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romanna
10 days ago
12

Driving down a quiet street on a summer day with the car windows down, you see a convertible approaching from the opposite direc

tion, with the driver playing her car stereo loudly. Having a well-trained musician’s ear, you notice that the song you hear is in a key one semitone (the interval between two adjacent keys on a piano) above the key in which the song is typically heard. You also know that to shift any note by one semitone, the frequencies must be related by a factor of 1.06. Assume both drivers are traveling at the same speed, what is their speed?
Physics
1 answer:
ValentinkaMS [2.4K]10 days ago
7 0
The ratio of the perceived frequency to the actual frequency is 1.06. Denote their speeds as v. Based on the Doppler effect formula applicable to a sound source and listener moving toward each other: n = n₀ (V + v) / (V - v) where V represents the speed of sound at 340 m/s. Thus, n / n₀ = (V + v) / (V - v), which leads to 1.08 = (360.4 - 1.06 v)/(340 + v); solving yields v = 9.9 m/s.
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Yuliya22 [2438]

Answer:

fcosθ + Fbcosθ  =Wtanθ

Explanation:

Refer to the provided diagram in the attachment

fx= fcosθ (fx: x-direction component of friction force; f: frictional force)

Fbx= Fbcosθ (Fbx: x-direction component of braking force; Fb: braking force)

Wx= Wtanθ (Wx: x-direction component of weight; W: weight of the semi)

Sum of forces in the x-direction = 0

fx + Fbx = Wx

fcosθ + Fbcosθ  =Wtanθ

7 0
1 month ago
Solenoid 2 has twice the diameter, twice the length, and twice as many turns as solenoid 1. How does the field B2 at the center
Sav [2230]

Complete Question

The entire question is displayed in the first uploaded image

Answer:

The right choice is option  3

Explanation:

The question informs us that

   The diameter of solenoid 1 is  d_1

   The length of solenoid 1 is   L_1

    The  number of turns of solenoid is  N_1

   The diameter of solenoid 2 is  d_2 = 2d_1

   The length of solenoid 2 is   L_2 = 2L_1

    The  number of turns of solenoid  2 is    N_2 = 2 N_1

Typically, the magnetic field in a solenoid can be expressed mathematically as

     B = \frac{\mu_o * N * I }{L}

According to this formula we find that

     B \ \alpha \ \frac{N}{L}

     B = C \frac{N}{L}

Here C denotes constant

=>   C = \frac{B * \frac{L}{N}

=>    \frac{B_1 * \frac{L_1}{N_1} = \frac{B_2 * \frac{L_2}{N_2}

=>  \frac{B_1}{B_2 } = \frac{N_1 L _2}{ N_2L_1}

=>   \frac{B_1}{B_2 } = \frac{N_1 * (2 L_1)}{ (2 N_2)L_1}

=>   \frac{B_1}{B_2 } = 1

=>   B_2 = B_1

4 0
14 days ago
Read each scenario below. Then select the answer that best completes each sentence.
Ostrovityanka [2208]

Answer:

The power used by raul's microwave must match the power consumed by katrina's because both microwaves took different durations to accomplish the same heating task.

Explanation:

The power output from a car engine is equivalent to that of a bicycle since both perform the same amount of work over time. Both raul and katrina shared a frozen meal, heating each portion in different microwaves. Katrina's portion was warm in one minute, whereas raul's portion required two minutes. Therefore, the power utilized by raul's microwave aligns with that of katrina's, given that it took longer to achieve the same result.

7 0
26 days ago
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inna [2210]

Answer:

Acceleration(a) = 0.75 m/s²

Explanation:

Given:

Force(F) = 3 N

Mass of object(m) = 4 kg

Find:

Acceleration(a)

Computation:

Force(F) = ma

3 = (4)(a)

Acceleration(a) = 3/4

Acceleration(a) = 0.75 m/s²

3 0
1 month ago
A locomotive is accelerating at 1.6 m/s2. it passes through a 20.0-m-wide crossing in a time of 2.4 s. after the locomotive leav
kicyunya [2264]

Response:

Once it has crossed, the locomotive requires 17.6 seconds to achieve a speed of 32 m/s.

Details:

  The locomotive's acceleration is 1.6 m/s^2

  The duration taken to pass the crossing is 2.4 seconds.

  We can apply the motion equation, v = u + at, where v represents final velocity, u indicates initial velocity, a denotes acceleration, and t signifies time.

  When the speed reaches 32 m/s, we have v = 32 m/s, u = 0 m/s, and a= 1.6 m/s^2.

   32 = 0 + 1.6 * t

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  Therefore, the locomotive attains a speed of 32 m/s after 20 seconds, and it passes the crossing in 2.4 seconds.

Thus, after clearing the crossing, it takes an additional 17.6 seconds to reach the speed of 32 m/s.

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26 days ago
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