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Katena32 [7]
2 years ago
6

Which of these processes uses a die and a press to form parts?

Engineering
1 answer:
Irina-Kira [14]2 years ago
4 0

The process of die forming that uses a die and a press to form parts from the given processes is called; C: Hydroforming

<h3>What are some of the steps in the die forming operations?</h3>

There are different types of die forming in sheet metal operations. Now, Hydroforming is a specialized type of die forming that uses a high pressure hydraulic fluid to press room temperature working material into a die.

Tailor Welding is a process of making welded blanks made from individual sheets of steel of different thickness, strength and even coating that are joined together by the use of laser welding.

Read more about Welding operations at; brainly.com/question/9450571

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3 years ago
A garden hose fills a 2-gallon bucket in 5 seconds. The number of gallons, g (y), is proportional to the number of seconds, t (x
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Answer:

0.4 gallons per second

Explanation:

A function shows the relationship between an independent variable and a dependent variable.

The independent variable (x values) are input variables i.e. they don't depend on other variables while the dependent variable (y values) are output variables i.e. they depend on other variables.

The rate of change or slope or constant of proportionality is the ratio of the dependent variable (y value) to the independent variable (x value).

Given that the garden hose fills a 2-gallon bucket in 5 seconds. The dependent variable = g = number of gallons, the independent variable = t = number of seconds.

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State three active materials of a lead acid cell​
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Read 2 more answers
6. Find the heat flow in 24 hours through a refrigerator door 30.0" x 58.0" insulated with cellulose fiber 2.0" thick. The tempe
Ilia_Sergeevich [38]

Answer:

The heat flow in 24 hours through the refrigerator door is approximately 1,608.57 BTU

Explanation:

The given parameters are;

The duration of the heat transfer, t = 24 hours = 86,400 seconds

The area of the refrigerator door, A = 30.0" × 58.0" = 1,740 in.² = 1.122578 m²

The material of the insulator in the door = Cellulose fiber

The thickness of the insulator in the door, d = 2.0" = 0.0508 m

The temperature inside the fridge = 38° F = 276.4833 K

The temperature of the room = 78°F = 298.7056 K

The thermal conductivity of cellulose fiber = 0.040 W/(m·K)

By Fourier's law, the heat flow through a by conduction material is given by the following formula;

\dfrac{Q}{t} = \dfrac{k \cdot A \cdot (T_2 - T_1) }{d}

Q = \dfrac{k \cdot A \cdot (T_2 - T_1) }{d} \times t

Therefore, we have;

Q = \dfrac{0.04 \times 1.122578 \times (298.7056 - 276.4833 ) }{0.0508} \times 86,400 =1,697,131.73522

The heat flow in 24 hours through the refrigerator door, Q = 1,697,131.73522 J = 1,608.5705140685 BTU

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