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3 years ago

8

A sample of an unknown liquid has a volume of 24.0 mL and a mass of 6 g. What is its density? Show your work or explain how you

determined this value.

2 answers:

jek_recluse [69]3 years ago

8 0

Answer: 0.25 grams per milliliter

Explanation:

The formula to calculate the density is given by :-

$\text{Density}=\dfrac{\text{Mass}}{\text{Volume}}$

We are given that the volume of liquid is 24.0 mL.

Mass = 6g

Now, the density of liquid is given by :-

$\text{Density}=\dfrac{6}{24}\\\\\Rightarrow\ \text{Density}=0.25\ g/ml$

Hence, its density = 0.25 grams per milliliter.

adoni [48]3 years ago

3 0

I think this might be the answer let me know 0.2 g/cm^3

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Read 2 more answers

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Masja [62]

Answer:

Wavelength of the incident wave in air = 1 m

Wavelength of the incident wave in medium 2 = 0.33 m

Intrinsic impedance of media 1 = 377 ohms

Intrinsic impedance of media 2 = 125.68 ohms

Check the explanation section for a better understanding

Explanation:

a) Wavelength of the incident wave in air

The frequency of the electromagnetic wave in air, f = 300 MHz = 3 * 10⁸ Hz

Speed of light in air, c = 3 * 10⁸ Hz

Wavelength of the incident wave in air:

$\lambda_{air} = \frac{c}{f} \\\lambda_{air} = \frac{3 * 10^{8} }{3 * 10^{8}} \\\lambda_{air} = 1 m$

Wavelength of the incident wave in medium 2

The refractive index of air in the lossless dielectric medium:

$n = \sqrt{\epsilon_{r} } \\n = \sqrt{9 }\\n =3$

$\lambda_{2} = \frac{c}{nf}\\\lambda_{2} = \frac{3 * 10^{6} }{3 * 3 * 10^{6}}\\\lambda_{2} = 1/3\\\lambda_{2} = 0.33 m$

b) Intrinsic impedances of media 1 and media 2

The intrinsic impedance of media 1 is given as:

$n_1 = \sqrt{\frac{\mu_0}{\epsilon_{0} } }$

Permeability of free space, $\mu_{0} = 4 \pi * 10^{-7} H/m$

Permittivity for air, $\epsilon_{0} = 8.84 * 10^{-12} F/m$

$n_1 = \sqrt{\frac{4\pi * 10^{-7} }{8.84 * 10^{-12} } }$

$n_1 = 377 \Omega$

The intrinsic impedance of media 2 is given as:

$n_2 = \sqrt{\frac{\mu_r \mu_0}{\epsilon_r \epsilon_{0} } }$

Permeability of free space, $\mu_{0} = 4 \pi * 10^{-7} H/m$

Permittivity for air, $\epsilon_{0} = 8.84 * 10^{-12} F/m$

ϵr = 9

$n_2 = \sqrt{\frac{4\pi * 10^{-7} *1 }{8.84 * 10^{-12} *9 } }$

$n_2 = 125.68 \Omega$

c) The reflection coefficient,r and the transmission coefficient,t at the boundary.

Reflection coefficient, $r = \frac{n - n_{0} }{n + n_{0} }$

You didn't put the refractive index at the boundary in the question, you can substitute it into the formula above to find it.

$r = \frac{3 - n_{0} }{3 + n_{0} }$

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Maximum amplitudes in the total field is given by:

$E = tE_{0}$ and $E = r E_{0}$

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Verdich [7]

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This Law is originally expressed as follows:

<h2> $T^{2} =\frac{4\pi^{2}}{GM}a^{3}$ (1) </h2>

Where;

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$a=4.22(10^{5})km=4.22(10^{8})m$ is the semimajor axis of the orbit Io describes around Jupiter (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

<h2> $T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}$ (2) </h2>

$T=\sqrt{\frac{4\pi^{2}}{6.674(10^{-11})\frac{m^{3}}{kgs^{2}}1.9(10^{27})kg}(4.22(10^{8})m)^{3}}$

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<h2> $T=152938.0934s$ (3) </h2>

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<h2> $T=42.482h$ </h2>

Therefore, the answer is:

The orbital period of Io is 42.482 h

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3 years ago