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ArbitrLikvidat [17]
3 years ago
12

The mass of one atom would best be measured in this unit:

Physics
1 answer:
trapecia [35]3 years ago
8 0

Answer:

mass of one atom is measured in the unit of 1 amu

Explanation:

As we know that the mass of one atom is very small and it is of order of 10^{-27}

here we know that 1 amu is defined as 1/12 part of mass of 1 atom of C-12 isotope

so here we will have

1 amu = 1.66 \times 10^{-27} kg

so here we can say that mass of one atom is measured in the unit of 1 amu

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Andrej [43]
Answer: A positively charged ion/ cation

Explanation: When an atom loses all of its electrons, it becomes a positively charged ion. Also referred to as the cation, the object will have a shortage of electrons, leading to a positive charge. Keep in mind that an atom will become positive if it loses electrons, and will become negative if it gains electrons.


Hope this helps! Comment below for more questions.
7 0
2 years ago
(a) If two sound waves, one in a gas medium and one in a liquid medium, are equal in intensity, what is the ratio of the pressur
GarryVolchara [31]

Answer:

(a) The ratio of the pressure amplitude of the waves is 43.21

(b) The ratio of the intensities of the waves is 0.000535

Explanation:

Given;

density of gas, \rho _g = 2.27 kg/m³

density of liquid, \rho _l = 972 kg/m³

speed of sound in gas, C_g = 376 m/s

speed of sound in liquid, C_l = 1640 m/s

The of the sound wave is given by;

I = \frac{P_o^2}{2 \rho C} \\\\P_o^2 = 2 \rho C I\\\\p_o = \sqrt{2 \rho CI}

Where;

P_o is the pressure amplitude

P_o_g= \sqrt{2 \rho _g C_gI} -------(1)\\\\P_o_l= \sqrt{2 \rho _l C_lI}---------(2)\\\\\frac{P_o_l}{P_o_g} = \frac{\sqrt{2 \rho _l C_lI}}{\sqrt{2 \rho _g C_gI}} \\\\\frac{P_o_l}{P_o_g} = \sqrt{\frac{2 \rho _l C_lI}{2 \rho _g C_gI} }\\\\ \frac{P_o_l}{P_o_g} = \sqrt{\frac{ \rho _l C_l}{ \rho _g C_g} }\\\\ \frac{P_o_l}{P_o_g} = \sqrt{\frac{ (972)( 1640)}{ (2.27)( 376)} }\\\\\frac{P_o_l}{P_o_g} = 43.21

(b) when the pressure amplitudes are equal, the ratio of the intensities is given as;

I = \frac{P_o^2}{2 \rho C}\\\\I_g = \frac{P_o^2}{2 \rho _g C_g}-------(1)\\\\I_l = \frac{P_o^2}{2 \rho _l C_l}-------(2)\\\\\frac{I_l}{I_g} = (\frac{P_o^2}{2 \rho _l C_l})*(\frac{2\rho_gC_g}{P_o^2} )\\\\\frac{I_l}{I_g} = \frac{\rho _gC_g}{\rho_lC_l} \\\\\frac{I_l}{I_g} = \frac{(2.27)(376)}{(972)(1640)}\\\\ \frac{I_l}{I_g} = 0.000535

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3 years ago
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