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Usimov [2.4K]
3 years ago
14

A sealed container holds 0.020 moles of ideal nitrogen (n2) gas, at a pressure of 1.5 atmospheres and a temperature of 290k. the

atomic mass of nitrogen is 14.0g/mol. what is approimate translational kinetic energy of nitrogen molecuole, in si units
Physics
1 answer:
avanturin [10]3 years ago
6 0
Translational kinetic energy is the kinetic energy of an object of finite size. It is equal to one half the product of the mass of the object and the square of the velocity. For an ideal gas, it is simplified into the expression:

Kinetic energy = 3nRT / 2   where n is the number of moles, R is the universal gas constant and T is the temperature of the gas.

We calculate as follows:
Kinetic energy = 3nRT / 2 
Kinetic energy = 3 (0.020 moles ) (8.314 J / mol-K) (290 K) / 2 
Kinetic energy = 72.33 J

Therefore, the translational kinetic energy of the ideal nitrogen gas is 72.33 J.
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An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate
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Answer:

time required after impact for a puck is 2.18 seconds

Explanation:

given data

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diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 ×10^{-4} m

dynamic viscosity = 1.75 ×10^{-5} Ns/m²

air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

so final velocity = 0.9v

so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ \frac{du}{dy}

so

= µ  \frac{v}{h}   ............1

put here value

= 1.75×10^{-5} × \frac{v}{10^{-4}}

= 0.175 v

and

area between air and puck is given by

Area = \frac{\pi }{4} d^{2}

area  =  \frac{\pi }{4} 0.1^{2}

area = 7.85 × \frac{v}{10^{-3}} m²

so

force on puck is express as

Force = × area

force = 0.175 v × 7.85 × 10^{-3}

force = 1.374 × 10^{-3} v    

and now apply newton second law

force = mass × acceleration

- force = mass \frac{dv}{dt}

- 1.374 × 10^{-3} v = 0.03 \frac{0.9v - v }{t}

t =  \frac{0.1 v * 0.03}{1.37*10^{-3} v}

time = 2.18

so time required after impact for a puck is 2.18 seconds

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1. A bag dropped from a helicopter falls with an acceleration of 9.8m/s2. What is its velocity after 5s?
Tanzania [10]

Answer:

The answer is 49m/s.

Explanation:

Given,

acceleration (a) =9.8m/s^2

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now, final velocity (v) after 5s =?

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Therefore, the velocity after 5s is 49m/s.

<em><u>Hope</u></em><em><u> </u></em><em><u>it helps</u></em><em><u>.</u></em><em><u>.</u></em>

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