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pochemuha
1 year ago
12

very fine smoke particles are suspended in air. the translational rms speed of a smoke particle is 2.45 10-3 m/s, and the temper

ature is 288 k. find the mass of a particle?
Physics
1 answer:
m_a_m_a [10]1 year ago
5 0

The mass of a particle is 2.2x10⁻¹⁵ kg

Consider smoke particles as an ideal gas

The translational RMS speed of the smoke particles is 2.45x10⁻³ m/s.

<em>v= √3kT/m</em>

<em>where k= 1.38x10⁻²³J/K, T is 288K, and m is the mass of the smoke particle</em>

<em>2.45x10⁻³ = √3x1.38x10⁻²³x288/m</em>

<em>m= 2.2x10⁻¹⁵ kg</em>

Therefore, the mass of a particle is 2.2x10⁻¹⁵ kg.

To learn more about the translational root mean square speed of gases, visit brainly.com/question/6853705

#SPJ4

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Given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Given the data in the question;

Hubble's constant; H_0 = 51km/s/Mly

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Age\ of\ Universe; t = \frac{1}{H_0}

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1\ light\ years = 9.46*10^{15}m

so

1\ Million\ light\ years = [9.46 * 10^{15}m] * 10^6 = 9.46 * 10^{21}m

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H_0 = 51\frac{km}{\frac{s}{Mly} } = 51000\frac{m}{s\ *\ Mly}  \\\\H_0 = 51000\frac{m}{s\ *\ (9.46*10^{21}m)} \\\\H_0 =  5.39 *10^{-18}s^{-1}\\

Now, we input this Hubble's constant value into our equation;

Age\ of\ Universe; t = \frac{1}{H_0}\\\\t = \frac{1}{ 5.39 *10^{-18}s^{-1}} \\\\t = 1.855 * 10^{17}s\\\\We\ convert\ to\ years\\\\t =  \frac{ 1.855 * 10^{17}}{60*60*24*365}yrs \\\\t = \frac{ 1.855 * 10^{17}}{31536000}yrs\\\\t = 5.88 *10^9 years

Therefore, given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Learn more: brainly.com/question/14019680

6 0
2 years ago
The kitchen in your house measures 10 ft by 10 ft. You need to have 50 footcandles of illumination in the room. How many 60 watt
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Answer:

<u>6 bulbs</u> are needed to illuminate the room.

Explanation:

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Now, one bulb emits = 834 lumes

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N=\frac{Number\ of\ lumens}{Number\ of\ lumens\ by\ 1\ bulb}\\\\N=\frac{5000}{834}=5.995\approx6

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3 years ago
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