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3 years ago

13

Define Plastic vs elastic deformation.

1 answer:

Snowcat [4.5K]3 years ago

3 0

Answer:

Plastic deformation, irreversible or permanent. Deformation mode in which the material does not return to its original shape after removing the applied load. This happens because, in plastic deformation, the material undergoes irreversible thermodynamic changes by acquiring greater elastic potential energy.

Elastic deformation, reversible or non-permanent. the body regains its original shape by removing the force that causes the deformation. In this type of deformation, the solid, by varying its tension state and increasing its internal energy in the form of elastic potential energy, only goes through reversible thermodynamic changes.

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Technicians have to determine the flow rate of water in a pipe with the aid of a venturi installation and a mercury au.oaeter. T

Schach [20]

Answer:

Manometric difference x=142.85 mm.

Explanation:

Given :

Pipe diameter $d_1=40 mm$

venturi meter $d_2=20 mm$

We can know that discharge through venturi meter is given as

$Q=C_d\dfrac{A_1A_2\sqrt{2gh}}{\sqrt {A_1^2-A_2^2}}$

$A_1=1.24\times 10^{-3},A_2=3.12\times 10^{-4}$

$Q=A_1V_1$

$Q=1.24\times 10^{-3}\times 1.5=0.00186 m^3/s$

$0.00186=0.97\dfrac{1.24\times 10^{-3}\times 3.12\times 10^{-4} \sqrt{2gh}}{\sqrt {(1.24\times 10^{-3})^2-(3.12\times 10^{-4})^2}}$

h=1.8 m

We know that $h=x\left (\dfrac{\rho_{hg}}{\rho_w}-1\right )$

Where x is the manometric deflection

⇒ $1.8=x\left (\dfrac{13600}{1000}-1\right )$

So x=14.28 mm

Manometric difference x=142.85 mm.

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3 years ago

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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

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sd = √10²+5²

sd = √100+25 = 11.18ohm

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

Z = ( X -u)/sd

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4 years ago

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4 years ago