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ICE Princess25 [194]
3 years ago
6

(4%) Problem 10: The standard frequency of the musical pitch called "middle C" is 261.6 Hz. show answer No Attempt How fast, in

meters per second, would a source emitting a tone of middle C have to move toward you so that you would hear the pitch C# (C sharp) above middle C, with a frequency of 277.2 Hz? Assume that takes place in air at a high altitude, where the speed of sound is 298 m/s.
Physics
1 answer:
bearhunter [10]3 years ago
6 0

Answer:

16.77 m/s

Explanation:

Given that

Frequency of middle pitch, Fo = 261.6 Hz

Frequency of C sharp, f = 277.2 Hz

Velocity of sound in air, v = 298 m/s

Speed of sound from the source, Vs = ? m/s

Using the formula

f = Fo•(V + Vr)/(V + Vs)

← Doppler

Vr would be +ve if the receiver is moving toward source;

Vs would be -ve if source is moving toward the receiver

277.2 Hz = 261.6Hz * (298 + 0) / (298 - Vs)

277.2 = 77956.8 / (298 - Vs)

298 - Vs = 77956.8 / 277.2

298 - Vs = 281.23

Vs = 298 - 281.23

Vs = 16.77 m/s

Thus, the speed needed is 16.77 m/s

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chubhunter [2.5K]

Answer:

Explanation:

Force = q ( v x B)

- 5.6 x 10⁻⁹ (v x - 1.25 k )

- 3.4x 10⁻⁷i + 7.4 x 10⁻⁷j

Let v = ai+bj +ck

Force = - 5.6 x 10⁻⁹ [(ai+bj +ck) x - 1.25 k )]

= - 5.6 x 10⁻⁹ ( 1.25aj - 1.25bi )

= - 7 a j + 7 b i

( 7bi - 7aj ) x 10⁻⁹

Comparing with given force

7b x 10⁻⁹ b = - 3.4 x 10⁻⁷

b = - 48.57

- 7 a x 10⁻⁹ = 7.4 x 10⁻⁷

a = - 105.7

velocity

= -105.7 i - 48.57 j + ck

b ) Component along k can not be obtained .

c ) v . F = ( -105.7 i - 48.57 j + ck ) . −(3.40×10−7N) ˆı +(7.40×10−7N) ˆȷ

= 105.7 x 3.4 x 10⁻⁷ - 48.57 x 7.4 x 10⁻⁷

= 359.38 x 10⁻⁷ - 359.38 x 10⁻⁷

=0

angle between v and F = 90 degree

4 0
3 years ago
Light shines through a single slit whose width is 5.6 × 10-4 m. A diffraction pattern is formed on a flat screen located 4.0 m a
Mrac [35]

Answer:

462 nm

Explanation:

Given: width of the slit, d = 5.6 × 10⁻⁴ m

Distance of the screen, D = 4.0 m

Fringe width, β = 3.3 mm = 3.3 × 10⁻³ m

First dark fringe means n =1

Wavelength of the light, λ = ?

\beta = \frac{\lambda D}{d}\\ \Rightarrow \lambda = \frac{d \beta}{D} =\frac{5.6\times 10^{-4} \times 3.3 \times 10^{-3}}{4.0} = 4.62 \times 10^{-7}m = 462 nm

5 0
3 years ago
Three different objects, all with different masses, are initially at rest at the bottom of a set of steps. Each step is of unifo
fgiga [73]

The definition of gravitational potential energy would gravitate allows to find the answers for the energy of each object on the highest rung of the ladder are:

 a) expression for gravitational potential energy is  U_p = m g \ \Delta y

 b) Energy value

        * Body 1    U_p = 82 J

        * Body 2   U_p = 38.6 J

        * Body  3  18.5 J

The gravitational potential energy is a configuration energy that represents the amount of work that the system can do for a given configuration, its expression is

           U_p = m g \  \Delta y

Where U_p is the gravitational potential energy, m the mass of the body and Δy the difference in height.

In this case it indicates that the minimum height is at the bottom and is equal to zero, the steps of a ladder have a height of approximately y = 15 cm = 0.15 m

They indicate the mass of the bodies are m₁ = 3.10 kg, the mass of the body 2 m₂ = 1.46 kg , the mass of the third object is not observed, suppose it is m₃ = 0.7 kg.

They indicate that the body is placed on the last step, suppose it is a standard 18-step ladder, therefore the total height is

          y = # _ steps    y_each step

          y = 18   0.15

          y = 2.70 m

the gravitational potential energy to the capacity of the body is

body 1

              U_p = m₁ g Δy₁

              U_p = 3.10  9.8 (2.70 -0)

               U_p = 82 J

body 2

              U_p = 1.46 9.8 ( 2.7-0)

              U_p = 38.6 J

body  3

             U_pUp = 0.70 9.8 (2.7-0)

             U_p = 18.5 J

In conclusion, using the definition of gravitational potential energy, we can find the answers for the energy of each object on the highest rung of the ladder are;

     a) expression for gravitational potential energy is  U_p = m g \ \Delta y

     b) Energy value

        * Body 1    U_p = 82.0 J

        * Body 2   U_p = 38.6 J

        * Body  3   U_p = 18.5 J

Learn more here: brainly.com/question/3884855

8 0
3 years ago
This subatomic particle adds the least amount of mass to an atom and is called the
sergey [27]
A nuetron is the lightest subatomic particle
8 0
3 years ago
PLEASE HELP ASAP!!!
I am Lyosha [343]

Answer:

B - Velocity

Explanation:

Velocity definition: “The speed of something in a given direction.”

6 0
3 years ago
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