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makvit [3.9K]
2 years ago
12

Why do waves with high frequencies have short wavelengths

Physics
2 answers:
dalvyx [7]2 years ago
7 0

Answer: because  ν = velocity/λ where ν and λ are the frequency and wavelegth of the wave.

Explanation: In order to explain this problem we have to consider the relationship between frequency and wavelengths which  are related by the velocity of the wave as follows ν*λ=v where ν and λ are the frequency and wavelegth of the wave. These parameters have an inverse proportionality.

Then, ν = velocity/λ

makvit [3.9K]2 years ago
3 0

Answer:

The relationship between the wavelength and the frequency of a wave comes from the equation:

v = λ*f

Where v is the velocity of the wave, λ is the wavelength and f is the frequency.

So if we keep the velocity constant, you can see that if we decrease the wavelength, then the frequency must increase (to contrast the decrease in the other variable)

And the same in the other direction, if the wavelength increases, then the frequency must decrease.

So for a fixed velocity, as highest is the frequency, lower will be the wavelength.

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Two identical loudspeakers are some distance apart. A person stands 5.80 m from one speaker and 3.90 m from the other. What is t
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Answer:

f = 632 Hz

Explanation:

As we know that for destructive interference the path difference from two loud speakers must be equal to the odd multiple of half of the wavelength

here we know that

\Delta x = (2n + 1)\frac{\lambda}{2}

given that path difference from two loud speakers is given as

\Delta x = 5.80 m - 3.90 m

\Delta x = 1.90 m

now we know that it will have fourth lowest frequency at which destructive interference will occurs

so here we have

\Delta x = 1.90 = \frac{7\lambda}{2}

\lambda = \frac{2 \times 1.90}{7}

\lambda = 0.54 m

now for frequency we know that

f = \frac{v}{\lambda}

f = \frac{343}{0.54} = 632 Hz

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3 years ago
What are the effects of gravity on earth?​
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Answer:

Without it, we could not survive on Earth. Earth orbits the sun due to the gravity of the sun, which maintains us at a convenient distance from it to enjoy the sun's warmth and light. It keeps the air we need to breathe and our atmosphere in place. Our planet is held together by gravity.

5 0
1 year ago
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Through both refraction and diffraction, the atmosphere alters the apparent speed and, to a lesser extent, the direction of the
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The Ionospheric Effect

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One of the largest errors in GPS positioning is attributable to the atmosphere. The long, relatively unhindered travel of the GPS signal through the virtual vacuum of space changes as it passes through the earth’s atmosphere. Through both refraction and diffraction, the atmosphere alters the apparent speed and, to a lesser extent, the direction of the signal. This causes an apparent delay in the signal's transit from the satellite to the receiver.

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A person is in a room whose walls are maintained at a temperature of 17 °C. The temperature of the person’s skin is 32 °C. The s
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The orbital period of a satellite is 2 × 106 s and its total radius is 2.5 × 1012 m. The tangential speed of the satellite, writ
LenaWriter [7]

The orbital period of the satellite[T] is given as 2*10^{6} S.

The radius of the satellite is given [R] 2.5*10^{12} m.

we are asked here to calculate the tangential speed of the satellite.

Before going to get the solution first we have understand the tangential speed.

The tangential speed of a satellite is given as the speed required to keep the satellite along the orbit. If satellite speed is less than tangential speed,there is the chance of it falling down towards earth. If it is more,then it will deviate from it orbit and can't stick to the orbit further.In a simple way  the tangential speed is the linear speed of an object in a circular path.

Now we have to calculate the tangential speed [V].

Mathematically the tangential speed [V]   written as -

                                V=\frac{2\pi R}{T}

where T is the time period of the satellite and R is the radius of the satellite.

                        V=\frac{2*3.14*10^{12} }{2*10^{6} }

                               = 7.85*10^{6} m/s

There is also another way through which we can get  the solution as explained below-

We know that the tangential speed of a satellite V=\sqrt{\frac{GM}{R^{2} } }

where G is the gravitational constant and M is the mas of central object.

But we know that g=\frac{GM}{R^{2} }

                               ⇒GM=gR^{2}  where g is the acceleration due to gravity of that central object.


Hence    V=\sqrt{\frac{gR^{2} }{R} }

               ⇒   V=\sqrt{gR}

By knowing the value of g due to that central object we can also calculate its tangential speed.

                           

 




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