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3 years ago

7

The speed of sound in room temperature (20°C) air is 343 m/s; in room temperature helium, it is 1010 m/s. The fundamental freque

ncy of an open-closed tube is 315 Hz when the tube is filled with air. What is the fundamental frequency if the air is replaced with helium?

1 answer:

Lera25 [3.4K]3 years ago

4 0

Answer: f = 927.55Hz

Explanation: Since the the tube is open-closed, the length of air and the wavelength of sound passing through the tube is given below

L = λ/4 where λ = wavelength.

speed of sound in air = v = 343m/s.

fundamental frequency of open closed tube = 315Hz

λ = 4L.

v = fλ

343 = 315 * 4L

343 = 1260 * L

L = 343/ 1260

L = 0.27m

In the same tube of length L = 0.27m but different medium ( helium), the speed of sound is 1010m/s.

The length of tube and wavelength are related by the formulae below

L = λ/4, λ=4L

λ = 4 * 0.27

λ = 1.087m.

v = fλ

1010 = f * 1.087

f = 1010/1.807

f = 927.55Hz

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Circular motion formulas

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3 years ago

3. A coil of 100 turns encloses an area of 100 cm2. It is placed at an angle of 700 with a

sasho [114]

Explanation:

Given that,

Number of turns in the coil, N = 100

Area of the coil, A = 100 cm² = 0.01 m²

It is placed at an angle of 70°.

Magnetic field, B = 0.1 Wb/m²

We need to find the magnetic flux through the coil and the emf is induced in the coil after 10⁻³ s.

Magnetic flux is given by :

$\phi =BA\cos\theta\\\\\text{For N turns},\\\phi =NBA\cos\theta \\\\\phi=100\times 0.1\times 0.01\times \cos(70)\\\\=0.034\ Wb$

So, the magnetic flux through the coil is 0.1 Wb.

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3 years ago

A 80 W light bulb (normally run at 120 V) is attached to a transformer. The voltage source in the transformer is 65 V and Np = 3

Marina CMI [18]

67.8 turns needed by the secondary coil to run the bulb.

<u>Explanation</u>:

We know that,

$\text { Electric power }(p)=\frac{V^{2}}{R}$

$\text { Hence, } \frac{P_{1}}{P_{2}}=\frac{V_{1}^{2} / R}{V_{2}^{2} / R}$

$\frac{P_{1}}{P_{2}}=\frac{V_{1}^{2}}{V_{2}^{2}}$

For calculating number of turns

$\frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}$

Given that,

$80 \mathrm{W}\left(P_{1}\right) \text { bulb with voltage } 120 \mathrm{V}\left(V_{1}\right) \text { is connected to a transformer. }$

$\text { The source voltage of a transformer is }\left(V_{P}\right) \text { is } 65 \mathrm{V}$

$\text { The number of turns in primary winding of transformer is }\left(N_{P}\right) \text { is } 30 .$

We need to find the number of turns in the secondary winding $\left(N_{S}\right)$ to run the bulb at 120W $\left(P_{2}\right)$

Firstly find the secondary voltage in the transformer use, $\frac{P_{1}}{P_{2}}=\frac{V_{1}^{2}}{V_{2}^{2}}$

$\frac{80}{120}=\frac{120^{2}}{V_{2}^{2}}$

$V_{2}^{2}=\frac{120^{2} \times 120}{80}$

$V_{2}^{2}=\frac{1728000}{80}$

$V_{2}^{2}=21600$

$V_{2}=\sqrt{21600}$

$V_{2}=146.9 \mathrm{V}=V_{S}$

Now, finding the number of turns in secondary coil. Use, $\frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}$

$\frac{30}{N_{S}}=\frac{65}{146.9}$

$N_{S}=\frac{30 \times 146.9}{65}$

$N_{S}=\frac{4407}{65}$ $N_{S}=67.8$

The number of turns in the secondary winding are 67.8 turns.

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3 years ago

Evgen [1.6K]

Answer:

Circuit one will have more current than circuit two

Explanation:

I am assuming that you have to see which circuit has the greater current in this case. Well, this is the perfect example of Ohm's Law, which states the following -

V = IR,

where V = voltage / potential difference, I = current, and R = resistance

If one circuit has twice the voltage and half the resistance of the second circuit, as voltage is directly proportional to the resistance -

2V = I( 1 / 2R ),

4V = IR,

I = 4V / R

Whereas in the second circuit -

V = IR,

I = V / R

As you can note, voltage is directly proportional to the current ( I ) as well as the resistance. The only difference between the two formulas I = 4V / R, and I = V / R is the difference in the voltage. With the voltage being 4 times greater in the first circuit, and current is 4 times greater in the first circuit as well.

<u><em>Hence, circuit one will have more current than circuit two</em></u>

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3 years ago

Read 2 more answers