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telo118 [61]
3 years ago
5

The speed of sound is 330.0 m/s and the wave length of a particular sound wave is 33.0 meters. Calculate the frequency of the so

und wave.
A)11,000 Hz B)10.0 Hz C) 0.10 Hz D)9.1 x 10-5 Hz
Physics
2 answers:
Alekssandra [29.7K]3 years ago
6 0
V: velocity of wave
f: frequency 
L: wavelenght

<span>v = fL => f = v/L => f = (330)/(33) => L = 10Hz</span>
Nataly [62]3 years ago
6 0

Answer:

B). f = 10.0 Hz

Explanation:

As we know the relation between frequency, wavelength and speed is given as

v = f \lambda

here we know that

v = 330 m/s

\lambda = 33 m

now in order to find the frequency of the sound we can use above relation as

f = \frac{v}{\lambda}

plug in all values in it

f = \frac{330}{33}

f = 10 Hz

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A circuit containing an inductor and a capacitor in series is designed to have a resonant frequency of 4511 Hz. If the inductor
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Answer:

(e)6.835\times 10^{-7}F

Explanation:

At resonance we know that X_l=X_C

That is \omega L=\frac{1}{\omega C}

\omega ^2=\frac{1}{LC}

\omega =\frac{1}{\sqrt{LC}}

f=\frac{1}{2\pi \sqrt{LC}}

We have given resonance frequency f =4511 Hz and inductance L=1.82 mH

So 4511=\frac{1}{2\pi \sqrt{LC}}

LC=\frac{1}{4\pi ^2\times 4511^2}

LC=1.244\times 10^{-9}

C=\frac{1.244\times 10^{-9}}{1.82\times 10^{-3}}=0.6835\times 10^{-6}=6.835\times 10^{-7}F

So option e is the correct answer

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3 years ago
Which one of the following substances is a liquid fuel used in rocket engines?
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The substance which is a liquid fuel used in rocket engines is liquid oxygen. The correct answer is A.
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The head of a grass string trimmer has 100 g of cord wound in a light, cylindrical spool with inside diameter 3.00 cm and outsid
Karolina [17]

Answer:

a).11.546J

b).2.957kW

Explanation:

Using Inertia and tangential velocity

a).

w=2250*2\pi *\frac{1}{60}\\ w=235.61

I=\frac{1}{2}*m*((\frac{d_{i} }{2})^{2} +(\frac{d_{e} }{2})^{2})\\m=100g *\frac{ikg}{1000g}=0.1kg\\ d_{i}=3cm*\frac{1m}{100cm}=0.03m \\ d_{e}=18cm*\frac{1m}{100cm}=0.18m\\I=\frac{1}{2}*0.1kg*((\frac{0.03m}{2})^{2} +(\frac{0.18m}{2})^{2})\\I=0.41625x10^{-3}kg*m^{2}

Now using Inertia an w

E=\frac{1}{2}*I*(w)^{2} \\ E=\frac{1}{2}*0.416x10^{-3}*(235.61)^{2} \\E=11.54J

average power=\frac{11.4J}{0.230s}=50.2 W

b).

power=t*w

P=11.5465*0.25*235.61

P=2.957 kW

8 0
2 years ago
Read 2 more answers
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