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podryga [215]
3 years ago
9

A) Find the wavelength of an electromagnetic wave with frequency 9.35 GHz = 9.35 109 Hz (G = giga = 109), which is in the microw

ave range. λ = m
(b) Find the speed of a sound wave in an unknown fluid medium if a frequency of 532 Hz has a wavelength of 2.00 m.
Physics
1 answer:
miss Akunina [59]3 years ago
4 0

Answer:

(a) \lambda=0.032\ m

(b) v = 1064 m/s                                                                

Explanation:

(a) Frequency of electromagnetic wave, f=9.35\ GHz=9.35\times 10^9\ Hz

Let \lambda is the wavelength of electromagnetic wave. It can be calculated as :

c=f\lambda

\lambda=\dfrac{c}{f}

\lambda=\dfrac{3\times 10^8\ m/s}{9.35\times 10^9\ Hz}

\lambda=0.032\ m

(b) Frequency of medium, f = 532 Hz

Wavelength, \lambda=2\ m

Let v is the speed of sound wave in an unknown fluid. Using the relation as :

v=f\lambda

v=532\times 2

v = 1064 m/s

Hence, this is the required solution.

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Answer: 5.72 x 10-3Ω

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Hi, to answer this question, first we have to calculate the cross sectional area of the cable:

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R = (2.82x10-8 x 5.87) / 0.000028937

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The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me
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Answer:

(a):  \rm meter/ second^2.

(b):  \rm meter/ second^3.

(c):  \rm 2ct-3bt^2.

(d):  \rm 2c-6bt.

(e):  \rm t=\dfrac{2c}{3b}.

Explanation:

Given, the position of the particle along the x axis is

\rm x=ct^2-bt^3.

The units of terms \rm ct^2 and \rm bt^3 should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of \rm ct^2=meter

Therefore, unit of \rm c= meter/ second^2.

(b):

Unit of \rm bt^3=meter

Therefore, unit of \rm b= meter/ second^3.

(c):

The velocity v and the position x of a particle are related as

\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.

(d):

The acceleration a and the velocity v of the particle is related as

\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.

(e):

The particle attains maximum x at, let's say, \rm t_o, when the following two conditions are fulfilled:

  1. \rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.
  2. \rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Applying both these conditions,

\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.

For \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c

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Thus, particle does not reach its maximum value at \rm t = 0\ s.

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\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.

Here,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Thus, the particle reach its maximum x value at time \rm t_o = \dfrac{2c}{3b}.

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