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Juliette [100K]
2 years ago
15

Does the length of a persons hair affect the power of static electricity

Physics
2 answers:
tamaranim1 [39]2 years ago
5 0

Answer:

Yes

Explanation:

goldenfox [79]2 years ago
3 0
I don’t think it’s does but i’m not 100% sure
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What is an astronomer? Answer in the simpliest way please.
Sphinxa [80]
They would be an expert in astronomy
astronomy:the branch of science that deals with celestial objects
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3 years ago
What is the purpose of an experiment design?
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An experimental design is used to assign variables for testing. In contrast to a control design where nothing is changed, the experimental design allows you to test various new inputs to see how they would vary from the original results.
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3 years ago
A point on the string of a violin moves up and down in simple harmonic motion with an amplitude of 1.24 mm and frequency of 875
mario62 [17]

Using

V = Amplitude x angular frequency(omega)

But omega= 2πf

= 2πx875

=5498.5rad/s

So v= 1.25mm x 5498.5

= 6.82m/s

B. .Acceleration is omega² x radius= 104ms²

5 0
3 years ago
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The stoplights on a street are designed to keep traffic moving at 26 mi/h. the average length of a street block between traffic ligh
bonufazy [111]

We use the formula,

v = \frac{d}{t}.

Here, v is velocity and its value given 26 mi/h ( in m/s,  \frac{26 \times 1.609 \times 10^{3} m}{60 \times 60 s} =11.62 \ m/s ) and d is distance and its value is given 80 m.

Substituting these values in above formula we get,

t = \frac{80 m}{11.62m/s } = 6.88 \ s

Thus, the time delay between green lights on successive blocks to keep the traffic moving continuously is 6.88 s


3 0
3 years ago
Determine the empirical formula for a compound that is 36.86% n and 63.14% o by mass.
olga2289 [7]

Answer:

NO_{1.499}

Explanation:

Let assume that 100 kg of the compound is tested. The quantity of kilomoles for each element are, respectively:

n_{N} = \frac{36.86\,kg}{14.006\,\frac{kg}{kmol} }

n_{N} = 2.632\,kmol

n_{O} = \frac{63.14\,kg}{15.999\,\frac{kg}{kmol} }

n_{O} = 3.946\,kmol

Ratio of kilomoles oxygen to kilomole nitrogen is:

n^{*} = \frac{3.946\,kmol}{2.632\,kmol}

n^{*}= 1.499

It means that exists 1.499 kilomole oxygen for each kilomole nitrogen.

The empirical formula for the compound is:

NO_{1.499}

8 0
3 years ago
Read 2 more answers
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