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2 years ago

If a system reliability of 0.998 is required, what reliability of two components in series is required?

Physics

1 answer:

Yakvenalex [24]2 years ago

3 0

Two components in the series have dependability of $0.996$ .

We need both two components to function in order to determine the engine's reliability. Since each component has dependability of $0.998$ , we must determine the likelihood that both components will function in order to determine the engine's reliability.

The possibility that a product, system, or service will function as intended for a predetermined amount of time, or that it will run faultlessly in a predetermined environment, is known as reliability.

By multiplying each of the $10$ reliability coefficients, we may calculate this probability:

$P = 0.998^2 = 0.996004$

P is equal to $0.9966$ after being rounded to three decimal places.

The engine has a dependability rating of $0.996$ .

Hence,

Two components connected in series have dependability of $0.996$ .

<h3>What is the compound of reliability ?</h3>

Reliability and failure probability are complimentary, or $F= 1 - R$ .

In various cases, the distinctive qualities of a series arrangement will be demonstrated. $R = R 1 R 2$ is the resultant reliability of two components.

A component's reliability is assessed in light of the application for which it will be utilised. An operational profile gives a description of the situation. Software reliability can be calculated or measured.

To learn more about compound of reliability, Visit:

brainly.com/question/14972427

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2 years ago

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2 years ago

A high-pass filter consists of a 1.66 μF capacitor in series with a 80.0 Ω resistor. The circuit is driven by an AC source with

Julli [10]

Explanation:

Given that,

Capacitor $C=1.66\ \mu F$

Resistor $R=80.0\ \Omega$

Peak voltage = 5.10 V

(A). We need to calculate the crossover frequency

Using formula of frequency

$f_{c}=\dfrac{1}{2\pi R C}$

Where, R = resistor

C = capacitor

Put the value into the formula

$f_{c}=\dfrac{1}{2\pi\times80.0\times1.66\times10^{-6}}$

$f_{c}=1198.45\ Hz$

(B). We need to calculate the $V_{R}$ when $f = \dfrac{1}{2f_{c}}$

Using formula of $V_{R}$

$V_{R}=V_{0}(\dfrac{R}{\sqrt{R^2+(\dfrac{1}{2\pi fC})^2}})$

Put the value into the formula

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times\dfrac{1}{2}\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=2.280\ Volt$

(C). We need to calculate the $V_{R}$ when $f = f_{c}$

Using formula of $V_{R}$

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=3.606\ Volt$

(D). We need to calculate the $V_{R}$ when $f = 2f_{c}$

Using formula of $V_{R}$

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times2\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=4.561\ Volt$

Hence, This is the required solution.

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3 years ago

A basketball player is 4.22 m from

max2010maxim [7]

Answer: The height above the release point is 2.96 meters.

Explanation:

The acceleration of the ball is the gravitational acceleration in the y axis.

A = (0, -9.8m/s^)

For the velocity we can integrate over time and get:

V(t) = (9.20m/s*cos(69°), -9.8m/s^2*t + 9.20m/s^2*sin(69°))

for the position we can integrate it again over time, but this time we do not have any integration constant because the initial position of the ball will be (0,0)

P(t) = (9.20*cos(69°)*t, -4.9m/s^2*t^2 + 9.20m/s^2*sin(69°)*t)

now, the time at wich the horizontal displacement is 4.22 m will be:

4.22m = 9.20*cos(69°)*t

t = (4.22/ 9.20*cos(69°)) = 1.28s

Now we evaluate the y-position in this time:

h = -4.9m/s^2*(1.28s)^2 + 9.20m/s^2*sin(69°)*1.28s = 2.96m

The height above the release point is 2.96 meters.

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3 years ago

Read 2 more answers