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Delicious77 [7]
3 years ago
6

The siren on an ambulance emits a sound of frequency 2.80×103Hz. If the ambulance is traveling at 26.0 m/s (93.6 km/h or 58.2 mi

/h) the speed of sound is 340 m/s and the air is still, what is the frequency that you hear if you are standing in front of the ambulance?
Physics
1 answer:
Norma-Jean [14]3 years ago
5 0

To solve this problem it is necessary to apply the concepts related to the described wavelength through frequency and speed. Mathematically it can be expressed as:

\lambda = \frac{v}{f}

Where,

\lambda = Wavelength

f = Frequency

v = Velocity

Our values are given as,

f = 2.8*10^3Hz

v = 340m/s \rightarrow Speed of sound

Keep in mind that we do not use the travel speed of the ambulance because we are in front of it. In case it approached or moved away we should use the concepts related to the Doppler effect:

Replacing we have,

\lambda = \frac{340}{2.8*10^3}

\lambda = 0.1214m

Therefore the frequency that you hear if you are standing in from of the ambulance is 0.1214m

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The lons entering the mass spectrometer have the same charges. After being accelerated through a potential difference of 8.20 kV
Ratling [72]

The calculated magnitude is  6.73 x 10³ V/m.

AMU is described as being one-twelfth the mass of a carbon-12 atom (12C). C makes up more than 98% of the carbon that can be found in nature, making it the most prevalent isotope. The magnitude of the field is the change in potential across a small distance in the indicated direction divided by that distance.

Potential difference = 8.20 kV= 8.20 x 10³ V

radius= 19.4/100=0.194 m

total distance that is circumference of the circle= 2πr =2 x 3.14 x 0.194

                                                                               = 1.218 m

therefore Magnitude= 8.20 x 10³ / 1.218

                                  =6.73 x 10³ V/m

Learn more about Magnitude here-

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4 0
1 year ago
Explain how atomic mass and molecular mass are determined
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4 0
3 years ago
Compare and contrast the terms vaporizing and condensation.
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5 0
3 years ago
You have a double slit experiment, with the distance between the two slits to be 0.025 cm. A screern is 120 cm behind the double
dimulka [17.4K]

Answer:

The wavelength of the light is 633 nm.

Explanation:

Given that,

Distance between the two slits d= 0.025 cm

Distance between the screen and slits D = 120 cm

Distance between the slits y= 1.52 cm

We need to calculate the angle

Using formula of double slit

\tan\theta=\dfrac{y}{D}

Where, y = Distance between the slits

D = Distance between the screen and slits

Put the value into the formula

\tan\theta=\dfrac{1.52}{120}

\theta=\tan^{-1}\dfrac{1.52}{120}

\theta=0.725

We need to calculate the wavelength

Using formula of wavelength

d\sin\theta=n\lambda

Put the value into the formula

0.025\times\sin0.725=5\times\lambda

\lambda=\dfrac{0.025\times10^{-2}\times\sin0.725}{5}

\lambda=6.326\times10^{-7}\ m

\lambda=633\ nm

Hence, The wavelength of the light is 633 nm.

4 0
3 years ago
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