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avanturin [10]
3 years ago
11

Consider an ideal monatomic gas of N particles with mass m in thermal equilibrium at a temperature T. The gas is contained in a

cube of side L whose top and bottom are parallel to the Earth’s surface, and the gravitational acceleration is a constant g throughout the box. (a) What is the average kinetic energy of an atom of this gas?
Physics
1 answer:
harina [27]3 years ago
3 0

Answer:

K.E.=\dfrac{3}{2}KT

Explanation:

Given that

Number of particle =N

Equilibrium temperature= T

Side of cube = L

Gravitational acceleration =g

The kinetic energy of an atom  given as

K.E.=\dfrac{3}{2}KT

Where

Equilibrium temperature= T

Boltzmann constant =K

        K =1.380649×10−23 J/K

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irina1246 [14]

Answer:

a  average load current = 11.33 A

b rms load current = 8.02A

c true power =962.64 W

d apparent power =962.64 W

e. power factor cosθ =1

Explanation:

Vs (t) = 170 sin(377t) v

Vm =170v

Vrms = 170/√2 =120.23 v

Im = Vm/R = 170/15 = 11.33 A

Irms = Im/ √2 = 11.33/√2 =8.02A

Resistors are electronic components that consume energy

the power in a resistor is given by  P =IVcosθ ; in a resistor cosθ =1

P =IV

The electrical power consumed by a resistance, (R) is called the true or real power

and is obtained by multiplying the rms voltage with the rms current.

P= Vrms × Irms

120.03×8.02

P= 962.64 W ; true power

apparent power = Vrms × Irms

=120.03×8.02

= 962.64W ; apparent power

power factor cosθ = true power/ apparent power

cosθ = 962.64/962.64

cosθ = 1

For the purely resistive circuit, the power factor is 1 , because the reactive power is equal to zero (0).

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Please find attached photograph for your answer.

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A 50.0-kg block is being pulled up a 16.0° slope by a force of 250 n that is parallel to the slope. the coefficient of kinetic
Drupady [299]
When dealing with multiple forces acting on a body, it is advisable to draw a free-body diagram like that shown in the picture. There are four forces acting on the box: weight (W) pointing straight down, normal force perpendicular to the slope denoted as Fn, force used to push the box upwards along the slope and the frictional force acting opposite to the direction of motion of the box denoted as Ff. Frictional force is equal to coefficient of kinetic friction (μk) multiplied with Fn.

∑Fy = Fn - mgcos30° = 0
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