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Annette [7]
4 years ago
9

Assume a form y1 = A sin(α) for the transverse displacement of the string. Enter an expression for α of a transverse wave on a s

tring traveling along the positive x-direction in terms of its wavenumber k, the position x, its angular frequency ω, and the time t?
Physics
1 answer:
maria [59]4 years ago
4 0

Answer:

α = 2Πft + 2Πx/¶

Explanation:

A transverse wave is one of the type of electromagnetic waves which occurs when the vibration of the medium is perpendicular to the direction of the wave. Mathematically, the wave equation according to the question

y1 = Asin(α).... (1)

since its characteristic wave gives a sine wave.

y1 is the vertical displacement

A is the amplitude.

The expression for α of the transverse wave on a string traveling along the positive x-direction in terms of its wavenumber k, the position x, its angular frequency ω, and the time t can be given as;

α = ωt+theta ... (2)

theta is the phase angle. This phase angle is a function of the position x of the wave and the wave number.

theta = kx

where k = 2Πx/¶... (3)

¶ is the wavelength.

ω which is the angular frequency is a function of the frequency (f) of the wave and it is given as;

ω = 2Πf... (4)

Substituting equation 3 and 5 into 2;

α = 2Πft + 2Πx/¶

Therefore,

y1 = Asin(2Πft + 2Πx/¶)

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3 years ago
What is the saturated zone
uranmaximum [27]

Answer:

Saturated zone is area below the water table in which the soil is completely saturated with groundwater.

Explanation:

The saturated zone lies below the ground. It is mainly the lower zone of rock along with the water table where pore spaces are completely filled with water. Even the saturated zone is sometimes separated into 2 subzones: the phreatic zone and the capillary fringe.

The area where pores spaces are not saturated with water is also unsaturated zone. Localized saturated zones can occur within the unsaturated zone. The unsaturated zone lies above the groundwater table.

6 0
3 years ago
A flywheel with radius of 0.400 mm starts from rest and accelerates with a constant angular acceleration of 0.600 rad/s2rad/s2.
charle [14.2K]

Answer: 0.00024\ m/s^2

Explanation:

Given

Radius of flywheel is r=0.4\ mm

Angular acceleration \alpha=0.6\ rad/s^2

For no change in radius, tangential acceleration is  given as

\Rightarrow a_t=a\lpha \times r

Insert the values

\Rightarrow a_t=0.6\times 0.4\times 10^{-3}\ m/s^2\\\Rightarrow a_t=2.4\times 10^{-4}\ m/s^2\ \text{or}\ 0.00024\ m/s^2

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An object at rest on a flat, horizontal surface explodes into two fragments, one seven times as massive as the other. The heavie
Veronika [31]

Answer:

Explanation:

Given that,

One fragment is 7 times heavier than the other

Let one fragment mass be M

Let this has a velocity v

And the other 7M

And this a velocity V

Initially the fragment is at rest u = 0

Applying conservation of momentum

Momentum is given as p=mv

Initial momentum = final momentum

Po = Pf

(M+7M) × 0 = 7M •V − Mv

0 = 7M•V - Mv

Divide both sides by M

0 = 7V -v

v = 7V

Since friction decelerates the masses to zero speed, we can calculate the NET work on the individual blocks and relate this quantity to the change in kinetic energy of each block

The workdone by the 7M mass is

Distance moved by 7M mass is 6.8m, Then, d =6.8m

W = fr × d

Where fr = µkN

When N=W =mg, where m=7M

N= 7Mg

fr = −µk × 7mg

Then, W(7m) = −7µk•Mg×d

W(7m) = −7µk•Mg×6.8

W(7m) = −47.6 µk•Mg

Then, same procedure,

Let distance move by the small mass be m

Work done by M mass

W(m) = −µk•Mg×d'

Since it is a wordone by friction, that is why we have a negative sign.

Using conservation of energy

Work done by 7M mass is equal to work done by M mass

W(7m) = W(m)

−47.6 µk•Mg = −µk•Mg×d

Then, M, g and µk cancels out

We are left with

-46.7 = -d

Then, d = 46.7m

7 0
3 years ago
Read 2 more answers
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vlabodo [156]

Answer: 90\°

Explanation:

The Compton Shift \Delta \lambda in wavelength when the photons are scattered is given by the following equation:

\Delta \lambda=\lambda_{c}(1-cos\theta)     (1)

Where:

\lambda_{c}=2.43(10)^{-12} m is a constant whose value is given by \frac{h}{m_{e}c}, being h the Planck constant, m_{e} the mass of the electron and c the speed of light in vacuum.

\theta) the angle between incident phhoton and the scatered photon.

We are told the maximum Compton shift in wavelength occurs when a photon isscattered through 180\°:

\Delta \lambda_{max}=\lambda_{c}(1-cos(180\°))     (2)

\Delta \lambda_{max}=\lambda_{c}(1-(-1))    

\Delta \lambda_{max}=2\lambda_{c}     (3)

Now, let's find the angle that will produce a fourth of this maximum value found in (3):

\frac{1}{4}\Delta \lambda_{max}=\frac{1}{4}2\lambda_{c}(1-cos\theta)      (4)

\frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}(1-cos\theta)      (5)

If we want \frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}, 1-cos\theta   must be equal to 1:

1-cos\theta=1   (6)

Finding \theta:

1-1=cos\theta

0=cos\theta  

\theta=cos^{-1} (0)  

Finally:

\theta=90\°    This is the scattering angle that will produce \frac{1}{4}\Delta \lambda_{max}      

7 0
3 years ago
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