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2 years ago

15

Refer to Figure 9-18. A #_____ electrode lead and workpiece lead should be used to carry 150 amperes of electricity 75 feet to t

he workpiece and 75 feet back to the welding machine.

1 answer:

Anni [7]2 years ago

8 0

Answer:

AC or

Explanation:

BC

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A cylindrical bar of steel 10.1 mm (0.3976 in.) in diameter is to be deformed elastically by application of a force along the ba

Nina [5.8K]

Answer:

given d= 10.2x10-3m

change in diameter d' =3.4x10-6 m

elastic modulus=207x109pa

1/m=0.30

we know that stress/strain=E

but poissons ratio 1/m = lateral starin/longitudinal starin

0.3= (d'/d)/(L'/L)

L'/L = 3.4x10-6/(0.3x10.2x10-3)

= 1.11*10-3

E= (f/A)/(L'/L)

force f= E*A*(L'/L)

f =(1.11*10-3)*207*109*(3.1415*(10.2*10-3)2)/4

= 18775.2N =18.775KN

Explanation:

The force that produce an elastic reduction is 18.775KN

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3 years ago

Complex poles cmd zeros. Sketch the asymptotes of the Bode plot magnitude and phase for each of the listed open-loop mmsfer fuoc

Roman55 [17]

Answer: e is the best answer

Explanation:

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3 years ago

Your new mobile phone business is now approaching its first anniversary and you are able to step back and finally take a deep br

Leya [2.2K]

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.

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3 years ago

Water is stored in a tank which has vent open to the atmosphere. The water level is 1.0 m below the top the tank and the water i

sp2606 [1]

Answer:

6.99 x 10⁻³ m³ / s

Explanation:

Th e pressure difference at the two ends of the delivery pipe due to atmospheric pressure and water column will cause flow of water.

h = difference in the height of water column at two ends of delivery pipe

6 - 1 = 5 m

Velocity of flow of water

v = √2gh

= √ (2 x 9.8 x 5)

= 9.9 m /s

Volume of water flowing per unit time

velocity x cross sectional area

= 9.9 x 3.14 x .015²

= 6.99 x 10⁻³ m³ / s

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3 years ago

A rectangular steel bar, with 8" x 0.75" cross-sectional dimensions, has equal and opposite moments applied to its ends.

denpristay [2]

Answer:

Part a: The yield moment is 400 k.in.

Part b: The strain is $8.621 \times 10^{-4} in/in$

Part c: The plastic moment is 600 ksi.

Explanation:

Part a:

As per bending equation

$\frac{M}{I}=\frac{F}{y}$

Here

M is the moment which is to be calculated
I is the moment of inertia given as

$I=\frac{bd^3}{12}$

Here

b is the breath given as 0.75"
d is the depth which is given as 8"

$I=\frac{bd^3}{12}\\I=\frac{0.75\times 8^3}{12}\\I=32 in^4$

y is given as

$y=\frac{d}{2}\\y=\frac{8}{2}\\y=4"\\$

Force is 50 ksi

$\frac{M_y}{I}=\frac{F_y}{y}\\M_y=\frac{F_y}{y}{I}\\M_y=\frac{50}{4}{32}\\M_y=400 k. in$

The yield moment is 400 k.in.

Part b:

The strain is given as

$Strain=\frac{Stress}{Elastic Modulus}$

The stress at the station 2" down from the top is estimated by ratio of triangles as

$F_{2"}=\frac{F_y}{y}\times 2"\\F_{2"}=\frac{50 ksi}{4"}\times 2"\\F_{2"}=25 ksi$

Now the steel has the elastic modulus of E=29000 ksi

$Strain=\frac{Stress}{Elastic Modulus}\\Strain=\frac{F_{2"}}{E}\\Strain=\frac{25}{29000}\\Strain=8.621 \times 10^{-4} in/in$

So the strain is $8.621 \times 10^{-4} in/in$

Part c:

For a rectangular shape the shape factor is given as 1.5.

Now the plastic moment is given as

$shape\, factor=\frac{Plastic\, Moment}{Yield\, Moment}\\{Plastic\, Moment}=shape\, factor\times {Yield\, Moment}\\{Plastic\, Moment}=1.5\times400 ksi\\{Plastic\, Moment}=600 ksi$

The plastic moment is 600 ksi.

3 0

3 years ago