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Fudgin [204]
3 years ago
9

A system consists of N very weakly interacting particles at a temperature T sufficiently high so that classical statistical mech

anics is applicable. Each particle has a mass m and is free to perform one-dimensional oscillations about its equilibrium position. Calculate the heat capacity of this system of particles at this temperature in each of the following cases:

Engineering
1 answer:
algol [13]3 years ago
7 0

Answer:

the restoring force is = 3/4NKT

Explanation:

check the attached files for answer.

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3 years ago
A 03-series cylindrical roller bearing with inner ring rotating is required for an application in which the life requirement is
-BARSIC- [3]

Answer:

\mathbf{C_{10} = 137.611 \ kN}

Explanation:

From the information given:

Life requirement = 40 kh = 40 40 \times 10^{3} \ h

Speed (N) = 520 rev/min

Reliability goal (R_D) = 0.9

Radial load (F_D) = 2600 lbf

To find C10 value by using the formula:

C_{10}=F_D\times \pmatrix \dfrac{x_D}{x_o +(\theta-x_o) \bigg(In(\dfrac{1}{R_o}) \bigg)^{\dfrac{1}{b}}} \end {pmatrix} ^{^{^{\dfrac{1}{a}}

where;

x_D = \text{bearing life in million revolution} \\  \\ x_D = \dfrac{60 \times L_h \times N}{10^6} \\ \\ x_D = \dfrac{60 \times 40 \times 10^3 \times 520}{10^6}\\ \\ x_D = 1248 \text{ million revolutions}

\text{The cyclindrical roller bearing (a)}= \dfrac{10}{3}

The Weibull parameters include:

x_o = 0.02

(\theta - x_o) = 4.439

b= 1.483

∴

Using the above formula:

C_{10}=1.4\times 2600 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{1}{\dfrac{10}{3}}}

C_{10}=3640 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{3}{10}}

C_{10} = 3640 \times \bigg[\dfrac{1248}{0.9933481582}\bigg]^{\dfrac{3}{10}}

C_{10} = 30962.449 \ lbf

Recall that:

1 kN = 225 lbf

∴

C_{10} = \dfrac{30962.449}{225}

\mathbf{C_{10} = 137.611 \ kN}

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