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lukranit [14]
4 years ago
13

On the planet Arrakis, a male ornithoid sings at a frequency of 1180 Hz. When flying toward her mate at a speed of 30 m/s, a fem

ale ornithoid hears the sound at a frequency of 1250 Hz. What is the speed of sound in the atmosphere of Arrakis?

Physics
2 answers:
melisa1 [442]4 years ago
8 0

Answer:

The speed of sound in the atmosphere of Arrakis is 505.7 m/s.

Explanation:

given information:

male frequency, f_{m} = 1180 Hz

female speed, v_{f} = 30 m/s

female frequency, f_{f} = 1250

to calculate the speed of sound in the atmosphere of Arrakis, we can use Doppler effect

f_{l}  = \frac{v + v_{l} }{v+v_{s} } f_{s}

where

f_{l} = the frequency of listener

f_{s} = the frequency of source

v = speed of sound wave

v_{l} = speed of listener

v_{s} = speed of source

in this case, the male ornithoid is the source while the female is the listener.

v_{l} or speed of listener is positive since the listener move toward the source, so

f_{l}  = \frac{v + v_{l} }{v+v_{s} } f_{s}

1250 = \frac{v +30}{v+0} 1180

\frac{1250}{1180} = \frac{v+30}{v}

\frac{125}{118} = \frac{v+30}{v}

125v = 118v+3540

(125-118) v = 3540

v = \frac{3540}{7}

  = 505.7 m/s

dangina [55]4 years ago
6 0

Answer:

775m/s

Explanation:

See attached file

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4 years ago
A jogger accelerates from rest to 3.0 m/s in 2.0 s. A car accelerates from 38.0 to 41.0 m/s also in 2.0 s. (a) Find the accelera
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Answer:

(a)  a₁:  jogger  acceleration= 1.5 m/s²

(b)  a₂:  car  acceleration = 1.5 m/s²

(b)  d= 76m : the car travels 76 meters longer than the jogger during the 2 seconds

Explanation:

we apply uniformly accelerated motion formulas:

vf= v₀+at Formula (1)

vf²=v₀²+2*a*d Formula (2)

d= v₀t+ (1/2)*a*t² Formula (3)

Where:  

d:displacement in meters (m)  

t : time in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s²

Nomenclature

d₁:  jogger displacement   

t₁ :  jogger time

v₀₁:  jogger initial speed

vf₁:  jogger  final speed

a₁:  jogger  acceleration

d₂: car displacement   

t₂ : car  time

v₀₂: car  initial speed

vf₂:  car  final speed

a₂:  car  acceleration

Data

v₀₁ = 0

vf₁ = 3 m/s

t₁ =2.0 s

v₀₂ = 38.0m/s

vf₂ = 41.0 m/s

t₂ = 2.0 s

Problem development

(a) Find the acceleration (magnitude only) of the jogger.

We apply the formula (1) for calculate acceleration :

vf₁= v₀₁+a₁*t₁

3 = 0 +(a₁)*(2)

a₁= (3)/(2)

a₁= 1.5 m/s²

(b) Determine the acceleration (magnitude only) of the car.

We apply the formula (1) for calculate acceleration :

vf₂= v₀₂+a₂*t₂

41 = 38 +(a₂)*(2)

a₂= (41 - 38)/(2)

a₂= 3 /2

a₂= 1.5 m/s²

(c) Does the car travel farther than the jogger during the 2.0 s? If so, how much farther?

We apply the formula (1) for calculate distance :

d₁= v₀₁*t₁+ (1/2)*a₁*t₁²= 0+ (1/2)*(1.5) *(2)² = 3 m

d₂= v₀₂*t₂+ (1/2)*a₂*t₂² =38*(2)+ (1/2)*(1.5) *(2)²= 79 m

d= 79 m-3 m

d= 76m : the car travels 76 meters longer than the jogger during the 2 seconds

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Answer:

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1200 = 25 + (a*0.0333)

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Making μ the subject of formula, we have:

u=\frac{F_k}{F_n}

Substituting the given parameters into the formula, we have;

u=\frac{4.5}{20}

u = 0.225 ≈ 0.23

Read more on kinetic coefficient of friction here: brainly.com/question/13940648

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3 years ago
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Estimated by the observer.

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