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vodka [1.7K]
3 years ago
14

An avant-garde composer wants to use the Doppler effect in his new opera. As the soprano sings, he wants a large bat to fly towa

rd her from the back of the stage. The bat will be outfitted with a microphone to pick up the singer's voice and a loudspeaker to rebroadcast the sound toward the audience. The composer wants the sound the audience hears from the bat to be, in musical terms, one half-step higher in frequency than the note they are hearing from the singer. Two notes a half-step apart have a frequency ratio of 21/12 = 1.059. With what speed must the bat fly toward the singer?
Physics
1 answer:
Brilliant_brown [7]3 years ago
6 0

Answer:

18.94  m/s

Explanation:

The bat becomes the source of sound . Source is moving towards observer

apparent frequency = actual frequency x (V / V - V_s)

V is velocity of sound and V_s is velocity of source or bat.

apparent frequency / actual frequency = V / (V - V_s)

21 / 12 = 340 / (340 - V_s)

1.059(340 - V_s) = 340

340 x .059 = 1.059 V_s

V_s = 18.94  m/s

=

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8 0
1 year ago
Please answer this question for me and explain why.
horsena [70]

Answer:

D.None of these

Explanation:

The derivation of acceleration formula:

Let us call the 5kg mass m_2 and the 4kg mass m_1. If the tension in the string is T then for the mass m_2

(1). T-m_2g=-m_2a <em>(the negative sign on the right side indicates that acceleration is downwards)</em>

And for the mass m_1

(2). T-m_1g =m_1a<em> (the acceleration is upwards, hence the positive sign)</em>

Solving for T in the 2nd equation we get:

T =m_1a+m_1g,

and putting this into the 1st equation we get:

m_1a+m_1g-m_2g=-m_2a\\\\m_1a+m_2a = m_2g-m_1g\\\\a(m_1+m_2)= (m_2-m_1)g

\boxed{a= \dfrac{(m_2-m_1)}{(m_1+m_2)} g}

Back to the question:

Using the formula for the acceleration we find

a= \dfrac{(5kg-4kg)}{(5kg+4kg)} g

a = \dfrac{g}{9},

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7 0
3 years ago
1.)Two objects, one of m=20,000 kg, and another of 12,500 kg, are placed at a distance of 5 meters apart. What is the force of g
Delvig [45]

1) 6.67\cdot 10^{-4} N

The force of gravitation between the two objects is given by:

F=G\frac{m_1 m_2}{r^2}

where

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m1 = 20,000 kg is the mass of the first object

m2 = 12,500 kg is the mass of the second object

r = 5 m is the distance between the two objects

Substituting the numbers inside the equation, we find

F=(6.67\cdot 10^{-11})\frac{(20,000 kg)(12,500 kg)}{(5 m)^2}=6.67\cdot 10^{-4} N


2)  2.7\cdot 10^{-3} N

From the formula in exercise 1), we see that the force is inversely proportional to the square of the distance:

F \sim \frac{1}{r^2}

this means that if we cut in a half the distance without changing the masses, the magnitude of the forces changes by a factor

F'\sim \frac{1}{(r/2)^2}=4 \frac{1}{r^2}=4F

So, the gravitational force increases by a factor 4. Therefore, the new force will be

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3)  12.5 Nm

The torque is equal to the product between the magnitude of the perpendicular force and the distance between the point of application of the force and the centre of rotation:

\tau=Fd

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d = 0.5 m is the distance between the force and the center

By using the equation, we find

\tau=(25 N)(0.5 m)=12.5 Nm


4) 0.049 kg m^2/s

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L=I\omega

In this problem, we have

I=0.007875 kgm^2

\omega=6.28 rad/s

So, the angular momentum is

L=I\omega=(0.007875 kgm^2)(6.28 rad/s)=0.049 kg m^2/s

6 0
2 years ago
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Natali5045456 [20]
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3 years ago
A small airplane flies 780 miles with an average speed of 260 miles per hour. 1.5 hours after the plane leaves, a Boeing 747 lea
Lera25 [3.4K]

Answer:

520 miles per hour

Explanation:

Let the speed of the Boeing 747 be x miles per hour.

The small airplane covers distance of 780 miles with speed 260 miles per hour.

Also,

After 1.5 hours the Boeing 747 leave the same place and reaches at same time. Both covered distance of 780 miles.

So,

<u>Time taken by Boeing 747 + 1.5 hours = Time taken by small plane.</u>

Also,

Time =  Distance/ speed

So,

780 / x + 1.5 = 780/ 260

Solving for x, we get:

<u>x = 520 miles per hour</u>

6 0
3 years ago
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