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ololo11 [35]
2 years ago
8

During shielded metal arc welding, a welder can receive a burn comparable to sunburn if proper protective clothing is not worn.

Engineering
1 answer:
aksik [14]2 years ago
6 0

Answer:

A. True

Explanation:

The statement is true

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A 300 mm long steel bar with a square cross section (25 mm per edge) is pulled in tension with a load of 83,051 N , and experien
umka2103 [35]

Answer:

the answer is attached with required units.

Explanation:

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3 years ago
Describe what V1-V4 is
photoshop1234 [79]

Answer:

it is wave inversion

Explanation:

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3 years ago
LUNES MARTES MIÉRCOLES JUEVES VIERNES SÁBADO DOMINGO
scZoUnD [109]

Answer:

si

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3 years ago
A 14 inch diameter pipe is decreased in diameter by 2 inches through a contraction. The pressure entering the contraction is 28
Delicious77 [7]

Answer:

5984.67N

Explanation:

A 14 inch diameter pipe is decreased in diameter by 2 inches through a contraction. The pressure entering the contraction is 28 psi and a pressure drop of 2 psi occurs through the contraction if the upstream velocity is 4.0 ft/sec. What is the magnitude of the resultant force (lbs) needed to hold the pipe in place?

from continuity equation

v1A1=v2A2

equation of continuity

v1=4ft /s=1.21m/s

d1=14 inch=.35m

d2=14-2=0.304m

A1=pi*d^2/4

0.096m^2

a2=0.0706m^2

from continuity once again

1.21*0.096=v2(0.07)

v2=1.65

force on the pipe

(p1A1- p2A2) + m(v2 – v1)

from bernoulli

p1 + ρv1^2/2 = p2 + ρv2^2/2

difference in pressure or pressure drop

p1-p2=2psi

13.789N/m^2=rho(1.65^2-1.21^2)/2

rho=21.91kg/m^3

since the pipe is cylindrical

pressure is egh

13.789=21.91*9.81*h

length of the pipe is

0.064m

AH=volume of the pipe(area *h)

the mass =rho*A*H

0.064*0.07*21.91

m=0.098kg

(193053*0.096- 179263.6* 0.07) + 0.098(1.65 – 1.21)

force =5984.67N

4 0
3 years ago
A cylindrical drill with radius 4 is used to bore a hole through the center of a sphere of radius 5. Find the volume of the ring
ANTONII [103]

Answer:

The volume of the ring shaped solid that remains is 21 unit^3.

Explanation:

The total volume of the sphere is given as:

Volume of Sphere = (4/3)πr^3

where, r = radius of sphere

Volume of Sphere = (4/3)(π)(5)^3

Volume of Sphere = 523.6 unit^3

Now, we find the volume of sphere removed by the drill:

Volume removed = (Cross-sectional Area of drill)(Diameter of Sphere)

Volume removed = (πr²)(D)

where, r = radius of drill = 4

D = diameter of sphere = 2*5 = 10

Therefore,

Volume removed = (π)(4)²(10)

Volume removed = 502.6 unit^3

Therefore, the volume of ring shaped solid that remains will be the difference between the total volume of sphere, and the volume removed.

Volume of Ring = Volume of Sphere - Volume removed

Volume of Ring = 523.6 - 502.6

<u>Volume of Ring = 21 unit^3</u>

5 0
3 years ago
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