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Helga [31]
3 years ago
6

What evidence do you that suggest water waves are transverse wave​

Physics
1 answer:
svp [43]3 years ago
5 0

Answer:

If you throw a pebble into a pond, ripples

spread out from where it went in. These

ripples are waves travelling through the

water. The waves move with a transverse

motion.

Explanation:

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Answer:

The pump puts more gas particles inside the ball

Explanation:

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Knowing the constant g what will the gravitational force between two masses be if the gravitational force between them is 36n an
charle [14.2K]
The gravitational force between two masses is given by:
F=G \frac{m_1 m_2}{r^2}
where
G is the gravitational constant
m1 and m2 are the two masses
r is the separation between the two masses

We see that the force is proportional to the inverse of the square of the distance: F \sim  \frac{1}{r^2}
therefore, if the distance is tripled:
r'=3r
The force decreases by a factor 1/9:
F \sim  \frac{1}{(3r)^2}= \frac{1}{9}  \frac{1}{r^2}

Since the original force was 36 N, the new force will be
F' =  \frac{1}{9} (36 N)= 4 N
6 0
3 years ago
A flat sheet of ice has a thickness of 1.4 cm. It is on top of a flat sheet of crown glass that has a thickness of 3.0 cm. Light
MAXImum [283]

Answer:

t = 2.13 10-10 s , d = 6.39 cm

Explanation:

For this exercise we use the definition of refractive index

        n = c / v

Where n is the refraction index, c the speed of light and v the speed in the material medium.

The refractive indices of ice and crown glass are 1.13 and 1.52, respectively, therefore the speed of the beam in the material medium is

        v = c / n

As the beam strikes perpendicularly, the beam path is equal to the distance of the leaves, there is no refraction, so we can use the uniform motion relationships

        v = d / t

        t = d / v

        t = d n / c

Let's look for the times on each sheet

Ice

        t₁ = 1.4 10⁻² 1.31 / 3 10⁸

        t₁ = 0.6113 10⁻¹⁰ s

Crown glass (BK7)

        t₂ = 3.0 10⁻² 1.52 / 3.0 10⁸

        t₂ = 1.52 10⁻¹⁰ s

Time is a scalar therefore it is additive

         t = t₁ + t₂

         t = (0.6113 + 1.52) 10⁻¹⁰

         t = 2.13 10-10 s

The distance traveled by this time in a vacuum would be

        d = c t

       d = 3 10⁸ 2.13 10⁻¹⁰

       d = 6.39 10⁻² m

       d = 6.39 cm

3 0
3 years ago
Question 4. A tuning fork ‘A’ produces 6 beats/sec with another fork ‘B’ of un-known frequency. On
8090 [49]

Clever problem.

We know that the beat frequency is the DIFFERENCE between the frequencies of the two tuning forks.  So if Fork-A is 256 Hz and the beat is      6 Hz, then Fork-B has to be EITHER 250 Hz OR 262 Hz.  But which one is it ?

Well, loading Fork-B with wax increases its mass and makes it vibrate SLOWER, and when that happens, the beat drops to 5 Hz.  That means that when Fork-B slowed down, its frequency got CLOSER to the frequency of Fork-A ... their DIFFERENCE dropped from 6 Hz to 5 Hz.

If slowing down Fork-B pushed it CLOSER to the frequency of Fork-A, then its natural frequency must be ABOVE Fork-A.

The natural frequency of Fork-B, after it gets cleaned up and returns to its normal condition, is 262 Hz.  While it was loaded with wax, it was 261 Hz.

4 0
3 years ago
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