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3 years ago

Cold forging makes metal more workable than hot forging. True False

Engineering

2 answers:

Whitepunk [10]3 years ago

7 0

Answer:

I don't really know but i have some info for you...

Explanation:

The cold forging manufacturing process increases the strength of a metal through strain hardening at a room temperature. On the contrary the hot forging manufacturing process keeps materials from strain hardening at high temperature, which results in optimum yield strength, low hardness and high ductility.

Pani-rosa [81]3 years ago

4 0

Answer:

false

Explanation:

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An insulated rigid tank is divided into two compartments of different volumes. Initially, each compartment contains the same ide

storchak [24]

Answer:

Explanation:

Given that

Mass of 1 = $m_1$

Mass of 2 = $m_2$

Temperature in 1 = $T_1$

Temperature in 2 = $T_2$

Pressure remains i the group apartment

The closed system and energy balance is

$E_{in}-E_{out}=\Delta E_{system}$

The kinetic energy and potential energy are negligible

since it is insulated tank ,there wont be eat transfer from the system

And there is no work involved

$\Delta U = 0$

Let the final temperature be final temperature

$m_1+c_v(T_3-T_1)+m_2c_v(T_3+T_2)=0\\\\m_1+c_v(T_3-T_1)=m_2c_v(T_2-T_3)---(i)$

Using mass balance

$m_3+m_2+m_1$

from eqn i

$m_1+c_v(T_3-T_1)=m_2c_v(T_2-T_3)m_1T_3-m_1T_1=m_2T_2-m_2T_3\\\\m_1T_3+m_2T_3=m_2T_2+m_1T_1\\\\(m_1+m_2)T_3=m_1T_1+m_2T_2\\\\(m_1)T_3=m_1T_1+m_2T_2\\\\T_3=\frac{m_1T_1_m_2T_2}{m_3}$

Therefore the final temperature can be express as

$\large \boxed {T_3=\frac{m_1}{m_3} T_1+\frac{m_2}{m_3}T_2 }$

8 0

4 years ago

A small submarine has a triangular stabilizing fin on its stern. The fin is 1 ft tall and 2 ft long. The water temperature where

Arturiano [62]

Answer:

$\mathbf{F_D \approx 1.071 \ lbf}$

Explanation:

Given that:

The height of a triangular stabilizing fin on its stern is 1 ft tall

and it length is 2 ft long.

Temperature = 60 °F

The objective is to determine the drag on the fin when the submarine is traveling at a speed of 2.5 ft/s.

From these information given; we can have a diagrammatic representation describing how the triangular stabilizing fin looks like as we resolve them into horizontal and vertical component.

The diagram can be found in the attached file below.

If we recall ,we know that;

Kinematic viscosity v = $1.2075 \times 10^{-5} \ ft^2/s$

the density of water ρ = 62.36 lb /ft³

$Re_{max} = \dfrac{Ux}{v}$

$Re_{max} = \dfrac{2.5 \ ft/s \times 2 \ ft }{1.2075 \times 10 ^{-5} \ ft^2/s}$

$Re_{max} = 414078.6749$

$Re_{max} = 4.14 \times 10^5$ which is less than < 5.0 × 10⁵

Now; For laminar flow; the drag on the fin when the submarine is traveling at 2.5 ft/s can be determined by using the expression:

$dF_D = (\dfrac{0.664 \times \rho \times U^2 (2-x) dy}{\sqrt{Re_x}})^2$

where;

$(2-x) dy$ = strip area

$Re_x = \dfrac{2.5(2-x)}{1.2075 \times 10 ^{-5}}$

Therefore;

$dF_D = (\dfrac{0.664 \times 62.36 \times 2.5^2 (2-x) dy}{\sqrt{ \dfrac{2.5(2-x)}{1.2075 \times 10 ^{-5}}}})$

$dF_D = 1.136 \times(2-x)^{1/2} \ dy$

Let note that y = 0.5x from what we have in the diagram,

so , x = y/0.5

By applying the rule of integration on both sides, we have:

$\int\limits \ dF_D = \int\limits^1_0 \ 1.136 \times(2-\dfrac{y}{0.5})^{1/2} \ dy$

$\int\limits \ dF_D = \int\limits^1_0 \ 1.136 \times(2-2y)^{1/2} \ dy$

Let U = (2-2y)

-2dy = du

dy = -du/2

$F_D = \int\limits^0_2 \ 1.136 \times(U)^{1/2} \ \dfrac{du}{-2}$

$F_D = - \dfrac{1.136}{2} \int\limits^0_2 \ U^{1/2} \ du$

$F_D = -0.568 [ \dfrac{\frac{1}{2}U^{ \frac{1}{2}+1 } }{\frac{1}{2}+1}]^0__2$

$F_D = -0.568 [ \dfrac{2}{3}U^{\frac{3}{2} } ] ^0__2$

$F_D = -0.568 [0 - \dfrac{2}{3}(2)^{\frac{3}{2} } ]$

$F_D = -0.568 [- \dfrac{2}{3} (2.828427125)} ]$

$F_D = 1.071031071 \ lbf$

$\mathbf{F_D \approx 1.071 \ lbf}$

8 0

3 years ago

Voltage-regulated channels can be found a. at the motor end plate. b. on the surfaces of dendrites. c. in the membrane that cove

USPshnik [31]

Answer:

Option C

In the membrane that covers axons

Explanation:

Voltage-regulated channels allow for selective passage of different beneficial ions such as potassium and are found on the surface of a wide variety of cells such as nerve, muscle, and secretory cells. They mainly regulate cell membrane excitability, repetitive low frequency firing in some neurons, and recover the nerve fiber membrane.

7 0

3 years ago

A vapor-compression refrigeration cycle operates at steady state with Refrigerant 134a as the working fluid. Saturated vapor ent

mart [117]

Answer:

(3) the compressor power, in kW. = 4.17KW

(4) the refrigeration capacity, in tons.= 4.905tons

(5) the coefficient of performance. = 4.1336

check attached files for other answers.

Explanation:

4 0

3 years ago

A. 50

Mademuasel [1]

Answer:ahahahahahah

Explanation:boyboy

6 0

3 years ago