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IgorLugansk [536]
3 years ago
8

A physics professor demonstrates the Doppler effect by tying a 600 Hz sound generator to a 1.0-m-long rope and whirling it aroun

d her heard in a horizontal circle at 100 rpm. What are the highest and lowest frequencies heard by a student in the classroom? Assume the room temperature is 20∘C.
Physics
1 answer:
nalin [4]3 years ago
8 0

Answer:

Therefore the the highest frequency is 620Hz and lowest frequency is 580Hz

Explanation:

Given data

Source Frequency fs=600Hz

Length r=1.0m

RPM=100 rpm

The speed of the generator is calculated as:

v_{s}=rw\\v_{s}=r(2\pi f)

Substitute the given values

v_{s}=(1.0m)2\pi (\frac{100}{60}rev/s )\\v_{s}=10.47m/s

For approaching generator the frequency is calculated as:

f_{+}=\frac{f_{s}}{1-\frac{v_{s}}{v} }\\f_{+}=\frac{600Hz}{1-\frac{10.47m/s}{343m/s} } \\f_{+}=620Hz

On the other hand,for the receding generator,Doppler's effect is expressed as:

 f_{-}=\frac{f_{s}}{1+\frac{v_{s}}{v} }\\f_{-}=\frac{600Hz}{1+\frac{10.47m/s}{343m/s} } \\f_{-}=580Hz

Therefore the the highest frequency is 620Hz and lowest frequency is 580Hz

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raketka [301]

Answer:

The maximum height reached is 4.63 m.

Explanation:

Given:

Initial speed of the man (u) = 31.0 m/s

Speed at the top of trajectory (u_x) = 29.5 m/s

Acceleration due to gravity (g) = 9.8 m/s²

When the man reaches the top of the trajectory, the vertical component of velocity becomes zero and hence only horizontal component of velocity acts on him.

Also, since there is no net force acting in the horizontal direction, the acceleration is zero in the horizontal direction from Newton's second law. Thus, the horizontal component of velocity always remains the same.

So, speed at the top of trajectory is nothing but the horizontal component of initial velocity.

Now, initial velocity can be rewritten in terms of its components as:

u^2=u_x^2+u_y^2

Where, u_x\ and\ u_y are the initial horizontal and vertical velocities of the man.

Now, plug in the given values and simplify. This gives,

(31.0)^2=(29.5)^2+u_y^2\\\\961=870.25+u_y^2\\\\u_y^2=961-870.25\\\\u_y^2=90.75\ m^2/s^2--------1

Now, we know that, for a projectile motion, the maximum height is given as:

H=\frac{u_y^2}{2g}

Plug in the value from equation (1) and 9.8 for 'g' to solve for 'H'. This gives,

H=\frac{90.75}{2\times 9.8}\\\\H=4.63\ m

Therefore, the maximum height reached is 4.63 m.

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sergejj [24]
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olasank [31]

Answer:

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Answer:

W = - 5.01 10¹⁰ J

Explanation:

Work is defined by the expression

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     W = G m₁ m₂2(-1 / r)

We evaluate between the lower limits r = Re and upper r = ∞

     W = G m₁m₂ (-1 / Re + 1 / ∞)

     W = - G m₁ m₂ / Re

Let's calculate

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Explanation:

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