Which equations represent the relationship between wavelength and frequency for a sound wave? Check all that apply.

5)

Zina [86]

Answer:

The current through the wire is equal to 0.8 A.

Explanation:

Given that,

The length of a copper wire = 2 m

Potential difference = 24 mV

The current through the wire is 0.40 A.

The new potential difference is 48 mV.

We need to find the current through the wire.

As the potential difference is doubled for second wire. So the new current will be :

I' = 2I

= 2 × 0.4

= 0.80 A

So, the current through the wire is equal to 0.8 A.

5 0

3 years ago

Ryan applied a force of 10N and moved a book 30 cm in the direction of the force. How much was the workdone by Ryan?

Orlov [11]

Force=10N
Displacement=30cm=0.3m

$\\ \sf\bull\longmapsto Wd=Force(Displacement)$

$\\ \sf\bull\longmapsto Work=10(0.3)$

$\\ \sf\bull\longmapsto Work=3J$

5 0

3 years ago

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Two sinusoidal waves, which are identical except for a phase shift, travel along in the same direction. The wave equation of the

Y_Kistochka [10]

Answer:

two sinusoidal waves, which are identical except for a phase shift, travel along in the same direction. The wave equation of the resultant wave is yR (x, t) = 0.70 m sin⎛ ⎝3.00 m−1 x − 6.28 s−1 t + π/16 rad⎞ ⎠ . What are the angular frequency, wave number, amplitude, and phase shift of the individual waves?

ω = 6.28 s − 1 ,

k = 3.00 m− 1 ,

φ = π rad,

A R = 2 A cos (φ 2 ) ,

A = 0.37 m

Explanation:

y1 ( x , t ) = A sin( k x − ω t +φ ) ,

y 2 ( x , t ) = A sin ( k x − ω t ) .

from the principle of superposition which states that when two or more waves combine, there resultant wave is the algebriac sum of the individual waves

y1 ( x , t ) = A sin( k x − ω t +φ ) , is generaL form of thw wave eqaution

A=amplitude

k=angular wave number

ω=angular frequency

φ =phase constant

k=2π/lambda

ω=2π/T

yR (x, t) = 0.70 m sin{3.00 m−1 x − 6.28 s−1 t + π/16 rad}....................*

two waves superposed to give the above, assuming they are moving in the +x direction

y1 ( x , t ) = A sin( k x − ω t +φ ) , .....................1

y 2 ( x , t ) = A sin ( k x − ω t ) ...........................2

adding the two equation will give

A sin( k x − ω t +φ )+A sin ( k x − ω t ) .................3

A( sin( k x − ω t +φ )+ sin ( k x − ω t ) ),......................4

similar to the following trigonometry identity

sina+sinb=2cos(a-b)/2sin(a+b)/2

let a= ( k x − ω t

b=k x − ω t +φ )

y(x,t)=2Acos(φ/2)sin(k x − ω t +φ/2)

k=3m^-1

lambda=2π/k=2.09m

ω=6.28= T=2π/6.28

T=1s

φ/2=π/16

φ=π/8rad

amplitude

2Acos(φ/2)=0.70 m

A=0.7/2cos(π/8)

A=0.37 m

6 0

4 years ago

A speaker fixed to a moving platform moves toward a wall, emitting a steady sound with a frequency of 205 Hz. A person on the pl

Arlecino [84]

Answer:

Explanation:

The question relates to Doppler effect and beat.

The observer is moving towards the reflected sound so apparent frequency will be increased

f = f₀ x (V + v₁) / (V - v₂)

f is apparent frequency , f₀ is real frequency , V is velocity of sound , v₁ is velocity of observer and v₂ is velocity of source . Here

v₁ = v₂ = vp as both observer and source have same velocity

f = f₀ x (V + v₁) / (V - v₂)

205 +5 = 205 x (344 +vp)/ ( 344 - vp)

1.0234 = (344 +vp)/ ( 344 - vp)

= 352 - 1.0234vp = 340+vp

12 / 2.0234vp

vp = 6 m /s approx.

6 0

3 years ago

A very long cylinder, of radius a, carries a uniform polarization P perpendicular to its axis. Find the electric field inside th

Juliette [100K]

Answer:

For electric field inside cylinder, check image 02 attached

For electric field outside cylinder, check image 03 attached

Explanation:

Let's consider the polarized cylinder as superposition of two cylinders with opposite,equal, uniform charge densities in a way shown in the figure in the "image 01"d attached ;

In general, if we have an object with polarization (P¬) , then we have to take two objects with similar shape to the system, with opposite, equal, and uniform charge densities and then we super-impose these two objects in such a way that the total dipole moment of this superimposed system is equal to the total dipole moment of original system.

Now, we can take the super- imposed system as equivalent to the original system for calculating electric field and potential.

Therefore,

For the electric field inside the cylinder, check the solution in "image 02" i attached

For the electric field outside the cylinder, check, "image 03" i attached.

4 0

3 years ago

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