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2 years ago

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From a vantage point very close to the track at a stock car race, you hear the sound emitted by a moving car. You detect a frequ

ency that is 0.86 times as small as the frequency emitted by the car when it is stationary. The speed of sound is 343 m/s. What is the speed of the car?

1 answer:

V125BC [204]2 years ago

6 0

Answer:

55.8 m/s

Explanation:

$f$ = Actual frequency of sound emitted by car

$f'$ = frequency observed when the car moves = $(0.86)f$

$V$ = Speed of sound = 343 m/s

$v$ = Speed of car

Frequency observed is given as

$f' = \frac{Vf}{V+v}$

$(0.86) = \frac{(343)}{343 + v}\\(0.86) = \frac{(343)}{343 + v}\\294.98 + (0.86) v = 343 \\v = 55.8 ms^{-1}$

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A physics professor demonstrates the Doppler effect by tying a 600 Hz sound generator to a 1.0-m-long rope and whirling it aroun

nalin [4]

Answer:

Therefore the the highest frequency is 620Hz and lowest frequency is 580Hz

Explanation:

Given data

Source Frequency fs=600Hz

Length r=1.0m

RPM=100 rpm

The speed of the generator is calculated as:

$v_{s}=rw\\v_{s}=r(2\pi f)$

Substitute the given values

$v_{s}=(1.0m)2\pi (\frac{100}{60}rev/s )\\v_{s}=10.47m/s$

For approaching generator the frequency is calculated as:

$f_{+}=\frac{f_{s}}{1-\frac{v_{s}}{v} }\\f_{+}=\frac{600Hz}{1-\frac{10.47m/s}{343m/s} } \\f_{+}=620Hz$

On the other hand,for the receding generator,Doppler's effect is expressed as:

$f_{-}=\frac{f_{s}}{1+\frac{v_{s}}{v} }\\f_{-}=\frac{600Hz}{1+\frac{10.47m/s}{343m/s} } \\f_{-}=580Hz$

Therefore the the highest frequency is 620Hz and lowest frequency is 580Hz

8 0

3 years ago

Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 kg), each having (a) a speed o

Ad libitum [116K]

Answer:

Part A:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 6.05x10^{-14}m$ < $\lambda_{electron} = 1.10x10^{-10}m$

Part B:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 1.29x10^{-13}m$ < $\lambda_{electron} = 5.525x10^{-12}m$

Explanation:

The wavelength of each particle can be determined by means of the De Broglie equation.

$\lambda = \frac{h}{p}$ (1)

Where h is the Planck's constant and p is the momentum.

$\lambda = \frac{h}{mv}$ (2)

Part A

Case for the electron:

$\lambda = \frac{6.624x10^{-34} J.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

But $J = Kg.m^{2}/s^{2}$

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

$\lambda = 1.10x10^{-10}m$

Case for the proton:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(6.55x10^{6}m/s)}$

$\lambda = 6.05x10^{-14}m$

Hence, the proton has a smaller wavelength than the electron.

<em>Part B </em>

For part b, the wavelength of the electron and proton for that energy will be determined.

First, it is necessary to find the velocity associated to that kinetic energy:

$KE = \frac{1}{2}mv^{2}$

$2KE = mv^{2}$

$v^{2} = \frac{2KE}{m}$

$v = \sqrt{\frac{2KE}{m}}$ (3)

Case for the electron:

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{9.11x10^{-31}Kg}}$

but $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{9.11x10^{-31}Kg}}$

$v = 1.316x10^{8}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(1.316x10^{8}m/s)}$

$\lambda = 5.525x10^{-12}m$

Case for the proton :

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{1.67x10^{-27}Kg}}$

But $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{1.67x10^{-27}Kg}}$

$v = 3.07x10^{6}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(3.07x10^{6}m/s)}$

$\lambda = 1.29x10^{-13}m$

Hence, the proton has a smaller wavelength than the electron.

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3 years ago

A 2-m long string is stretched between two supports with a tension that produces a wave speed equal to vw=50.00m/s. What are the

svetoff [14.1K]

Answer

given,

Length of the string, L = 2 m

speed of the wave , v = 50 m/s

string is stretched between two string

For the waves the nodes must be between the strings

the wavelength is given by

$\lambda = \dfrac{2L}{n}$

where n is the number of antinodes; n = 1,2,3,...

the frequency expression is given by

$f = n\dfrac{v}{2L}$

now, wavelength calculation

n = 1

$\lambda_1 = \dfrac{2\times 2}{1}$

λ₁ = 4 m

n = 2

$\lambda_2 = \dfrac{2\times 2}{2}$

λ₂ = 2 m

n =3

$\lambda_3 = \dfrac{2\times 2}{3}$

λ₃ = 1.333 m

now, frequency calculation

n = 1

$f = n\dfrac{v}{2L}$

$f_1 =1\times \dfrac{50}{2\times 2}$

f₁ = 12.5 Hz

n = 2

$f = n\dfrac{v}{2L}$

$f_2 =2\times \dfrac{50}{2\times 2}$

f₂= 25 Hz

n = 3

$f = n\dfrac{v}{2L}$

$f_3 =3\times \dfrac{50}{2\times 2}$

f₃ = 37.5 Hz

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3 years ago

If a current-carrying wire is in a magnetic field, in what direction will a force be exerted on the wire?.

astraxan [27]

Answer:

The magnetic force on a current-carrying wire is perpendicular to both the wire and the magnetic field with direction given by the right hand rule.

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Pablo preto lift a barbell. During which stage of the lift does Pablo do work on the barbell

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While he is lifting the dumbell as the definition of work done = moving a mass through a distance = F x d

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