The element that does not normally form chemical bonds is Xenon.
Reasoning: Xenon is a noble gas and is located in group 18. Noble gases already has a full outer shell of a electron therefore they do not bond, unless with each other.

HOPE THIS HELPS! ^_^
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Two sound waves are traveling through the same medium. They have the same amplitude, wavelength, and direction of travel. If the
Sloan [31]

Answer:

Constructive interference

Explanation:

Here, the medium is same, same wavelength, same frequency, same amplitude and same direction of propagation.

Let the intensity of waves be I which is same for both

The formula for the net intensity is

I^{2}=I_{1}^{2}+I_{2}^{2}+2I_{1}I_{2}Cos\phi

where, Ф be the phase difference

So, I_{R}^{2}=I^{2}+I^{2}+2IICos6\pi

Here, IR is maximum so the interference is constructive in nature.

7 0
4 years ago
How can a magnetic field be used to create an electric current
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Electric current produces a magnetic field. This magnetic field can be visualized as a pattern of circular field lines surrounding a wire. One way to explore the direction of a magnetic field is with a compass, as shown by a long straight current-carrying wire in.

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3 years ago
What is the frequency of highly energetic ul-
Mnenie [13.5K]

Frequency = (speed) / (wavelength)

Frequency = (3 x 10⁸ m/s) / (124 x 10⁻⁹ m)

Frequency = <em>2.42 x 10¹⁵ Hz</em>

8 0
3 years ago
Two loudspeakers are placed on a wall 3.00 m apart. A listener stands 3.00 m from the wall directly in front of one of the speak
Mashutka [201]

Answer:

Part a)

\Delta \phi = 2.2 \pi

Part b)

f = 411.3 Hz

Explanation:

As we know that the observer is standing in front of one speaker

So here the path difference of the two sound waves reaching to the observer is given as

\Delta x = 3\sqrt2 - 3

\Delta x = 1.24 m

now phase difference is related with path difference as

\Delta \phi = \frac{2\pi}{\lambda}(\Delta x)

\Delta \phi = \frac{2\pi}{\lambda}(1.24)

here in order to find the wavelength

\lambda = \frac{c}{f}

\lambda = \frac{340}{300} = 1.13

now we have

\Delta \phi = \frac{2\pi}{1.13}(1.24) = 2.2\pi

Part b)

Now we know that when phase difference is odd multiple of \pi

then in that case the the sound must be minimum

So nearest value for minimum intensity would be

\Delta \phi = 3\pi

so we have

3\pi = \frac{2\pi}{\lambda}(1.24)

so we have

\lambda = 0.827

now we have

\frac{340}{f} = 0.827

f = 411.3 Hz

4 0
3 years ago
How do you relate mas and acceleration?
Shtirlitz [24]

Answer:

yes i relate mass but not acceleration

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3 years ago
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