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3 years ago

Is there anyone who can help me with welding?

Engineering

1 answer:

Artist 52 [7]3 years ago

8 0

Answer:

I do know some things on welding. Though I don't know what kind of welding you mean ill give you some simple tips.

Explanation:

When welding make sure there are no air pockets or the whole piece of material will break. Always watch your piece closely, letting your eyes off for a minute could be the end. Always weld your piece onto the handle as strongly as possible. If you don't have I good handle it will fall off. Also always wear goggles so nothing gets in your eye. Hope this helps!

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1. A glass window of width W = 1 m and height H = 2 m is 5 mm thick and has a thermal conductivity of kg = 1.4 W/m*K. If the inn

emmasim [6.3K]

Answer:

1. $\dot Q=19600\ W$

2. $\dot Q=120\ W$

Explanation:

Given:

height of the window pane, $h=2\ m$

width of the window pane, $w=1\ m$

thickness of the pane, $t=5\ mm= 0.005\ m$

thermal conductivity of the glass pane, $k_g=1.4\ W.m^{-1}.K^{-1}$

temperature of the inner surface, $T_i=15^{\circ}C$

temperature of the outer surface, $T_o=-20^{\circ}C$

<u>According to the Fourier's law the rate of heat transfer is given as:</u>

$\dot Q=k_g.A.\frac{dT}{dx}$

here:

A = area through which the heat transfer occurs = $2\times 1=2\ m^2$

dT = temperature difference across the thickness of the surface = $35^{\circ}C$

dx = t = thickness normal to the surface = $0.005\ m$

$\dot Q=1.4\times 2\times \frac{35}{0.005}$

$\dot Q=19600\ W$

air spacing between two glass panes, $dx=0.01\ m$

area of each glass pane, $A=2\times 1=2\ m^2$

thermal conductivity of air, $k_a=0.024\ W.m^{-1}.K^{-1}$

temperature difference between the surfaces, $dT=25^{\circ}C$

<u>Assuming layered transfer of heat through the air and the air between the glasses is always still:</u>

$\dot Q=k_a.A.\frac{dT}{dx}$

$\dot Q=0.024\times 2\times \frac{25}{0.01}$

$\dot Q=120\ W$

5 0

3 years ago

Given a mass-spring-damper system. The impulse response of strength 1 can be obtained from a unit step response by: ______

Alina [70]

Answer:

Multiplying impulse response by t ( option D )

Explanation:

We can obtain The impulse response of strength 1 considering a unit step response by Multiplying impulse response by t .

When we consider the Laplace Domain, and the relationship between unit step and impulse, we can deduce that the Impulse response will take the inverse Laplace transform of the function ( transfer ) . Hence Multiplying impulse response by t will be used .

5 0

3 years ago

Select the best answer for the question.

dalvyx [7]

I think the Acid level

5 0

3 years ago

The process _______ boosted the production of fertilizers and strengthened the agriculture industry.

ANTONII [103]

Answer:

Process of Green Revolution

Explanation:

Green Revolution is a process which started its initiatives between the 1950s and 1960s, a technological research process, that increase agricultural productions through various systematic approach, which are different from the traditional methods. These technological approach includes:

1. Utilization of new varieties with a high yield potential, in addition with adequate water supply, pesticides and fertilizers.

2. New methods of cultivation and mechanization

Hence, the process of Green Revolution boosted production of fertilizers and strengthen the agriculture industry as a whole.

6 0

3 years ago

Race car is accelerating and has a velocity of 10 m/s @ t=0. It completes a lap on a circular track of 400 m in 14 seconds. Calc

wariber [46]

Answer:

component of acceleration are a = 3.37 m/s² and ar = 22.74 m/s²

magnitude of acceleration is 22.98 m/s²

Explanation:

given data

velocity = 10 m/s

initial time to = 0

distance s = 400 m

time t = 14 s

to find out

components and magnitude of acceleration after the car has travelled 200 m

solution

first we find the radius of circular track that is

we know distance S = 2πR

400 = 2πR

R = 63.66 m

and tangential acceleration is

S = ut + 0.5 ×at²

here u is initial speed and t is time and S is distance

400 = 10 × 14 + 0.5 ×a (14)²

a = 3.37 m/s²

and here tangential acceleration is constant

so velocity at distance 200 m

v² - u² = 2 a S

v² = 10² + 2 ( 3.37) 200

v = 38.05 m/s

so radial acceleration at distance 200 m

ar = $\frac{v^2}{R}$

ar = $\frac{38.05^2}{63.66}$

ar = 22.74 m/s²

so magnitude of total acceleration is

A = $\sqrt{a^2 + ar^2}$

A = $\sqrt{3.37^2 + 22.74^2}$

A = 22.98 m/s²

so magnitude of acceleration is 22.98 m/s²

8 0

3 years ago