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loris [4]
3 years ago
9

Sonar signals and infrared light are are used to send messages to submarines deep under water. If we compare the two signals, wh

ich statement describes a difference between the two?
Physics
1 answer:
WARRIOR [948]3 years ago
6 0
What are the statement choices?
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Describe and explain the motion of a small ball floating on a pond when waves travel across the pond
laila [671]
I think the answer is periodic motion.
4 0
3 years ago
A student notices that an inflated balloon gets larger when it is warmed by a lamp. Which best describes the mass of the balloon
nirvana33 [79]
It gets larger because
well let me give you an example
so today in class we looked at a lava lamp with wax inside and there was a lightbulb at the bottom.
we watched as the wax floated up because the molecules inside the wax spreads out and makes the wax less dense.
the wax floats up because (which is related to the balloon getting bigger) the wax is getting less dense and the particles get bigger which ALSO makes the wax less dense.
hope this helps and hope you can relate it to your problem! say thanks if I did help AT ALL! :)
7 0
3 years ago
Read 2 more answers
The charge of an electron is
AlladinOne [14]
Proton positive; electron negative; neutron no charge<span>. </span>The charge<span> on the proton and </span>electron<span> are exactly the same size but opposite. The same number of protons and </span>electrons<span> exactly cancel one another in a neutral atom.
</span> 
hoped it helped
6 0
3 years ago
Radio waves transmitted through space at 3.00 ✕ 108 m/s by the Voyager spacecraft have a wavelength of 0.137 m. What is their fr
fredd [130]

Answer:

Frequency, f=2.18\times 10^9\ Hz

Explanation:

We have,

Speed of radio waves is 3\times 10^8\ m/s

Wavelength of radio waves is \lambda=0.137\ m

It is required to find the frequency of the radio waves. The speed of a wave is given by :

v=f\lambda\\\\f=\dfrac{v}{\lambda}\\\\f=\dfrac{3\times 10^8}{0.137}\\\\f=2.18\times 10^9\ Hz

So, the frequency of the radio wave is 2.18\times 10^9\ Hz.

8 0
2 years ago
A student measures the speed of yellow light in water to be 2.00x10^8
max2010maxim [7]

NOTE: The given question is incomplete.

<u>The complete question is given below.</u>

A student measures the speed of yellow light in water to be 2.00 x 10⁸ m/s. Calculate the speed of light in air.

Solution:

Speed of yellow light in water (v) = 2.00 x 10⁸ m/s

Refractive Index of water with respect to air (μ) = 4/3

Refractive Index = Speed of yellow light in air / Speed of yellow light in water

Or,  The speed of yellow light in air = Refractive Index × Speed of yellow light in water

or,                                           = (4/3) × 2.00 x 10⁸ m/s

or,                                           = 2.67 × 10⁸ m/s ≈ 3.0 × 10⁸ m/s

Hence, the required speed of yellow light in the air will be 3.0 × 10⁸ m/s.

7 0
3 years ago
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