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Virty [35]
2 years ago
5

A baseball hits a car, breaking its window and triggering its alarm which sounds at a frequency of 1210 Hz. What frequency (in H

z) is heard by a boy on a bicycle riding away from the car at 8.25 m/s? Take the speed of sound to be 343 m/s.
Physics
1 answer:
IRINA_888 [86]2 years ago
8 0

Answer:

The frequency of sound heard by the boy is 1181 Hz.

Explanation:

Given that,

Frequency of sound from alarm  f_{0} = 1210\ Hz

Speed = -8.25 m/s

Negative sign show the boy riding away from the car

Speed of sound = 343

We need to calculate the heard frequency

Using formula of frequency

f = f_{0}(\dfrac{v+v_{0}}{v-v_{s}})

Where, f_{0} = frequency of source

v_{0} = speed of observer

v_{s} = speed of source

v = speed of sound

Put the value into the formula

f=1210\times\dfrac{343+(-8.25)}{343-0}

here, source is at rest

f=1180.8\ Hz

f=1181\ Hz

Hence, The frequency of sound heard by the boy is 1181 Hz.

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Two swift canaries fly toward each other, each moving at 15.0 m/s relative to the ground, each warbling a note of frequency 1750
lesya692 [45]

Answer:

Part a)

f = 1911.5 Hz

Part b)

\lambda = 0.186 m

Explanation:

Here the source and observer both are moving towards each other

so we know that the apparent frequency is given as

f' = f_0 (\frac{v + v_o}{v - v_s})

here we know that

f_0 = 1750 Hz

v_o = 15 m/s

v_s = 15 m/s

now we will have

f' = (1750)(\frac{340 + 15}{340 - 15})

f' = 1911.5 Hz

Part b)

Apparent wavelength is given by the formula

\lambda = \frac{v_{relative}}{f_{app}}

here we will have

\lambda = \frac{340 + 15}{1911.5}

\lambda = 0.186 m

3 0
2 years ago
Eac of the two Straight Parallel Lines Each of two very long, straight, parallel lines carries a positive charge of 24.00 m C/m.
Cloud [144]

Answer:

The magnitude of the electric field at a point equidistant from the lines is 4.08\times10^{5}\ N/C

Explanation:

Given that,

Positive charge = 24.00  μC/m

Distance = 4.10 m

We need to calculate the angle

Using formula of angle

\theta=\sin^{-1}(\dfrac{\dfrac{d}{2}}{2d})

\theta=\sin^{-1}(\dfrac{1}{4})

\theta=14.47^{\circ}

We need to calculate the magnitude of the electric field at a point equidistant from the lines

Using formula of electric field

E=\dfrac{2k\lambda}{r}\times2\cos\theat

Put the value into the formula

E=\dfrac{2\times9\times10^{9}\times24.00\times2\times10^{-6}\cos14.47}{2.05}

E=408094.00\ N/C

E=4.08\times10^{5}\ N/C

Hence, The magnitude of the electric field at a point equidistant from the lines is 4.08\times10^{5}\ N/C

6 0
2 years ago
Determine the cutting force f exerted on the rod s in terms of the forces p applied to the handles of the heavy-duty cutter.
Alexus [3.1K]
If you use the next formula with the data given in the exercise you are asking:
Ey[3.4] - F[1.7] = 0 
<span>Ey = F/2 
</span>and after that what you need to do is sum the moments of the handle about D to zero asumming it is a positive moment and we proceed like this
Ey[1.5sin19] – P[21 – 1.5sin19] = 0 
<span>(F/2)[1.5sin19] = P[21 – 1.5sin19] </span>
<span>F = 2P[21 – 1.5sin19] / [1.5sin19] </span>
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5 0
2 years ago
Read 2 more answers
Two tiny, spherical water drops, with identical charges of -6.19 × 10-16 C, have a center-to-center separation of 1.22 cm. (a) W
Lynna [10]

Answer:

a

  F  =  2.32*10^{-17} \  N

b

n  =3869 \  electrons

Explanation:

From the question we are told that

   The  charge on each water drop is  q_1=q_2=q =  - 6.19*10^{-16} \ C

   The  distance of separation is  d =  1.22\ cm  = 0.0122 \ m  

   

Generally the electrostatic force between the water drops is mathematically represented as

      F  =  \frac{k *  q_1  *  q_2 }{ d^ 2}

Here  k is the coulombs constant with value  k  =  9*10^9 \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.

So  

      F  =  \frac{9*10^9  * -6.19 *10^{-16} *  (-6.19*10^{-16}) }{ 0.0122^ 2}

       F  =  2.32*10^{-17} \  N

Generally the quantity of charge is mathematically represented as

        q =  n * e

Here n is the number of electron present

and  e is the charge on one electron with value  e =  1.60*10^{-19} \  C

    So

         n  = \frac{6.19 *10^{-16}}{1.60*10^{-19}}

         n  =3869 \  electrons

   

4 0
2 years ago
What's your normal body temperature? it may not be 98.6°f, the oft-quoted average that was determined in the nineteenth century.
harkovskaia [24]

(98.6 - 98.2)ºF = .4ºF

.4ºF* 5ºC/9ºF = 0.222 ºC

6 0
2 years ago
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