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Usimov [2.4K]
2 years ago
10

45. What is the wovelength of a 30. Hertz periodic

Physics
1 answer:
Leviafan [203]2 years ago
4 0

The wavelength is 2m.

Hence, Option c) 2m is the correct answer

Given that;

Frequency; f = 30Hz

Speed; v = 60m/s

Wavelength; \lambda =\ ?

using the expression for the relations between wavelength, frequency and speed of wave:

\lambda = \frac{v}{f}

Where \lambda is wavelength, f is frequency and v is speed.

We substitute our given values into the equation

\lambda = \frac{60m/s}{30Hz}\\\\\lambda = \frac{60m/s}{30s^{-1}}\\\\\lambda = 2m

The wavelength is 2m.

Hence, Option c) 2m is the correct answer.

To learn more about wavelength, click here: brainly.com/question/1347107

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Answer:

B. changing by a constant amount each second

Explanation:

thats my answer

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3 years ago
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If a substance is found to be reactive flammable soluble and explosive what observation is also a physical property
Ainat [17]

Answer:

  • <u><em>soluble</em></u>

Explanation:

Chemical properties only manifest when a chemical reaction occurs. Being reactive, flammable and explosive are chemical properties, because they involve chemical reactions: the substances are changed; the chemical bonds of some substances, called reactants, are broken, and the chemical bonds are created, forming other substances, called products.

Solubility is a<em> physical property</em> because during dissolution no new substances are formed. You can prove it when the solvent evaporates leaving behind the same original substance.

The the observation that the substance is <em>soluble</em> is describing a <em>physical property.</em>

3 0
3 years ago
The latent heat of fusion for Aluminium is 3.97 x 105. How much energy would be required to melt 0.75 kg of it?
RoseWind [281]

Answer:

E = 2.9775\times10^5 J

Explanation:

Given:  The latent heat of fusion for Aluminum is L = 3.97\times10^5  J/Kg

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Energy require to melt E = mL

E = 3.97\times10^5\times0.75 = 2.9775\times10^5 J

Therefore, energy required to melt 0.75 Kg aluminum

E = 2.9775\times10^5 J

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3 years ago
How can a magnetic field be produced, using a wire, a battery, and and a nail?
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Step 2 - Remove some Insulation
Step 3 - Wrap the Wire Around the Nail
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5 0
3 years ago
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