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Two resistors, with resistances R1 and R2, are connected in series. R1 is normally distributed with mean 65 and standard deviati

Sedbober [7]

Answer:

n this question, we are asked to find the probability that

R1 is normally distributed with mean 65 and standard deviation 10

R2 is normally distributed with mean 75 and standard deviation 5

Both resistor are connected in series.

We need to find P(R2>R1)

the we can re write as,

P(R2>R1) = P(R2-R1>R1-R1)

P(R2>R1) = P(R2-R1>0)

P(R2>R1) = P(R>0)

Where;

R = R2 - R1

Since both and are independent random variable and normally distributed, we can do the linear combinations of mean and standard deviations.

u = u2-u1

u = 75 - 65 = 10ohm

sd = √sd1² + sd2²

sd = √10²+5²

sd = √100+25 = 11.18ohm

Now we will calculate the z-score, to find P( R>0 )

Z = ( X -u)/sd

the z score of 0 is

z = 0 - 10/11.18

z= - 0.89

4 0

4 years ago

A steel cylindrical pressure vessel has an inside diameter of 2m and a wall thickness of 2cm. If the steel has a yield stress Y

Feliz [49]

Answer:

The pressure that will cause yielding in the tank in the longitudinal direction is 1.7 Mpa

Explanation:

We are dealing with a thin-walled pressure vessel here.

the formula for calculating the hoops stress of this pressure vessel is $\sigma =P d/4t$

$\sigma Y =$ longitudinal yield stress = $425 \times 10^{6} Pa$

Thickness = 2/1000 =0.002m

Outside diameter = inside siameter + thickness = 2m + 0.002m= 2.002m

Dm = Mean diameter = Outside diameter - thickness = 2.002 - 0.002 = 2m

$P= 425 \times 10^{6} \times 4 \times 0.002 /2=1.7 Mpa$

4 0

3 years ago

The atomic radii of a divalent cation and a monovalent anion are 0.77 nm and 0.136 nm, respectively.1- Calculate the force of at

MrMuchimi

Answer:

A) attractive force between ions = 5.09x10^-19 N of attractive force

B) there will be no repulsion since both ions have opposite charges.

Explanation:

Detailed explanation and calculation is shown in the image below.

5 0

4 years ago

A milling operation was used to remove a portion of a solid bar of square cross section. Forces of magnitude P = 19 kN are appli

lakkis [162]

Answer:

Complete question for your reference is below.

A milling operation was used to remove a portion of a solid bar of square cross section. Forces of magnitude P = 19 kN are applied at the centers of the ends of the bar. Knowing that a = 28 mm and sigma_all = 130 MPa, determine the smallest allowable depth d of the milled portion of the bar

Explanation:

Please find attached file for complete answer solution and explanation.

3 0

4 years ago

Design complementary static CMOS circuits with minimized number of transistors to realize the following Boolean functions (hint:

Pie

Answer:

as pull up network. the metteing point of pull down and pull up is the point where we take the output

note 1: if two n-mos are connected in series it gives logical AND and p-mos paralle gives logical-AND

note 2: if two n-mos are connected in parallel it gives logical OR and p-mos series gives logical-OR

note 3: output is always complement of what we implement

example Y= (AB)'

image attached

A) F = (ABC + D(A+B) )'

pulldown:

this can be realize by takeing three n-mos in series which gives ABC ,two n-mos are parallel which in series with another n-mos whic gives D(A+B), now connect ABC and D(A+B) in parallel

pull up

this can be realize by takeing three p-mos in parallel which gives ABC ,two p-mos are series which is in serires with

another p-mos whic gives D(A+B), now connect ABC and D(A+B) in series

the out put will be (ABC + D(A+B) )'

so we require total 6-mos and 6-pmos total 12mos transistors

B) F = AC + BD

pull down

this can be realize by takeing two n-mos in series which gives AB ,two n-mos are in series

which whic gives BD, now connect AC and BD in parallel

pull up

this can be realize by takeing two p-mos in parallel which gives Ac ,two p-mos are in parallel

which whic gives BD, now connect AC and BD in series

the output is (AC+BD)'

to avoid the complement we have to connect the output to c-mos inverter then we get AC+BD

so we require 5-nmos, 5-pmos total 10 mos transistors

8 0

3 years ago

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