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12345 [234]
3 years ago
11

Only an outer panel is being replaced. Technician A says that removing the spot welds by drilling through both panels allows the

plug welds to be made through the replacement outer panel. Technician B says that placing a metal tab between panels may help prevent cutting into the flange. Who is right
Engineering
1 answer:
Angelina_Jolie [31]3 years ago
5 0

Answer:

6e66363636633747747363637737373737337374

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What is the weight density of a 2.24 in diameter titanium sphere that weights 0.82 lb?
nasty-shy [4]

Answer:

0.14\ lb/in^{3}

Explanation:

Density is defined as mass ler unit volume, expressed as

\rho=\frac {m}{v}

Where m is mass, \rho is density and v is the volume. For a sphere, volume is given as

v=\frac {4\pi r^{3}}{3}

Replacing this into the formula of density then

\rho=\frac {m}{\frac {4\pi r^{3}}{3}}

Given diameter of 2.24 in then the radius is 1.12 in. Substituting 0.82 lb for m then

\rho=\frac {0.82}{\frac {4\pi 1.12^{3}}{3}}=0.13932044952427\approx 0.14 lb/in^{3}

4 0
3 years ago
A gas metal arc welder is also known as a _____ welder.<br> A) TIGB) GTAWC) GMAWD) Resistance spot
nignag [31]

Answer:

GMAW

Explanation:

It's literally the initials of that type of welding

7 0
3 years ago
Which career best fits the group of the words below? High voltage, lines
Llana [10]

Answer:

I forget the word for it, but probably the guys who set up the power lines in the city.

Explanation:

5 0
3 years ago
Read 2 more answers
The total solids production rate in an activated sludge aeration tank is 7240 kg/d on a dry mass basis. It is necessary to maint
snow_lady [41]

Answer:

volume of biological sludge = 28.566 m³ per day

Explanation:

given data

mass of solid = 7240 kg/day

initial moisture content = 78%

solution

here percentage of solid will be

% of solid = 100 - initial moisture content

% of solid = 100 - 78 = 22 %

so that

mass of sludge produced = \frac{100}{100 - P} M kg  per day

put her value

mass of sludge produced = \frac{100}{100 - 78} 7240 kg

mass of sludge produced = 32909.09 kg

so

specific gravity of sludge =  \frac{\rho sludge}{\rho water }

and as we know that

\frac{100}{S sludge} = \frac{solid percentage}{S solid} = \frac{water percentage}{S water}

\frac{100}{S sludge} = \frac{22}{2.5} = \frac{78}{1}

S sludge = 1.152

so that

density of sludge = S sludge × density of water

density of sludge = 1.152 × 1000

density of sludge = 1152 kg/m³

so that

volume of biological sludge = \frac{mass sludge produce}{\rho sludge}

volume of biological sludge = \frac{32909.09}{1152}

volume of biological sludge = 28.566 m³ per day

6 0
3 years ago
Anyone help me please ?
Degger [83]

Answer:

I can help but I need to know what it looking for

5 0
3 years ago
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