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Dominik [7]
2 years ago
12

Technician A says that a circuit with continuity reads 0 ohms. Technician B says that an open circuit reads 0 ohms. Who is corre

ct
Engineering
1 answer:
patriot [66]2 years ago
8 0

Answer:

  A

Explanation:

An open circuit has infinite resistance, not 0. A circuit with continuity usually measures a few ohms or less, depending on the length and gauge of the wire and the condition of the connections. Often, on a general-purpose ohmmeter, the reading is very near 0 ohms.

Technician A is correct.

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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

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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}

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The Weibull parameters include:

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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

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1 kN = 225 lbf

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

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

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