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3 years ago

What is another term for discontinuity? this is for WELDING, not engineering.

Engineering

1 answer:

lesantik [10]3 years ago

8 0

disconnectedness, disconnection, break, lack of unity, disruption, interruption, lack of coherence, disjointedness.

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In a semiconductor manufacturing process, three wafers from a lot are tested. Each wafer is classified as pass or fail. Assume t

gladu [14]

Answer:

F(x) = 0 ; x < 0

0.064 ; 0 ≤ x < 1

0.352 ; 1 ≤ x < 2

0.784 ; 2 ≤ x < 3

1 ; x ≥ 3

Explanation:

Each wafer is classified as pass or fail.

The wafers are independent.

Then, we can modelate X : ''Number of wafers that pass the test'' as a Binomial random variable.

X ~ Bi(n,p)

Where n = 3 and p = 0.6 is the success probability

The probatility function is given by :

$P(X=x)=f(x)=nCx.p^{x}.(1-p)^{n-x}$

Where $nCx$ is the combinatorial number

$nCx=\frac{n!}{x!(n-x)!}$

Let's calculate f(x) :

$f(0)=3C0.(0.6^{0}).(0.4^{3})=0.4^{3}=0.064$

$f(1)=3C1.(0.6^{1}).(0.4^{2})=0.288$

$f(2)=3C2.(0.6^{2}).(0.4^{1})=0.432$

$f(3)=3C3.(0.6^{3}).(0.4^{0})=0.6^{3}=0.216$

For the cumulative distribution function that we are looking for :

$P(X\leq x)=F(x)$

$F(0)=f(0)\\F(1)=f(0)+f(1)\\F(2)=f(0)+f(1)+f(2)\\F(3)=f(0)+f(1)+f(2)+f(3)=1$

$F(0)=0.064\\F(1)=0.064+0.288=0.352\\F(2)=0.064+0.288+0.432=0.784\\F(3)=0.064+0.288+0.432+0.216=1$

The cumulative distribution function for X is :

F(x) = 0 ; x < 0

0.064 ; 0 ≤ x < 1

0.352 ; 1 ≤ x < 2

0.784 ; 2 ≤ x < 3

1 ; x ≥ 3

5 0

3 years ago

A laboratory in the Y building keep a vacuum pressure of 0.1 kPa abs. What is the net force acting on the door considering the a

seropon [69]

Answer:

net force acting on the floor is 100 kN

Explanation:

Given data:

$P_{vaccum} = 0.1 kPa$

$P_{atm} = 101.325 kPa$

dimension of floor = 2 m \times 0.5 m

we know that

Net force can be calculated as follow

$f_{net} = P_{vaccum} \times area$

$f_{net} = 0.1\times 10^3 \times 2\times 0.5$

$f_{net} = 0.1\times 10^3 \times 1$

$f_{net} = 100 kN$

Therefore net force acting on the floor is 100 kN

7 0

4 years ago

When the grounded conductor is to be spliced in a 12-inch by 12-inch junction box, there shall be at least ? of free conductor l

slamgirl [31]

There should be a 6" free conductor left for splicing

Why there should be 6" splicing?

You must leave the junction box with at least six inches of free conductor wiring when running electrical cables from the box to the box for connecting needs.

Actually, 6-8 inches can be left when running the electric cables, additionally even if the grounded conductor is to be spliced in a 12-inch by 12-inch.

Conductor splicing: A splice, which can be finished using either the crimping or soldering method, is the joining of two or more conductors in a way that produces a permanent electrical termination and mechanical link.

Hence we can conclude that 6 inches is fine for the conductor splicing

To learn more about splicing
brainly.com/question/7668842
#SPJ4

7 0

1 year ago

Question 9 Two cells are connected in parallel. All cells are rated 1.5 V at 10 A. Et = ? (V)

Eddi Din [679]

Parallel hookups increase amp hour capacity but voltage remains the same... 1.5 volts

Series the two and voltage is 3.0 volts.

7 0

4 years ago

LINKS GET BODIED ON SITE! RAWR

bulgar [2K]

Answer:

False

Explanation:

MRK ME BRAINLIEST PLZZZZZZZZZZZZZZZZZZ

7 0

3 years ago