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3 years ago

Copper is generally considered easy to weld because of its high thermal conductivity: (a) true or (b) false?

Engineering

1 answer:

inn [45]3 years ago

8 0

Answer:

b)False

Explanation:

Those material have high thermal it is very difficult to weld because due to high thermal conductivity it transmit the heat in to the surrounding and can not reach a particular temperature required to melt the material.And when material does not melt then there is no possibility to weld the material.

So from above we can say that it is very difficult to weld the copper material due to high thermal conductivity.Generally welding of copper done by usiong gas welding technique.

So our option b is right.

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A large prime number isP = 232582657 - 1

sattari [20]

Answer:

32582657

Explanation:

If any number is in the form of 2^{n}-1 is known as Mersenne prime. Here, n is a prime number.

For example:

If n is 3 then the corresponding binary number is as follows:

P=2^{3}-1

P=7

Here, the binary representation of P is (111)₂.

The number of binary digits is 3 which are equal to n.

Consider the given expression

P=2^32582657-1

This is also in the form 2^{n}-1

Here the value of n is 32582657.

Hence, the number of binary digits for the given prime number is 32582657

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3 years ago

Oil, with density of 900 kg/m3 and kinematic viscosity of 0.00001 m2/s, flows at 0.2 m3/s through 500 m of 200-mm-diameter cast-

Vsevolod [243]

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2 years ago

An insulated tank having a total volume of 0.6 m3 is divided into two compartments. Initially one compartment contains 0.4 m3 of

IRISSAK [1]

Answer:

a. The final temperature is 69.8°C

b. The final pressure is 2.67bar

c. The amount of entropy produced is 0.38568kJ/K

Explanation:

Pls, check the attached files for detail step by step explanation as typing the solution here can not be explicit.

5 0

3 years ago

Most equipment is cooled by bringing cold air in the front and ducting the heat out of the back. What is the term for where the

vitfil [10]

Answer: Hot aisle

Explanation:

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2 years ago

The solid cylinders AB and BC are bonded together at B and are attached to fixed supports at A and C. The modulus of rigidity is

romanna [79]

Answer:

a) 0.697*10³ lb.in

b) 6.352 ksi

Explanation:

For cylinder AB:

Let Length of AB = 12 in

$c=\frac{1}{2}d=\frac{1}{2} *1.1=0.55in\\ J=\frac{\pi c^4}{2}=\frac{\pi}{2}0.55^4=0.1437\ in^4\\$

$\phi_B=\frac{T_{AB}L}{GJ}=\frac{T_{AB}*12}{3.3*10^6*0.1437} =2.53*10^{-5}T_{AB}$

For cylinder BC:

Let Length of BC = 18 in

$c=\frac{1}{2}d=\frac{1}{2} *2.2=1.1in\\ J=\frac{\pi c^4}{2}=\frac{\pi}{2}1.1^4=2.2998\ in^4\\$

$\phi_B=\frac{T_{BC}L}{GJ}=\frac{T_{BC}*18}{5.9*10^6*2.2998} =1.3266*10^{-6}T_{BC}$

$2.53*10^{-5}T_{AB}=1.3266*10^{-6}T_{BC}\\T_{BC}=19.0717T_{AB}$

$T_{AB}+T_{BC}-T=0\\T_{AB}+T_{BC}=T\\T_{AB}+T_{BC}=14*10^3\ lb.in\\but\ T_{BC}=19.0717T_{AB}\\T_{AB}+19.0717T_{AB}=14*10^3\\20.0717T_{AB}=14*10^3\\T_{AB}=0.697*10^3\ lb.in\\T_{BC}=13.302*10^3\ lb.in$

b) Maximum shear stress in BC

$\tau_{BC}=\frac{T_{BC}}{J}c=13.302*10^3*1.1/2.2998=6.352\ ksi$

Maximum shear stress in AB

$\tau_{AB}=\frac{T_{AB}}{J}c=0.697*10^3*0.55/0.1437=2.667\ ksi$

8 0

3 years ago