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GalinKa [24]
3 years ago
13

On a violin, the highest note Mandy can play is an A-note, which produces a sound wave with a high frequency. The lowest note sh

e can play is a G-note, which has a low frequency.
How does the wavelength of a G-note sound wave compare to the wavelength of an A-note sound wave?
The G-note's wavelength is the same size.
The G-note's wavelength is shorter.
The G-note's wavelength is longer.
There is not enough information to tell.
Physics
2 answers:
lora16 [44]3 years ago
6 0
I think the answer would be: The G-note's wavelength is longer

Here are the formula to calculate wavelength

Wavelength = Wave speed/Frequency

Which indicates that the wavelength will become larger as the frequency became smaller.

polet [3.4K]3 years ago
4 0

Its the G-note wavelength is longer

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A proton moves along the x-axis with vx=1.0×107m/s. As it passes the origin, what are the strength and direction of the magnetic
Sunny_sXe [5.5K]

Answer:

Magnetic field will be ZERO at the given position

Explanation:

As we know that the magnetic field due to moving charge is given as

B = \frac{\mu_0 qv sin\theta}{4\pi r^2}

so here we know that for the direction of magnetic field we will use

\hat B = \hat v \times \hat r

so we have

\hat B = \hat i \times (\hat i + 0\hat j + 0\hat k)

so magnetic field must be ZERO

So whenever charge is moving along the same direction where the position vector is given then magnetic field will be zero

3 0
3 years ago
A particle moves with acceleration function a(t) = 5 + 4t - 2t^2. Its initial velocity v(0) = 3 m/s and its initial displacement
eimsori [14]

Answer:

Its position after 4 seconds is 62 meters.

Explanation:

It is given that,

The acceleration of the particle is given by equation :

a(t)=5+4t-2t^2

Also, a=\dfrac{dv}{dt}

v=\int\limits {a.dt}

v=\int\limits {(5+4t-2t^2).dt}

v=5t+2t^2-\dfrac{2}{3}t^3+c

At t = 0, v(0)=3\ m/s. So, c = 3

v=5t+2t^2-\dfrac{2}{3}t^3+3

Also, v=\dfrac{ds}{dt}, s is the position

s=\int\limits {v.dt}

s=\int\limits {(5t+2t^2-\dfrac{2}{3}t^3+3).dt}

s=\dfrac{5}{2}t^2+\dfrac{2}{3}t^3-\dfrac{t^4}{6}+3t+c'

At t = 0, s(0)=10\ m. So, c' = 10

s=\dfrac{5}{2}t^2+\dfrac{2}{3}t^3-\dfrac{t^4}{6}+3t+10

At t = 4 s

s=\dfrac{5}{2}(4)^2+\dfrac{2}{3}(4)^3-\dfrac{(4)^4}{6}+3(4)+10

s = 62 m

So, at t = 4 seconds the position of the particle is 62 meters. Hence, this is the required solution.

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3 years ago
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Sedaia [141]

Hello. You did not present the combinations the question refers to, which makes it impossible for this question to be answered accurately. However, I will try to help you in the best possible way.

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To calculate the acceleration you must use the formula: Force / mass.

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