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S_A_V [24]
3 years ago
8

Name some technical skills that are suitable for school leavers .​

Engineering
1 answer:
Natalka [10]3 years ago
7 0

Answer:

Welding, carpentry, masonry, construction worker, barber

Explanation:

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Consider the 'scope trace showing the input signal to an rc filter (the one having the larger amplitude) and the output of the f
yulyashka [42]

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6 0
2 years ago
A gas pressure difference is applied to the legs of a U-tube manometer filled with a liquid with specific gravity of 1.7. The ma
MArishka [77]

Answer:

5320.6 Pascal

Explanation:

Manometer is a pressure measuring device use to measure gas pressure .

Pressure difference in Manometer is a function of density,gravity and the height difference of the liquid.

Pressure difference = density x acceleration due to gravity x difference in height of liquid

Density of liquid = specific gravity of object x density of water.

Density of water = 997 kg/m^3

Specific gravity of liquid = 1.7

Density of liquid = 997 x 1.7 =1694.9kg/m^3

g= 9.81 m/s^2

h =320mm = 0.320m

Pressure difference = 1694.9 x 9.81 x 0.320 = 5320.6 Pascal

3 0
4 years ago
Read 2 more answers
A gas cylinder is connected to a manometer that contains water. The other end of the manometer is open to the atmosphere, which
horrorfan [7]

Answer: the absolute static pressure in the gas cylinder is 82.23596 kPa

Explanation:

Given that;

patm = 79 kPa, h = 13 in of H₂O,

A sketch of the problem is uploaded along this answer.

Now

pA = patm + 13 in of H₂O ( h × density × g )

pA= 79 + (13 × 0.0254 × 9.8 × 1000/1000)

pA = 82.23596 kPa

the absolute static pressure in the gas cylinder is 82.23596 kPa

4 0
3 years ago
A smelter emits SO2 at a rate of 10,000 kg/day. The stack is 200 m tall and the plume rise is 100 m. The windspeed is 2 m/s at a
Olegator [25]

Answer:

lol

Explanation:

lol

5 0
3 years ago
A rectangular car-top carrier of 1.7-ft height, 5.0-ft length (front to back), and 4.2-ft width is attached to the top of a car.
Nataliya [291]

Answer:

\Delta P =1.2 \frac{1.3}{2}(26.822m/s)^2 (4.2*1.7*(0.3048)^2)=13.88 hp

Explanation:

We can assume that the general formula for the drag force is given by:

D= C_D \frac{\rho}{2}V^2 A

And we can see that is proportional to the area. On this case we can calculate the area with the product of the width and the height. And we can express the grad force like this:

D_1 = C_{D1} \frac{\rho}{2}V^2 (wh)

Where w is the width and h the height.

The last formula is without consider the area of the carrier, but if we use the area for the carrier we got:

D_2 = C_{D2} \frac{\rho}{2}V^2 (wh+ A_{carrier})

If we want to find the additional power added with the carrier we just need to take the difference between the multiplication of drag force by the velocity (assuming equal velocities for both cases) of the two cases, and we got:

\Delta P = C_{D2} \frac{\rho}{2}V^2 (wh+ A_{carrier}) V-  C_{D1} \frac{\rho}{2}V^2 (wh) V

We can assume the same drag coeeficient C_{D1}=C_{D2}=C_{D} and we got:

\Delta P = C_{D} \frac{\rho}{2}V^2 (wh+ A_{carrier}) V-  C_{D} \frac{\rho}{2}V^2 (wh) V

\Delta P = C_{D} \frac{\rho}{2}V^3 (A_{carrier})

1.7 ft =0.518 m

60 mph = 26.822 m/s

In order to find the drag coeffcient we ned to estimate the Reynolds number first like this:

R_E= \frac{Vl}{v}= \frac{26.822m/s*0.518 m}{1.58x10^{-4} Pa s}= 8.79 x10^{4}

And the value for the kinematic vicosity was obtained from the table of physical properties of the air under standard conditions.

Now we can find the aspect ratio like this:

\frac{l}{h}=\frac{5}{1.7}2.941

And we can estimate the calue of C_D = 1.2 from a figure.

And we can calculate the power difference like this:

\Delta P =1.2 \frac{1.3}{2}(26.822m/s)^2 (4.2*1.7*(0.3048)^2)=13.88 hp

8 0
3 years ago
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