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Alja [10]
3 years ago
7

Two trumpet players are trying to tune their instruments. When they are in tune, they will both be playing the same note and no

beats will be heard. The more experienced player plays a note with a frequency of 520 Hz. When the second musician begins to play, the note sounds lower and they can hear 20 beats in 4.0 s. With what frequency is the second trumpeter playing?
Physics
1 answer:
marusya05 [52]3 years ago
8 0

Answer:

The second trumpeter will be playing at frequency = 515 Hz

Explanation: Given that the note sounds lower and they can hear 20 beats in 4.0 s. 

Beat frequency = 20/4 = 5 Hz

Beat frequency = F2 - F1

5 = 520 - F1

F1 = 520 - 5

F1 = 515 Hz

Since the note sound lower, the second trumpeter will be playing at 515 Hz frequency

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Select all the answers that apply. Scientists think past changes in climate have been caused by _____.
Natasha2012 [34]
<span>C. plate tectonics....</span>
8 0
4 years ago
n isolated charged soap bubble of radius R0=7.45 cmR0=7.45 cm is at a potential of V0=307.0 volts.V0=307.0 volts. If the bubble
Gnesinka [82]

Complete Question

An isolated charged soap bubble of radius R0 = 7.45 cm  is at a potential of V0=307.0 volts. V0=307.0 volts. If the bubble shrinks to a radius that is 19.0%19.0% of the initial radius, by how much does its electrostatic potential energy ????U change? Assume that the charge on the bubble is spread evenly over the surface, and that the total charge on the bubble r

Answer:

The difference is    U_f -U_i = 16 *10^{-7} J

Explanation:

From the question we are told that

     The radius of the soap bubble  is  R_o =  7.45 \ cm =  \frac{7.45}{100} =  0.0745 \ m

      The potential of the soap bubble is  V_1  =307.0 V

      The new radius of the soap bubble  is R_1 =  0.19 * 7.45=1.4155\ cm = 0.014155 \ m

The initial electric potential is mathematically represented as

     U_i  = \frac{V_1^2 R_o }{2k }

The final  electric potential is mathematically represented as

    U_f  = \frac{V_2^2 R_1 }{2k }

The initial potential is mathematically represented as

     V_1 =  \frac{kQ}{R_o}

The final  potential is mathematically represented as

        V_2 =  \frac{kQ}{R_1}

Now  

         \frac{V_2}{V_1}  =  \frac{R_o}{R_1}

substituting values

        \frac{V_2}{V_1}  =  \frac{7.45}{1.4155} =   \frac{1}{0.19}

=>      V_2 =  \frac{V_1}{0.19}

    So

         U_f  = \frac{V_1^2 R_2 }{2k * 0.19^2}

Therefore

        U_f -U_i = \frac{V_1^2 R_2 }{2k * 0.19^2} - \frac{V_1^2 R_o }{2k }

       U_f -U_i =     \frac{V_1^2}{2k} [\frac{ R_1 }{ * 0.19^2} - R_o]

where k is the coulomb's constant with value 9*10^{9} \  kg\cdot m^3\cdot s^{-4}\cdot A^2.

substituting values

       U_f -U_i =     \frac{307^2}{9 * 10^{9}} [\frac{ 0.014155 }{ 0.19^2} - 0.0745]

       U_f -U_i = 16 *10^{-7} J

           

     

8 0
3 years ago
Suppose you are driving a car, and a cup of coffee is on the seat beside you. Choose the frame with respect to which the cup is
Reil [10]

We want to find the frame of reference (viewpoint) where the cup of coffee inside your car moves the fastest.

The correct option will be A, the astronaut in the ISS.

---------------------------------

Let's see how to get the correct option, first, remember that the frame of reference means the "viewpoint" that you are taking. So we need to find the viewpoint where the cup of coffee moves faster.

Now just let's analyze all the given options to <u>see in which one the cup moves fastest.</u>

A) Here the astronaut sees the car moving, so here the cup has the speed of the car, and the astronaut is also moving. Remember that the International Space Station orbits with a speed of 7.6km/s (this is really fast). So in the frame of reference of the astronaut, he/she is at a stop, so this velocity is assigned to all the other objects. Thus, he sees the cup of coffee moving really fast.

B) In it's own frame of reference, the cup does not move.

C) The observer will see the car moving, and the cup is inside the car, so in this frame of reference the <u>velocity of the cup is the same as the one of the car.</u>

D) The incoming car has a given velocity V', from this frame, the car with the cup of coffee will have its own velocity plus the V' of the incoming car, so from this frame <u>the velocity of the cup of coffee is larger than the velocity of the car with the cup.</u>

E) You are driving the car with the cup of coffee, so the cup of coffee is always next to you, so in this frame, <u>the cup does not move.</u>

<u></u>

Concluding, is easy to see that the frame of reference where the cup moves the fastest is in the first one, for the nature of the <u>astronaut's frame of reference velocity</u>, he/she will see the cup moving the fastest.

If you want to learn more, you can read:

brainly.com/question/12222532

8 0
2 years ago
What kind of food service is the most expensive and requires the largest staff of skilled servers?
MrMuchimi
Your answer is french
4 0
3 years ago
Se deja caer una pelota inicialmente en reposo desde una altura de 50m sobre el nivel del suelo. ¿cuanto tiempo requiere para ll
umka2103 [35]

Answer:

a) t = 3.2 s

b) v_{f} = -32 m/s

Explanation:

a) El tiempo requerido para llegar al suelo se puede calcular usando la siguiente fórmula:

t = \sqrt{\frac{2y_{0}}{g}}

En donde:

y_{0}: es la altura inicial = 50 m

g: es la gravedad = 10 m/s²

t = \sqrt{\frac{2y_{0}}{g}} = \sqrt{\frac{2*50 m}{10 m/s^{2}}} = 3.2 s

Entonces, el tiempo requerido para llegar al suelo es 3.2 s.

b) La rapidez de la pelota justo antes del choque es el siguiente:

v_{f} = v_{0} - gt

En donde:

v_{0}: es la velocidad inicial = 0 (dado que se deja caer en resposo)

v_{f} = v_{0} - gt = 0 - 10 m/s^{2}*3.2 s = -32 m/s

Por lo tanto, la rapidez de la pelota justo en el momento anterior del choque es -32 m/s (el signo negativo es porque la pelota está cayendo).

Espero que te sea de utilidad!

6 0
3 years ago
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