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2 years ago

A fire engine is moving south at 35 m/s while blowing its siren at a frequency of 400 Hz.

1 answer:

8 0

To solve this problem we will apply the concepts related to the Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. Mathematically it can be described as

$f_d= f_s \frac{(v+v_d)}{(v-v_s)}$

Here,

$f_d$ =frequency received by detector

$f_s$ =frequency of wave emitted by source

$v_d$ =velocity of detector

$v_s$ =velocity of source

v=velocity of sound wave

Replacing we have that,

$f_d = 400(\frac{(343+18)}{(343-35)})$

$f_d=422 Hz$

Therefore the frequencty that will hear the passengers is 422Hz

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Two basketball teams are resting during halftime. While they are resting, a truck driver asks them for help pushing his broken d

viva [34]

Answer:

It is a superordinate goal because both teams could have helped with the task.

Explanation:

If both teams pushed then they could have made it happened

7 0

2 years ago

Which letter correctly identifies the part of the hydrologic cycle that is most directly affected by impervious building materia

masha68 [24]

Infiltration

Explanation:

The component of the hydrologic cycle affected by impervious building such as concrete and asphalt is infiltration.

Water infiltration is a major component of the hydrologic cycle.
Concretes and other materials can prevent water from going down into the earth.
This affects the ground water system in place.
It leads to increase in surface run off and might cause inundation of an area.
Infiltration is a very important component of water cycle.
It takes water to plant root and recharges groundwater systems.
Impervious structures takes this capability away.

learn more:

Biogeochemical cycle brainly.com/question/3509510

#learnwithBrainly

3 0

2 years ago

Two workhorses tow a barge along a straight canal. Each horse exerts a constant force of magnitude F, and the tow ropes make an

morpeh [17]

Answer:

a) Fvt cosθ

b) Fv cosθ

Explanation:

Each horse exerts a force = F

the rope is inclined at an angle = θ

speed of each horse = v

a) In time t, the distance traveled d = speed x time

i.e d = v x t = vt

also, the resultant force = F cosθ

Work done W = force x distance

W = F cosθ x vt = Fvt cosθ

b) Power provided by the horse P = force x speed

P = F cosθ x v

P = Fv cosθ

8 0

1 year ago

A car travels 10 m/s east. Another car travels 10 m/s north. The relative speed of the first car with respect to the second is:

Thepotemich [5.8K]

Answer:

d. less than 20m/s

Explanation:

To the 2nd car, the first car is travelling 10m/s east and 10m/s south. So the total velocity of the first car with respect to the 2nd car is

[tex]\sqrt{10^2 + 10^2} =10\sqrt{2}=14.14m/s

As 14.14m/s is less than 20m/s. d is the correct selection for this question.

3 0

2 years ago

A proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the capacitor and reaches the neg

WINSTONCH [101]

Answer:

$=2,012,319.36 \ m/s$

Explanation:

-The only relevant force is the electrostatic force

-The formula for the electrostatic force is:

$F = Eq$

E is the electric field and q is the magnitude of the charge.

#Since the electric field is the same in both cases, and the charge of the protons and electrons have the same magnitude, you can state that the magnitude of the electric forces acting in both proton and electron are the same.

$F_e = F_p\\\\F_e= Force \ on \ electron\\F_p = Force \ on \ proton$

-Applying Newton's 2nd Law:

$F=ma$

$F_e=M_ea_e$

$F_p=M_pa_p$

#equate the two forces:

$F_e = F_p\\\\M_ea_e=M_pa_p\\\\a_e=\frac{M_pa_p}{M_e}$

#The equations for velocity in uniform acceleration:

$V_f^2=V_o^2+2ad\\\\V_o^2=0\\\\\therefore V_f^2=2ad$

#For the proton:

$V_f^2=2a_pd\\\\a_p=\frac{V_f^2}{2d}\\\\a_p=\frac{47000m/s)^2}{2d}$

#For the electron:

$V_f^2=2{a_e}^2\times 2d\\\\A_e=M_p\times A_p/M_e\\\\V_f^2=M_p\times (47000m/s)^2/2d\times2d/M_e\\\\V_f^2=M_p\times (47000m/s)^2/M_e\\\\V_f=47000m/s\times\sqrt{\frac{M_p}{M_e}}$

The mass values of the proton and electron are:

$M_p=1.67\times 10^{-27} kg\\\\M_e=9.11\times10^{-31}kg$

The speed of the ion is therefore calculated as:

$V_f=47000m/s\times\sqrt{\frac{M_p}{M_e}}\\\\=47000m/s\times\sqrt{\frac{1.67\times10^{-27}}{9.11\times10^{-31}}\\\\=2,012,319.36 \ m/s$

Hence, the ion's speed at the negative plate is $=2,012,319.36 \ m/s$

7 0

2 years ago