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guapka [62]
3 years ago
7

Whenever a musician plays a guitar, they pluck one of the guitar strings to produce a standing wave in the string. Then pinch do

wn on the string at the other end at the guitar in different places along the neck of the instrument. In doing so, by changing where they pinch the string, the same standing wave on that string can produce multiple different frequencies, and therefore multiple different notes. Using the knowledge you have learned about harmonic frequencies set up on strings, explain why simply changing where you pinch the string can change the frequency.
Physics
1 answer:
UkoKoshka [18]3 years ago
7 0

Answer:

It's due to the distance from either ends of strings origin...

Explanation:

As we know that waves behave moving in a flow from one side to another side and this gives a prospective of motion. Suppose a wave is pinched from the near one end of a guitar then due to the distortion created by the point of tie of strings the wave super imposes and moves with a velocity v and produces a wave frequency f. as we the pinching go down to the center the wave stabilizes itself to a stationary origin right at the center and the frequency then changes accordingly as moving down on the string.

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Consider an electron in an infinite well of width 0.7 nm . What is the wavelength of a photon emitted when the electron in the i
ollegr [7]

Answer:

 λ  = 538.0 nm

Explanation:

The solution of the Schrödinger equation for the inner part of the well gives energy

      E_{n} = (h² / 8mL²) n²

Where n is an integer and L is the length of the well

They ask for the transition from the first excited state n = 2 to the base state n = 1

       E₂ - E₁ = = (h² / 8mL²) (n₂² - n₁²)

Let's calculate

       E₂-E₁ = (6.63 10⁻³⁴)² / (8 9.1 10⁻³¹ (0.7 10⁻⁹)²) (2² -1²)

       E₂ –E₁ = 3.6968 10⁻¹⁹ J

Let's use the Planck equation

      E = h f

      c = λ f

      E = h c / λ

      E = E₂ ₂- E₁

      h c / λ  = 3.6968 10⁻¹⁹

      λ  = h c / (E₂-E₁)

      λ  = 6.63 10⁻³⁴ 3 10⁸ / 3.6968 10⁻¹⁹

      λ  = 5.380 10⁻⁷ m

Let's reduce

      λ  = 5.380 10⁻⁷ m (10 9 nm / 1 m)

      λ  = 538.0 nm

4 0
4 years ago
Each plate of a parallel‑plate capacitor is a square of side 4.19 cm, 4.19 cm, and the plates are separated by 0.407 mm. 0.407 m
alexandr1967 [171]

Answer:

The electric field strength inside the capacitor is 49880.77 N/C.

Explanation:

Given:

Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m

Separation between the plates (d) = 0.407 mm = 0.407\times 10^{-3}\ m

Energy stored in the capacitor (U) = 7.87\ nJ=7.87\times 10^{-9}\ J

Assuming the medium to be air.

So, permittivity of space (ε) = 8.854\times 10^{-12}\ F/m

Area of the square plates is given as:

A=a^2=(0.0419\ m)^2=1.75561\times 10^{-3}\ m^2

Capacitance of the capacitor is given as:

C=\dfrac{\epsilon A}{d}\\\\C=\frac{8.854\times 10^{-12}\ F/m\times 1.75561\times 10^{-3}\ m^2 }{0.407\times 10^{-3}\ m}\\\\C=3.819\times 10^{-11}\ F

Now, we know that, the energy stored in a parallel plate capacitor is given as:

U=\frac{CE^2d^2}{2}

Rewriting in terms of 'E', we get:

E=\sqrt{\frac{2U}{Cd^2}}

Now, plug in the given values and solve for 'E'. This gives,

E=\sqrt{\frac{2\times 7.87\times 10^{-9}\ J}{3.819\times 10^{-11}\ F\times (0.407\times 10^{-3})^2\ m^2}}\\\\E=49880.77\ N/C

Therefore, the electric field strength inside the capacitor is 49880.77 N/C

8 0
4 years ago
as a mercury atom absorbs a photon of energy as electron in the atom changes from energy level B to energy level E. calculate th
Setler79 [48]

Answer:

2.00x 10 14th Hz

Explanation:

8 0
3 years ago
Read 2 more answers
A researcher predicted that talking to plants enhances their growth. She gave 24 plants the same amount of water and sunlight. S
Ne4ueva [31]

Answer:

Growth of the plants

Explanation:

In an experiment a logical set of attributes of an object is changed and the logical set of attributes which is affected by the other attribute change is noted. The set of attributes which are logical are called variables.

There are two types of variables are

The independent variable is the input. The output is observed with respect to the input.

Dependent variables are the output which is the result of the change in the independent variables

Here, the researcher wants to know how talking to plants affects their growth. The input is talking to plants and the output is the growth of the plants.

Hence, the independent variable is the growth of the plants.

3 0
3 years ago
Describe manipulated Variable
Delicious77 [7]
A manipulated variable is the independent variable in an experiment. It’s called “manipulated” because it’s the one you can change. In other words, you can decide ahead of time to increase it or decrease it. In an experiment you should only have one manipulated variable at a time. The manipulated variable is the independent variable in an experiment.
6 0
3 years ago
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