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2 years ago

6.3. A __________ is used to indicate the base material that needs to be beveled.

Engineering

1 answer:

Ivan2 years ago

4 0

Answer:

A <u>break in the arrow</u> is used to indicate the base material that needs to be beveled

A. Break in the arrow

Explanation:

In a weld that involves the welding of a joint that has a, a bevel, flare bevel, or J-groove, the side that requires beveling (chamfering) preparation in the welded joint before welding is indicated by an arrow having a definitive before pointing to the beveled member

However, the break in the arrow could excluded when the side to be chamfered is unmistakable

Therefore, the break in the arrow is the option that is used to indicate which of the material that is to be beveled.

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If a heat engine has an efficiency of 30% and its power output is 600 W, what is the rate of heat input from the combustion phas

jarptica [38.1K]

Answer:

The heat input from the combustion phase is 2000 watts.

Explanation:

The energy efficiency of the heat engine ( $\eta$ ), no unit, is defined by this formula:

$\eta = \frac{\dot W}{\dot Q}$ (1)

Where:

$\dot Q$ - Heat input, in watts.

$\dot W$ - Power output, in watts.

If we know that $\dot W = 600\,W$ and $\eta = 0.3$ , then the heat input from the combustion phase is:

$\eta = \frac{\dot W}{\dot Q}$

$\dot Q = \frac{\dot W}{\eta}$

$\dot Q = \frac{600\,W}{0.3}$

$\dot Q = 2000\,W$

The heat input from the combustion phase is 2000 watts.

8 0

2 years ago

If the car passes point A with a speed of 20 m>s and begins to increase its speed at a constant rate of at = 0.5 m>s 2 , d

sattari [20]

Answer:

a = 1.68m/ $S^{2}$

Explanation:

Please kindly find the attached file for explanations

3 0

3 years ago

A closed system consists of 0.3 kmol of octane occupying a volume of 5 m³. Determine (a) the weight of the system, in N, and (b)

Leni [432]

Answer:

a) m=336.18N

b) Vn=16.67m/kmol

Vm=0.1459m^3/kg

Explanation:

To calculate the mass of the octane(m):

Number of mole of octane (n) =0.3kmol(given)

Molarmass of octane (M) =114.23kg/kmol

m=n*M

m=(0.3kmol)*(114.23kg/kmol)

m=34.269kg

To calculate for the weight of octane(W):

W=g*m

W=(9.81m/s^2)*(34.269kg)

W=336.18N

b) For specific volumes of Vn and Vm:

Given volume of octane (V) =5m^3

Vm=V/m

Vm=5m^3/34.269kg

Vm=0.1459m^3/kg

And Vn will be :

Vn=V/m=5m^3/0.3kmol

Vn=16.67m/Kmol

Therefore, the answers are:

a) m=336.18N

b) Vn=16.67m/kmol

Vm=0.1459m^3/kg

7 0

3 years ago

A thick steel slab (rho= 7800 kg/m3 , cp= 480 J/kg K, k= 50 W/m K) is initially at 300 °C and is cooled by water jets impinging

dimaraw [331]

Answer:

t = 2244.3 sec

Explanation:

calculate the thermal diffusivity

$\alpha = \frac{k}{\rho c}$

$= \frac{50}{7800\times 480} = 1.34 \times 10^{-5} m^2/s$

Temperature at 28 mm distance after t time = = 50 degree C

we know that

$\frac[ T_{28} - T_s}{T_i -T_s} = erf(\frac{x}{2\sqrt{at}})$

$\frac{ 50 -25}{300-25} = erf [\frac{28\times 10^{-3}}{2\sqrt{1.34\times 10^{-5}\times t}}]$

$0.909 = erf{\frac{3.8245}{\sqrt{t}}}$

from gaussian error function table , similarity variable w calculated as

erf w = 0.909

it is lie between erf w = 0.9008 and erf w = 0.11246 so by interpolation we have

w = 0.08073

$erf 0.08073 = erf[\frac{3.8245}{\sqrt{t}}]$

$0.08073 = \frac{3.8245}{\sqrt{t}}$

solving fot t we get

t = 2244.3 sec

3 0

2 years ago

What might cause birds in population to immigration To the island

loris [4]

Mass forest destruction

temperatures

7 0

3 years ago

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