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irga5000 [103]
3 years ago
8

A rectangular workpiece has the following original dimensions: 2a=100mm, h=25mm, and width=20mm. The metal has a strengh coeffic

ient of 400MPa and strain-hardening exponent of 0.5. It is being forged in plane strain with mu=0.2. Calculate the force required when the height is reduced by 20 percent. Do not use average pressure formulas.
Engineering
1 answer:
Elena-2011 [213]3 years ago
4 0

Answer:

See attachment for detailed answer.

Explanation:

Download pdf
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Determine if the fluid is satisfied​
guapka [62]
Is there a picture or something?
8 0
3 years ago
______________ help protect the lower legs and feet from heat hazards like molten metal and welding sparks. A) Safety shoesB) Le
vodka [1.7K]

Answer:

It's leggings. They help protect the lower legs and feet from heat hazards like molten metal and welding sparks.

6 0
3 years ago
On July 23, 1983, Air Canada Flight 143 required 22,300 kg of jet fuel to fly from Montreal to Edmonton. The density of jet fuel
Natasha2012 [34]

Answer:

20, 083 L

Explanation:

The mistake was the result of not using units when converting the 7862 l to Kg. They used the density in pounds hence they multiplied by 1.77 Lb/L and obtained 13597 Lb not Kg as they assumed.

To obtain the amount needed to refuel they subtracted this quantity from the 22,300 Kg required for the trip again obtaining the wrong quantity of 8703 Kg and they converted this to liters by dividing the density to get 4916 L and then placed then 5000 L of fuel

The quantity required was

7862 L * 1.77 Lb/L = 13915.74 Lb (pounds not kilos)

then converting this pounds to Kg by multiplying by  0.454 Kg/L one gets

6173 Kg on board

Amount Required

( 22,300 -6173)  :  16127 Kg

16127 Kg/ 0.803 Kg/L =  20083 L

5 0
3 years ago
A pump is used to deliver water from a lake to an elevated storage tank. The pipe network consists of 1,800 ft (equivalent lengt
Nataly_w [17]

Answer:

h_f = 15 ft, so option A is correct

Explanation:

The formula for head loss is given by;

h_f = [10.44•L•Q^(1.85)]/(C^(1.85))•D^(4.8655))

Where;

h_f is head loss due to friction in ft

L is length of pipe in ft

Q is flow rate of water in gpm

C is hazen Williams constant

D is diameter of pipe in inches

We are given;

L = 1,800 ft

Q = 600 gpm

C = 120

D = 8 inches

So, plugging in these values into the equation, we have;

h_f = [10.44*1800*600^(1.85)]/(120^(1.85))*8^(4.8655))

h_f = 14.896 ft.

So, h_f is approximately 15 ft

7 0
3 years ago
B1) 20 pts. The thickness of each of the two sheets to be resistance spot welded is 3.5 mm. It is desired to form a weld nugget
kap26 [50]

Answer:

minimum current level required =  8975.95 amperes

Explanation:

Given data:

diameter = 5.5 mm

length = 5.0 mm

T = 0.3

unit melting energy = 9.5 j/mm^3

electrical resistance = 140 micro ohms

thickness of each of the two sheets = 3.5mm

Determine the minimum current level required

first we calculate the volume of the weld nugget

v = \frac{\pi }{4} * D^2 * l = \frac{\pi }{4} * 5.5^2 * 5 = 118.73 mm^3

next calculate the required melting energy

= volume of weld nugget * unit melting energy

= 118.73 * 9.5 = 1127.94 joules

next find the actual required electric energy

= required melting energy / efficiency

= 1127 .94 / ( 1/3 )  = 3383.84 J

TO DETERMINE THE CURRENT LEVEL REQUIRED  use the relation below

electrical energy =  I^2 * R * T

3383.84 / R*T = I^2

3383.84 / (( 140 * 10^-6 ) * 0.3 ) = I^2

therefore  8975.95 = I ( current )

4 0
3 years ago
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