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ella [17]
2 years ago
7

Rolling and Shearing are the types of a)-Bulk Deformation Process b)- Sheet Metal Process c)- Machining Process d)- Both a &

c d)- None of them
Engineering
1 answer:
Margarita [4]2 years ago
5 0

Answer:

a)Bulk deformation process  

Explanation:

<u>Rolling</u>

Rolling is a metal forming process.In rolling work piece passes through two moving rollers and get compressed.in rolling thickness of work piece will reduces and length of work piece will increase for maintaining the constant area.Due to compression bulk deformation takes place.

<u>Shearing</u>

In shearing one surface slides on another surface and deformation take place.shearing is a machining process.This is also a bilk motion deformation process.

So from above we can say that option a is right.

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The ???? − i relationship for an electromagnetic system is given by ???? = 1.2i1/2 g where g is the air-gap length. For current
Artemon [7]

Answer:

a) The mechanical force is -226.2 N

b) Using the coenergy the mechanical force is -226.2 N

Explanation:

a) Energy of the system:

\lambda =\frac{1.2*i^{1/2} }{g} \\i=(\frac{\lambda g}{1.2} )^{2}

\frac{\delta w_{f} }{\delta g} =\frac{g^{2}\lambda ^{3}  }{3*1.2^{2} }

f_{m}=- \frac{\delta w_{f} }{\delta g} =-\frac{g^{2}\lambda ^{3}  }{3*1.2^{2} }

If i = 2A and g = 10 cm

\lambda =\frac{1.2*i^{1/2} }{g} =\frac{1.2*2^{1/2} }{10x10^{-2} } =16.97

f_{m}=-\frac{g^{2}\lambda ^{3}  }{3*1.2^{2} }=-\frac{16.97^{3}*2*0.1 }{3*1.2^{2} } =-226.2N

b) Using the coenergy of the system:

f_{m}=- \frac{\delta w_{f} }{\delta g} =-\frac{1.2*2*i^{3/2}  }{3*g^{2} }=-\frac{1.2*2*2^{3/2} }{3*0.1^{2} } =-226.2N

8 0
2 years ago
Consider a W21x93. Determine the moment capacity of the beam. Assume the compression flange is not laterally braced and that the
OLga [1]

Answer:

The answer is "828.75"

Explanation:

Please find the correct question:

For W21x93 BEAM,

Z_x = 221.00 in^3 \\\\\to \frac{b_t}{2t_f} =4.53\\\\\to \frac{h}{t_w}=32.3

For A992 STREL,

F_y= 50\  ks

Check for complete section:

\to \frac{b_t}{2t_f} =4.53 < \frac{65}{\sqrt{f_y = 9.19}}\\\\\to \frac{h}{t_w} =32.3 < \frac{640}{\sqrt{f_y = 90.5}}

Design the strength of beam =\phi_b Z_x F_y\\\\

                                                =0.9 \times 221 \times 50\\\\=9945 \ in \ \ kips\\\\=\frac{9945}{12}\\\\= 828.75 \ft \ kips \\

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3 years ago
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Answer:

import java.util.Scanner;

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Um, I do not know what to say ?

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