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seraphim [82]
3 years ago
6

Chaplets are used to support a sand core inside a sand mold cavity. The design of the

Engineering
1 answer:
ziro4ka [17]3 years ago
4 0

Answer:

a) 2 chaplets

b) 9 chaplets

Explanation:

Before we can determine the minimum number of chaplets that should be placed beneath and above the core, we must know the mass of the sand used to make the surface of the mold cavity as well as the mass of the steel metal poured inside the mold.

Density is defined as the ratio of mass of a substance to its density.

Density = Mass/Volume

For the STEEL:

Density of steel = 7.82g/cm³

Volume = volume of the core = 5450cm³

Mass = Density × Volume

Mass of steel = 7.82×5450

Mass of steel = 42619g

Mass of steel in kg = 42.619kg

For the SAND:

Density of sand = 1.6g/cm³

Volume = volume of the core = 5450cm³

Mass = Density × Volume

Mass of sand = 1.6×5450

Mass of sand = 8720g

Mass of sand in kg = 8.72kg

a) Since the the chaplet support the sand from beneath the core, and each chaplet weighs 45N, we need to know the amount of force possessed by the sand.

Since the mass of the sand is 87.2kg

Weight = mass × acceleration due to gravity

Weight = 8.72×9.81

Weight of sand used to mold the core = 85.54N

Since 1 chaplet weighs 45N

This means that (85.54N/45N) i.e 1.9 which is approximately 2 chaplets must be placed beneath the core to sustain it before the steel metal is poured.

b) Since the metal poured in the core is steel, this means that the chaplet placed above the core must be able to withstand the strength of the steel.

Weight of steel = mass of steel × acceleration due to gravity

Weight of steel = 42.619×9.81

Weight of steel = 418.09N

Since one chaplet weigh 45N, the amount of chaplets that must be placed above the core is;

418.09/45

= 9.3

Therefore the minimum number of chaplets that should be placed above the core after pouring steel metal is 9chaplets.

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Which word from the passage best explains what the web in the passage symbolizes
frez [133]

Answer:

I cant see the passage

Explanation:

I cant see the passage

6 0
3 years ago
A ball bearing has been selected with the bore size specified in the catolog as 35.000 mm to 35.020 mm. Specify appropriate mini
Fofino [41]

Answer:

the minimum shaft diameter is 35.026 mm

the maximum shaft diameter is 35.042mm

Explanation:

Given data;

D-maximum = 35.020mm and d-minimum = 35.000mm

we have to go through Tables "Descriptions of preferred Fits using the Basic Hole System" so from the table, locational interference fits H7/p6

so From table, Selection of International Trade Grades metric series

the grade tolerance are;

ΔD = IT7(0.025 mm)

Δd = IT6(0.016 mm)

Also from Table "Fundamental Deviations for Shafts" metric series

Sf = 0.026

so  

D-maximum

Dmax = d + Sf + Δd

we substitute

Dmax = 35 + 0.026 + 0.016

Dmax = 35.042 mm

therefore the maximum diameter of shaft is 35.042mm

d-minimum

Dmin = d + Sf

Dmin = 35 + 0.026

Dmin = 35.026 mm

therefore the minimum diameter of shaft is 35.026 mm

8 0
3 years ago
Technician A says that a voltage drop of 0.8 volts on the starter ground circuit is within specifications. Technician B says tha
Romashka-Z-Leto [24]

Answer:

Technician A is wrong

Technician B is right

Explanation:

voltage drop of 0.8 volts on the starter ground circuit is not within specifications. Voltage drop should be within the range of 0.2 V to 0.6 V but not more than that.

A spun bearing can seize itself around the crankshaft journal causing it not to move. As the car ignition system is turned on, the stater may draw high current in order to counter this seizure.

8 0
4 years ago
A single lane highway has a horizontal curve. The curve has a super elevation of 4% and a design speed of 45 mph. The PC station
andreyandreev [35.5K]

Answer: 112 + 19.27

Explanation:

Super elevation is an inward transverse slope provided through out the length of the horizontal curve which ends up serving as a counteract to the centrifugal force and checks tendency of overturning. It changes from infinite radius to radius of a transition curve.

Super curve elevation (e) = 4%

4/100= 0.04

e= V^2/gR

Make R the subject of the formula.

egR= V^2

R= V^2/eg

V= 45mph

=45 × 0.44704m/s

=20.1168m/s

g (force due to gravity) =9.81

Therefore,

R= (20.1168)^2/9.81 × 0.04

= 1031.31m

Tangent Length( T) = PI - PC

Tangent Length= 10875 - 10500

=375m

T= R Tan(I/2)

375= 1031.31 × Tan(I/2)

I= 39.96

Also,

L= πRI/180

= 719.27m

Station PT= Stat PC+ L

10500 + 719.27

=11219.27

=112 + 19.27

6 0
3 years ago
Help please and please
nataly862011 [7]
So people can tell what your think
8 0
3 years ago
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