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nata0808 [166]
3 years ago
13

Compared to arc welding, which of the following statements are true about gas welding?

Engineering
1 answer:
UNO [17]3 years ago
4 0

Show than arc welding

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Nataly_w [17]
Outdoor advertising
8 0
4 years ago
This is the first half of the other question I asked
Elodia [21]

Answer: 8/3

Explanation:

6 0
3 years ago
53. The plan of a building is in the form of a rectangle with
schepotkina [342]

Answer: 150m

Explanation:

The following can be depicted from the question:

Dimensions of outer walls = 9.7m × 14.7m.

Thickness of the wall = 0.30 m

Therefore, the plinth area of the building will be:

= (9.7 + 0.30/2 + 0.30/2) × (14.7 × 0.30/2 + 0.30/2)

= 10 × 15

= 150m

7 0
3 years ago
. For each of the following situations, determine if the diode below is in foward or reversebias(a) (2 points) V1 = 0 V, V2= 2V(
Mazyrski [523]

Assuming V1 is the anode and v2 the cathode (Anode is P region and Cathode is N)

Answer:

a) Reverse bias

b) Forward bias

c) Forward bias

Explanation:

Forward bias: It happens whenever the N region of the diode is more positive than the P region. Hence, the depletion zone increase ceasing the current through the circuit -> V1 -V2 < 0  

Reverse bias: It happens whenever the P region of the diode is more positive than the N region. In this case, the depletion zone begins to shrink, if enough voltage is applied current could go through the circuit -> V1 - V2 > 0

a) V = V1 - V2 = 0 - 2 = -2 -> -2 is smaller than zero therefore, we have reverse bias

b) V = V1 - V2 = 4.5 - 2.8 = 1.7 -> 1.7 is greater than zero therefore, we have forward bias

c9 V = V1 - V2 = -1 - -1.3 = 0.3 -> 0.3 is greater than zero therefore, we have forward bias

6 0
3 years ago
A motor vehicle has a mass of 1.8 tonnes and its wheelbase is 3 m. The centre of gravity of the vehicle is situated in the centr
seropon [69]

Answer:

1) The normal reactions at the front wheel is 9909.375 N

The normal reactions at the rear wheel is 8090.625 N

2) The least coefficient of friction required between the tyres and the road is 0.625

Explanation:

1) The parameters given are as follows;

Speed, u = 90 km/h = 25 m/s

Distance, s it takes to come to rest = 50 m

Mass, m = 1.8 tonnes = 1,800 kg

From the equation of motion, we have;

v² - u² = 2·a·s

Where:

v = Final velocity = 0 m/s

a = acceleration

∴ 0² - 25² = 2 × a × 50

a = -6.25 m/s²

Force, F =  mass, m × a = 1,800 × (-6.25) = -11,250 N

The coefficient of friction, μ, is given as follows;

\mu =\dfrac{u^2}{2 \times g \times s} = \dfrac{25^2}{2 \times 10 \times 50} = 0.625

Weight transfer is given as follows;

W_{t}=\dfrac{0.625 \times 0.9}{3}\times \dfrac{6.25}{10}\times 18000 = 2109.375 \, N

Therefore, we have for the car at rest;

Taking moment about the Center of Gravity CG;

F_R × 1.3 = 1.7 × F_F

F_R + F_F = 18000

F_R + \dfrac{1.3 }{1.7} \times  F_R = 18000

F_R = 18000*17/30 = 10200 N

F_F = 18000 N - 10200 N = 7800 N

Hence with the weight transfer, we have;

The normal reactions at the rear wheel F_R  = 10200 N - 2109.375 N = 8090.625 N

The normal reactions at the front wheel F_F =  7800 N + 2109.375 N = 9909.375 N

2) The least coefficient of friction, μ, is given as follows;

\mu = \dfrac{F}{R} =  \dfrac{11250}{18000} = 0.625

The least coefficient of friction, μ = 0.625.

3 0
4 years ago
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