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2 years ago

9

A steel plate has a hole drilled through it. The plate is put into a furnace and heated. What happens to the size of the inside

diameter of a hole as its temperature increases

1 answer:

djverab [1.8K]2 years ago

4 0

Answer:

The diameter increases

Explanation:

The expansion in the metal is uniform in every dimension

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a sprue is 12 in long and has a diameter of 5 in at the top. The molten metal level in the pouring basing is taken to be 3 in fr

vampirchik [111]

Answer:

See explaination

Explanation:

We can describe Aspiration Effect as a phenomenon of providing an allowance for the release of air from the mold cavity during the metal pouring.

See the attached file for detailed solution of the given problem.

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2 years ago

Read 2 more answers

A PMMA plate with a 25 mm (width) x 6.5 mm (thickness) cross-section has a contained crack of length 2c = 0.5 mm in the center o

victus00 [196]

Answer:

LAOD = 6669.86 N

Explanation:

Given data:

width $= 25 mm = 25\times 10^{-3} m$

thickness $= 6.5 mm = 6.5\times 10^{-3} m$

crack length 2c = 0.5 mm at centre of specimen

$\sigma _{applied} = 1000 N/cross sectional area$

stress intensity factor = k will be

$\sigma_{applied} = \frac{1000}{25\times 10^{-3}\times 6.5\times 10^{-3}}$

$= 6.154\times 10^{6} Pa$

we know that

$k =\sigma_{applied} (\sqrt{\pi C})$

$=6.154\sqrt{\pi (2.5\times 10^{-04})}$ [c =0.5/2 = 2.5*10^{-4}]

K = 0.1724 Mpa m^{1/2} for 1000 load

if $K_C = 1.15 Mpa m^{1/2}$ then load will be

$Kc = \sigma _{frac}(\sqrt{\pi C})$

$1.15 MPa = \sigma _{frac}\times \sqrt{\pi (2.5\times 10^{-04})}$

$\sigma _{frac} = 41.04 MPa$

$load = \sigma _{frac}\times Area$

$load = 41.04 \times 10^6 \times 25\times 10^{-3}\times 6.5\times 10^{-3} N$

LAOD = 6669.86 N

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2 years ago

Find At Least Three Instances Of Set Constructions In Your CS Courses And Use Them To Exhibit All Set Operations Discussed In Cl

hoa [83]

An expertly designed format for arranging, processing, accessing, and storing data is called a data structure.

Data structures come in both simple and complex forms, all of which are made to organize data for a certain use. Users find it simple to access the data they need and use it appropriately thanks to data structures. The organizing of information is framed by data structures in a way that both machines and people can better grasp. A data structure may be chosen or created in computer science and computer programming to store data in order to be used with different methods. In some circumstances, the design of the data structure and the algorithm's fundamental operations are closely related. Each data structure comprises information about the data values, relationships between the data and — in some situations — functions that can be applied to the data. For instance, in an object-oriented programming language, the data structure and its related methods are tied together as part of a class description. Although they may be designed to operate with the data structure in non-object oriented languages, these functions are not considered to be a part of the data structure. A data structure may be chosen or created in computer science and computer programming to store data in order to be used with different methods. In some circumstances, the design of the data structure and the algorithm's fundamental operations are closely related. Each data structure comprises information about the data values, relationships between the data and — in some situations — functions that can be applied to the data.

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1 year ago

When would working with machinery be a common type of caught-in and caught-between<br> hazard?

tigry1 [53]

Answer:

A working with machinery be a common type of caught-in and caught-between hazard is described below in complete detail.

Explanation:

“Caught in-between” accidents kill mechanics in a variety of techniques. These incorporate cave-ins and other hazards of tunneling activity; body parts extracted into unconscious machinery; reaching within the swing range of cranes and other installation material; caught between machine & fixed objects.

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2 years ago

A surveyor knows an elevation at Catch Basin to be elev=2156.77 ft. The surveyor takes a BS=2.67 ft on a rod at BM Catch Basin a

fenix001 [56]

Answer:

the elevation at point X is 2152.72 ft

Explanation:

given data

elev = 2156.77 ft

BS = 2.67 ft

FS = 6.72 ft

solution

first we get here height of instrument that is

H.I = elev + BS ..............1

put here value

H.I = 2156.77 ft + 2.67 ft

H.I = 2159.44 ft

and

Elevation at point (x) will be

point (x) = H.I - FS .............2

point (x) = 2159.44 ft - 6.72 ft

point (x) = 2152.72 ft

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3 years ago