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3 years ago

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"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

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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Answer:

(a) 12 × 10⁻³ C = 12 mC (b) The lower plate

Explanation:

Given

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k = 1/4πε₀ = 9.0 × 10⁹ Nm²/C²

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charge on oil drop, q = 15e

charge on plates, Q = ?

First, we find the charge density of the plates, D = Q/A where Q = charge on plates and A = area of plates. Since the plates are circular, the area is given by A=πr² where r = radius of plates. D=Q/πr²

Also, the electric field, E between the plates is given by E = D/ε =Q/Aε = Q/ε₀πr².

The force on the oil drop due to the electric field between the plates is given by F = qE = qQ/ε₀πr².

Since the oil drop is suspended between the plates, it means that the electric force due to the field on the oil drop balances the weight of the oil drop. So, since weight of oil drop W = mg where g = 9.8 m/s². F =W (for oil drop suspension).

So, qQ/ε₀πr²=mg

So, Q=mgε₀πr²/q

From k = 1/4πε₀, ε₀=1/4πk

So, Q = mgπr²/4πkq = mgr²/4kq = (0.025 × 10⁻⁶ × 9.8 × (6.5 × 10⁻²)²)÷(4 × 9 × 10⁹ × 15 × 1.6 × 10⁻¹⁹)= 0.012 C = 12 mC

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3 years ago

Three identical train cars, coupled together, are rolling east at speed v0. A fourth car traveling east at 2v0 catches up with t

aliina [53]

Answer: $v_o$

Explanation:

It is given that three cars has same mass m with speed $v_o$

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$P_f=5m\times v$

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$v=v_o$

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4 years ago

What would we need to know to calculate both work and power? (2 points) Select one: a. energy, force, and time b. force, distanc

goblinko [34]

B. force, distance, and time

Take a look at the definition of a Joule (SI unit of work) and the definition of a Watt (SI unit of power). They're (kg*m^2)/s^2 for work and (kg*m^2)/s^3 for power. Another definition for work is Newton Meter which is force times distance, and since you can define work as force times distance, then power is work per second. So it looks like you need force and distance to calculate work, and then time since power is work over time. So of the 4 choices, we've been given, let's see if any of them allow us to calculate both work and power.
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a. energy, force, and time
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b. force, distance, and time
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c. force, mass, and distance
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3 years ago

Read 2 more answers

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3 years ago

6. A 15 kg box is given an initial push so that it slides across the floor and comes to a stop.

Illusion [34]

Answer:

(a) Frictional force = 44.145 N

(b) Acceleration = -2.943 m/s²

(c) Distance covered = 1.529 m

Explanation:

Given:

Mass of the box is, $m=15\ kg$

Coefficient of friction is, $\mu =0.30$

Acceleration due to gravity is, $g=9.81\ m/s^2$

Normal force acting on the box is, $N=mg=(15)(9.81) = 147.15\ N$

(a)

Frictional force is given as:

$f=\mu N=0.30\times 147.15=44.145\ N$

Therefore, the frictional force is 44.145 N.

(b)

The acceleration of the box is given using Newton's second law as:

$a=-\frac{f}{m}=-\frac{44.145}{15}=-2.943\ m/s^2$

Therefore, the acceleration of the box is -2.943 m/s².

(c)

Initial velocity is, $u=3.0\ m/s$

Final velocity is, $v=0\ m/s$

Acceleration of the box is, $a=-2.943\ m/s^2$

The displacement of the box can be determined using equation of motion as:

$v^2=u^2+2ad\\0=3^2+2\times (-2.943)d\\0=9-5.886d\\5.886d=9\\d=\frac{9}{5.886}=1.529\ m$

Therefore, the displacement of the box is 1.529 m.

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4 years ago