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3 years ago

What is the mathematical equation for calculating the internal energy of a diatomic gas? -

Physics

1 answer:

Mariulka [41]3 years ago

7 0

Answer:

U = 5/2 NkT.

Explanation:

A diatomic molecule = H₂, O₂, N₂ .....etc

A diatomic molecule has 5 degrees of freedom.

3 transnational modes = x, y, z
2 rotational modes = ωx , ωy

So, the internal energy for diatomic gases is :

U = 5/2 NkT.

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HELPPP WILL GIVE BRAINLIEST!!!!

marishachu [46]

Explanation:

Thermal energy is basically the energy derived from heat when there is a temperature change in a body.

Kinetic energy is the energy due to the motion of a body.

Potential energy is the energy at rest or due to the position of a body.

Thermal energy is highly related to kinetic and potential energy.
Thermal energy is a type of kinetic energy because it causes the motion of the particles in a body.
An increase in thermal energy gives the particles more energy to move and then, their kinetic energy increases.
Kinetic energy is directly related to thermal energy.
A low thermal energy increases the potential energy in a body.
A body at rest will exert its maximum potential energy.

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4 years ago

Samantha walked 5km/h . how long did it take to her to travel 15 km?

valkas [14]

Answer:

3hrs is the answer

Explanation:

time=d/t

t=15/5 hrs

t=3 hrs

4 0

3 years ago

A square wave has amplitude 0 V for the low voltage and 4 V for the high voltage. Calculate the average voltage by integrating o

Margarita [4]

Answer:

V_{average} = $\frac{1}{2} V_o$ , V_{average} = 2 V

Explanation:

he average or effective voltage of a wave is the value of the wave in a period

V_average = ∫ V dt

in this case the given volage is a square wave that can be described by the function

V (t) = $\left \{ {{V=V_o \ \ \ t< \tau /2} \atop {V=0 \ \ \ \ t> \tau /2 } } \right.$

to substitute in the equation let us separate the into two pairs

V_average = $\int\limits^{1/2}_0 {V_o} \, dt + \int\limits^1_{1/2} {0} \, dt$

V_average = $V_o \ \int\limits^{1/2}_0 {} \, dt$

V_{average} = $\frac{1}{2} V_o$

we evaluate V₀ = 4 V

V_{average} = 4 / 2)

V_{average} = 2 V

6 0

3 years ago

A crate of mass 9.2 kg is pulled up a rough incline with an initial speed of 1.58 m/s. The pulling force is 110 N parallel to th

kari74 [83]

Given that,

Mass = 9.2 kg

Force = 110 N

Angle = 20.2°

Distance = 5.10 m

Speed = 1.58 m/s

(A). We need to calculate the work done by the gravitational force

Using formula of work done

$W_{g}=mgd\sin\theta$

Where, w = work

m = mass

g = acceleration due to gravity

d = distance

Put the value into the formula

$W_{g}=9.2\times(-9.8)\times5.10\sin20.2$

$W_{g}=-158.8\ J$

(B). We need to calculate the increase in internal energy of the crate-incline system owing to friction

Using formula of potential energy

$\Delta U=-W$

Put the value into the formula

$\Delta U=-(-158.8)\ J$

$\Delta U=158.8\ J$

(C). We need to calculate the work done by 100 N force on the crate

Using formula of work done

$W=F\times d$

Put the value into the formula

$W=100\times5.10$

$W=510\ J$

We need to calculate the work done by frictional force

Using formula of work done

$W=-f\times d$

$W=-\mu mg\cos\theta\times d$

Put the value into the formula

$W=-0.4\times9.2\times9.8\cos20.2\times5.10$

$W=-172.5\ J$

We need to calculate the change in kinetic energy of the crate

Using formula for change in kinetic energy

$\Delta k=W_{g}+W_{f}+W_{F}$

Put the value into the formula

$\Delta k=-158.8-172.5+510$

$\Delta k=178.7\ J$

(E). We need to calculate the speed of the crate after being pulled 5.00m

Using formula of change in kinetic energy

$\Delta k=\dfrac{1}{2}m(v_{2}^2-v_{1}^{2})$

$v_{2}^2=\dfrac{2\times\Delta k}{m}+v_{1}^2$

Put the value into the formula

$v_{2}^2=\dfrac{2\times178.7}{9.2}+1.58$

$v_{2}=\sqrt{\dfrac{2\times178.7}{9.2}+1.58}$

$v_{2}=6.35\ m/s$

Hence, (A). The work done by the gravitational force is -158.8 J.

(B). The increase in internal energy of the crate-incline system owing to friction is 158.8 J.

(C). The work done by 100 N force on the crate is 510 J.

(D). The change in kinetic energy of the crate is 178.7 J.

(E). The speed of the crate after being pulled 5.00m is 6.35 m/s

8 0

3 years ago

The plates of a parallel plate capacitor are separated by d = 1.6 cm. The potential of the negative plate is 0 V, and the potent

yaroslaw [1]

Answer:

The electric field between the plates is 1875 V/m.

Explanation:

Given that,

Separated d=1.6 cm

Potential of negative plate = 0 V

Potential halfway between the plates = 15 V

We need to calculate the total potential

Using formula of potential

$V=2\times\text{Potential halfway between the plates}$

Put the value into the formula

$V=2\times15$

$V=30\ V$

We need to calculate the electric field between the plates

Using formula of electric field

$E=\dfrac{V}{d}$

$E=\dfrac{30}{1.6\times10^{-2}}$

$E=1875\ V/m$

Direction is negative as the field always points from positive to negative plate

Hence, The electric field between the plates is 1875 V/m.

5 0

3 years ago