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SpyIntel [72]
3 years ago
12

A whistle of frequency 564 Hz moves in a circle of radius 71.2 cm at an angular speed of 17.1 rad/s. What are (a) the lowest and

(b) the highest frequencies heard by a listener a long distance away, at rest with respect to the center of the circle? (Take the speed of sound in air to be 343 m/s.)
Physics
1 answer:
trapecia [35]3 years ago
3 0

Answer:

a) f'=544.66 \textup{Hz}

b) f'=584.75 \textup{Hz}

Explanation:

Given:

Frequency of the whistle, f = 564 Hz

Radius of the circle, r = 71.2 cm = 0.712 m

Angular speed, ω = 17.1 rad/s

speed of source, v_s = rω = 0.712 × 17.1 = 12.1752 m/s

speed of sound, v = 343 m/s

Now, applying the Doppler's effect formula, we have

f'=f\frac{v\pm v_d}{v\pm v_s}

where,

v_d = relative speed of the detector with respect to medium = 0

a) for lowest frequency, we have the formula as:

f'=f\frac{v}{v+v_s}

on substituting the values, we get

f'=564\times\frac{343}{343+12.1752}

or

f'=544.66 \textup{Hz}

b) for maximum frequency, we have the formula as:

f'=f\frac{v}{v-v_s}

on substituting the values, we get

f'=564\times\frac{343}{343-12.1752}

or

f'=584.75 \textup{Hz}

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Answer:

Explanation:

Considering non - relativistic approach : ----

Speed of electron = 1 % of speed of light

= .01 x 3 x 10⁸ m /s

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V x e = kinetic energy of electron

V x 1.6 x 10⁻¹⁹ = 40.95 x 10⁻¹⁹

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An electron has a negative electric charge.

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A thin rod of length 1.4 m and mass 140 g is suspended freely from one end. It is pulled to one side and then allowed to swing l
m_a_m_a [10]

Answer:

a The kinetic energy is  KE = 0.0543 J

b The height of the center of mass above that position is  h = 1.372 \ m    

Explanation:

From the question we are told that

  The length of the rod is  L = 1.4m

   The mass of the rod m = 140 = \frac{140}{1000} = 0.140 \ kg  

   The angular speed at the lowest point is w = 1.09 \ rad/s

Generally moment of inertia of the rod about an axis that passes through its one end is

                   I = \frac{mL^2}{3}  

Substituting values

               I = \frac{(0.140) (1.4)^2}{3}

               I = 0.0915 \ kg \cdot m^2

Generally the  kinetic energy rod is mathematically represented as

             KE = \frac{1}{2} Iw^2

                    KE = \frac{1}{2} (0.0915) (1.09)^2

                           KE = 0.0543 J

From the law of conservation of energy

The kinetic energy of the rod during motion =  The potential energy of the rod at the highest point

   Therefore

                   KE = PE = mgh

                        0.0543 = mgh

                             h = \frac{0.0543}{9.8 * 0.140}

                                h = 1.372 \ m    

                 

5 0
3 years ago
Read 2 more answers
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