Answer:
(a) 20 MHz
(b) 1.025 KW
(c) 3.33 ns
(d) 33 pF
Explanation:
(a) 20,000,000 Hz = 20 x 10^6 Hz = 20 Mega Hz = <u>20 MHz</u>
(b) 1025 W = 1.025 x 10^3 W = 1.025 Kilo W = <u>1.025 KW</u>
(c) 0.333 x 10^(-8) s = 3.33 x 10^(-9) s = 3.33 nano s = <u>3.33 ns</u>
(d) 33 x10^(-12)F = 33 pico F = <u>33 pF</u>
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Answer:
The distance between the station A and B will be:
Explanation:
Let's find the distance that the train traveled during 60 seconds.
We know that starts from rest (v(0)=0) and the acceleration is 0.6 m/s², so the distance will be:
Now, we need to find the distance after 25 min at a constant speed. To get it, we need to find the speed at the end of the first distance.
Then the second distance will be:
The final distance is calculated whit the decelerate value:
The final velocity is zero because it rests at station B. The initial velocity will be v(1).
Therefore, the distance between the station A and B will be:
I hope it helps you!
Answer:
(b) Given the Weibull parameters of example 11-3, the factor by which the catalog rating must be increased if the reliability is to be increased from 0.9 to 0.99.
Equation 11-1: F*L^(1/3) = Constant
Weibull parameters of example 11-3: xo = 0.02 (theta-xo) = 4.439 b = 1.483
Explanation:
(a)The Catalog rating(C)
Bearing life:
Catalog rating:
From given equation bearing life equation,
we Dividing eqn (2) with (1)
The Catalog rating increased by factor of 1.26
(b) Reliability Increase from 0.9 to 0.99
Now calculating life adjustment factor for both value of reliability from Weibull parametres
Similarly
Now calculating bearing life for each value
Now using given ball bearing life equation and dividing each other similar to previous problem
Catalog rating increased by factor of 0.61