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aalyn [17]
3 years ago
14

"a horn emits a frequency of 1000 HZ. A 14 m/s wind is blowing toward a listener. What is the frequency of the sound heard by th

e listener
Physics
1 answer:
Semenov [28]3 years ago
7 0

Answer:

 f_L = 1000 Hz

Explanation:

given,

speed of wind, = 14 m/s

frequency of horn,f_s = 1000 Hz

speed of sound,V = 344 m/s

frequency heard by the listener

using Doppler effect

f_L = \dfrac{v+v_L}{v+v_s}f_s

f_L is the frequency of the sound heard by the listener

f_s is the frequency of sound emitted by the listener

V is the speed of sound

v_L is the speed of listener

v_s is the speed of source

now,

considering the frame of reference in which wind is at rest now, both listener and the source will be moving at 14 m/s

 f_L = \dfrac{v+14}{v+14}\times 1000

now on solving we will get

 f_L = 1000 Hz

hence, the frequency heard by the listener is equal to  1000 Hz

 

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2. A Se lanza un electrón con rapidez inicial v0 = 1.60×106 m/s hacia el interior de un campo uniforme entre las placas paralela
Vanyuwa [196]

Answer:

A)     E = 145.6 N / C , B)  y= 2,8 10-7 m with a downward direction

C) he shape of the trajectory of the two particles is to simulate a parabola,

D)     F_{e} /F_{g} = 10³⁴

Explanation:

A) For this exercise we use Newton's second law to find the acceleration of the electron, where the force is electric

           F = m a  

           - e E = m a

          a = - e E / m

with the field directed downward, the acceleration is in the vertical upward direction.

We look for how much the electron moves with kinematics, in the x direction there is no acceleration,

x axis (parallel to plates)

           x = v₀ t

           t = x / v₀

y axis (perpendicular to plates)

          y = y₀ + v_{oy} t + ½ a t²

Let's take the zero of the system in the middle of the plates y₀ = 0, also the initial vertical velocity is zero (v_{oy} = 0) the width of the plate is known

          y = ½ a t²

we substitute

         y = ½ (e E /m)  (x / v₀)²

         y = ½ e x2 /m v₀²   E

we look for the electric field

        E = 2 m y v₀² / e x²

where to use this expression the length and width of the condenser must be known, suppose that the length is x = l = 1 cm = 1 10⁻² m and the width is y = 0.5 mm = 0.5 10⁻³ m

let's calculate

         E = 2  9.1 10⁻³¹ 0.5 10⁻³ (1.6 10⁶)² / (1.6 10⁻¹⁹ (1 10⁻²)²)

         E = 145.6 N / C

B) The electron is exchanged for a proton

Let's look for the vertical displacement, in this case as the proton has a positive charge it moves towards the bottom of the plates

          y = ½ e x² / m v₀² E

          y = ½ 1.6 10⁻¹⁹ 1 10⁻⁴ / (1.67 10⁻²⁷ (1.6 10⁶)²   145.6

          y = 28.4375 10⁻⁸ m

since the distance between the plates is 0.5 10-3 m, the proton passes the condensate because its deflection is very small

In summary, its displacement is y= 2,8 10-7 m and with a downward direction (the same direction of the electric field)

C) The shape of the trajectory of the two particles is to simulate a parabola, but one for having a negative charge (electron) the force is upwards and the other for having a positive charge (proton) the trajectory is downwards

D) The force of gravity

           F_{g} = G m M / R²

electron

          Between the electron and the positive charges of the conducting plate

           F_{g}= 6.67 10⁻¹¹ 1.67 10⁻²⁷ 9.1 10⁻³¹ / (0.5 10⁻³)²

           F_{g} = 4.1 10⁻⁵¹ N

           

electric force

           F_{e} = -e E

           F_{e} = - 1.6 10⁻¹⁹ 145.6

           F_{e} = 2.3 10⁻¹⁷ N

let's look for the reason between these two forces

         F_{e} / F_{g} = 2.3 10⁻¹⁷ / 4.1 10⁻⁵¹

          F_{e} /F_{g} = 10³⁴

We see that the electric force is many orders of magnitude higher than the gravitational force.

5 0
3 years ago
A ball is dropped from rest at the top of a 6.10 m
natita [175]

Answer:

n = 5 approx

Explanation:

If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back

\frac{v_1}{v} = e ( coefficient of restitution ) = \frac{1}{\sqrt{10} }

and

\frac{v_1}{v} = \sqrt{\frac{h_1}{6.1} }

h₁ is height up-to which the ball bounces back after first bounce.

From the two equations we can write that

e = \sqrt{\frac{h_1}{6.1} }

e = \sqrt{\frac{h_2}{h_1} }

So on

e^n = \sqrt{\frac{h_1}{6.1} }\times \sqrt{\frac{h_2}{h_1} }\times... \sqrt{\frac{h_n}{h_{n-1} }

(\frac{1}{\sqrt{10} })^n=\frac{2.38}{6.1}= .00396

Taking log on both sides

- n / 2 = log .00396

n / 2 = 2.4

n = 5 approx

3 0
3 years ago
arzan, who weighs 700 N, swings from a cliff at the end of a convenient vine thatis 20 m long. From the top of the cliff to the
vekshin1

Answer:

= 7.07 m

Explanation:

The Tarzan reaches bottom of swing after descending 2.5 m,

change in his potential energy equals his kinetic energy at bottom of swing

m g h = (1/2) m v²   ,  

hence speed v of Tarzan at bottom of swing is given as  

v = ( 2 g h )1/2

= ( 2 × 9.8 × 2.5 )1/2

= 7 m/s

At the bottom of swing, if the vine breaks, then he is moving with horizontal velocity 7 m/s in gravitational field.  

If vertical distance from ground to bottom of swing is 5 m, then time t for Tarzan to reach ground is given by

S = (1/2)g t2 or   t = (2S/g)1/2

= ( 2 × 5 / 9.8 )1/2

= 1.01 s

Horizontal distance traveled by Tarzan = 1.01 × 7

= 7.07 m

7 0
3 years ago
Explain why the atomic mass of an element is weighted average mass
madreJ [45]

Explanation:

The mass written on the periodic table is an average atomic mass taken from all known isotopes of an element. This average is a weighted average, meaning the isotope's relative abundance changes its impact on the final average. The reason this is done is because there is no set mass for an element.

3 0
2 years ago
Iron-rich minerals in rock pointed in one direction and then switched to the exact opposite direction. This supports what idea?
NISA [10]
Iron rich minerals in rock pointed in one direction the switch to the exact opposite direction. I'd say that what supports this idea is that Earth's magnetic field goes through pole reversals.<span>
</span>
4 0
3 years ago
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