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3 years ago

What is it called after data is collected they are often arranged in a what?

Physics

2 answers:

Sveta_85 [38]3 years ago

8 0

Arranged in a chart or displayed in a graph...Pretty vague question...

valentinak56 [21]3 years ago

3 0

They are often arranged in a category

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A strong lightning bolt transfers an electric charge of about 31 C to Earth (or vice versa). How many electrons are transferred?

olchik [2.2K]

Answer:

$m=5.78\times 10^{-3}\ g$

$n_e=1.935\times 10^{20}$ is the no. of electrons

Explanation:

Given:

quantity of charge transferred, $Q=31\ C$

No. of electrons in the given amount of charge:

As we have charge on one electron $1.602\times 10^{-19}\ C$

so,

$n_e=\frac{Q}{e}$

$n_e=\frac{31}{1.602\times 10^{-19}}$

$n_e=1.935\times 10^{20}$ is the no. of electrons

Now if each water molecules donates one electron:

Then we require $n=1.935\times 10^{20}$ molecules.

Now the no. of moles in this many molecules:

$n_m=\frac{n}{N_A}$

where

$N_A=6.022\times 10^{23}$ Avogadro No.

$n_m=\frac{1.935\times 10^{20}}{6.022\times 10^{23}}$

$n_m=3.213\times 10^{-4}\ moles$

We have molecular mass of water as M=18 g/mol.

So, the mass of water in the obtained moles:

$n_m=\frac{m}{M}$

where:

m = mass in gram

$3.213\times 10^{-4}=\frac{m}{18}$

$m=5.78\times 10^{-3}\ g$

7 0

3 years ago

The latent heat of fusion of alcohol is 50 kcal/kg and its melting point is -54oC. It has a specific heat of 0.60 in its liquid

sashaice [31]

Answer:

Explanation:

To melt the alcohol

Heat needed = M . L = 2 . 25 = 50 kcal

To warm up the alcohol

Heat needed = M . sp. ht. . ∆t = 2 . 0.6 . 100 = 120 kcal

Total heat needed = 170 kcal

Assuming that 0.6 kcal/ kg / ˚C is the specific heat and that the answer is wanted in kcal ( a rather odd unit to be in use here.)

5 0

3 years ago

A student standing in a canyon yells "echo", and her voice produces a sound wave of frequency of f = 0.61 kHz. The echo takes t

klasskru [66]

Answer:

$d=627.9\ m$ is the distance from the obstacle of reflection.

wavelength $\lamb=0.5279\ m$

Explanation:

Given that:

frequency of sound, $f=610\ Hz$

time taken for the echo to be heard, $t=3.9\ s$

speed of sound, $v=322\ m.s^{-1}$

We know,

$\rm distance = speed \times time$

During an echo the sound travels the same distance back and forth.

$2d=v.t$

$2d=322\times 3.9$

$d=627.9\ m$ is the distance from the obstacle of reflection.

Now the wavelength of sound waves:

$\lambda=\frac{v}{f}$

$\lambda=\frac{322}{610}$

$\lamb=0.5279\ m$

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Answer:

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Explanation:

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