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oee [108]
3 years ago
7

A system has 4 independent and identical motors, in order for the system to work, at least one of the motors must operate. Each

motor's failure rate is 0.002 failures per hour. What is the system's Mean Time to Failure (MTTF)?
Physics
1 answer:
grin007 [14]3 years ago
8 0

Answer:

The system's Mean Time to Failure is 1,041.\bar 6 hours

Explanation:

The number of independent identical motors = 4

The failure rate of each motor, λ = 0.002 failures per hour

Given that in order for the system to work, at least one of the motors must operate, therefore, the system are in parallel

The Mean Time to Failure for a parallel system of four identical components is given as follows;

MTTF = \dfrac{1 }{\lambda} \times \left (1 + \dfrac{1}{2} +  \dfrac{1}{3} + \dfrac{1}{4} \right)

Which gives;

MTTF = \dfrac{1 }{0.002} \times \left (1 + \dfrac{1}{2} +  \dfrac{1}{3} + \dfrac{1}{4} \right) = 1,041.\bar 6

The system's Mean Time to Failure = 1,041.\bar 6 hours.

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v=\dfrac{d}{t}\\\\v=\dfrac{500\ miles}{3.217\ h}\\\\v=155.42\ mph

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Two Earth satellites, A and B, each of mass m = 940 kg , are launched into circular orbits around the Earth's center. Satellite
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Answer:

The required work done is 6.5\times10^{9}\ J

Explanation:

Given that,

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Using formula of potential

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Put the value into the formula

U_{A}=-\dfrac{6.67\times10^{-11}\times940\times5.98\times10^{24}}{6.38\times10^{6}+4.50\times10^{6}}

U_{A}=-3.44\times10^{10}\ J

We need to calculate the potential energy

Using formula of potential

U_{B}=-\dfrac{Gm_{B}m_{E}}{r_{A}}

Put the value into the formula

U_{B}=-\dfrac{6.67\times10^{-11}\times940\times5.98\times10^{24}}{6.38\times10^{6}+11.10\times10^{6}}

U_{B}=-2.14\times10^{10}\ J

We need to calculate the value of k_{A}

Using formula of k_{A}

k_{A}=-\dfrac{1}{2}U_{A}

Put the value into the formula

k_{A}=\dfrac{1}{2}\times3.44\times10^{10}

k_{A}=1.72\times10^{10}\ J

We need to calculate the value of k_{B}

Using formula of k_{B}

k_{B}=-\dfrac{1}{2}U_{B}

Put the value into the formula

k_{B}=\dfrac{1}{2}\times2.14\times10^{10}

k_{B}=1.07\times10^{10}\ J

We need to calculate the work done

Using formula of work done

W=\Delta K+\Delta U

W=(k_{B}-k_{A})+(U_{B}-U_{A})

W=(-\dfrac{U_{B}}{2}+\dfrac{U_{A}}{2})+(U_{B}-U_{A})

W=\dfrac{1}{2}(U_{B}-U_{A})

Put the value into the formula

W=\dfrac{1}{2}\times(-2.14\times10^{10}+3.44\times10^{10})

W=6.5\times10^{9}\ J

Hence, The required work done is 6.5\times10^{9}\ J

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