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QveST [7]
3 years ago
14

If the compound swivel base is set on 60 degrees at the lathe centerline index, how many degrees will the reading be at the cros

s slide index? A. 45 B. 150 C. 30 D. 90
Engineering
1 answer:
Effectus [21]3 years ago
6 0

Answer:

C.30°

Explanation:

Given that compound swivel base is set on 60° at the lathe center line index.

We need to find reading on cross slide index

We know that relationship between center line index and cross slide index in angle 2∝=β

Where ∝ Angle of swivel and β is the reading on cross line index.

So by using above  relationship between center line index and cross slide index  

2∝=60°⇒∝=30°

So our option is C.

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A hemispherical shell with an external diameter of 500 mm and a thickness of 20 mm is going to be made by casting, located entir
Olenka [21]

Solution :

Given :

External diameter of the hemispherical shell, D = 500 mm

Thickness, t = 20 mm

Internal diameter, d = D - 2t

                                 = 500 - 2(20)

                                 = 460 mm

So, internal radius, r = 230 mm

                                 = 0.23 m

Density of molten metal, ρ = $7.2 \ g/cm^3$

                                                  = $7200 \ kg/m^3$

The height of pouring cavity above parting surface is h = 300 mm

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So, the metallostatic thrust on the upper mold at the end of casting is :

$F=\rho g A h$

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            $=0.3324 \ m^2$

$F=\rho g A h$

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2 years ago
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Answer:

The pressure reduces to 2.588 bars.

Explanation:

According to Bernoulli's theorem for ideal flow we have

\frac{P}{\gamma _{w}}+\frac{V^{2}}{2g}+z=constant

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\frac{P_{u}}{\gamma _{w}}+\frac{V-{u}^{2}}{2g}+z_{u}=\frac{P_{L}}{\gamma _{w}}+\frac{V{L}^{2}}{2g}+z_{L}

Now by continuity equation we have

A_{u}v_{u}=A_{L}v_{L}\\\\\therefore v_{L}=\frac{A_{u}}{A_{L}}\times v_{u}\\\\v_{L}=\frac{d^{2}_{u}}{d^{2}_{L}}\times v_{u}\\\\\therefore v_{L}=\frac{2500}{900}\times 3.5\\\\\therefore v_{L}=9.72m/s

Applying the values in the Bernoulli's equation we get

\frac{P_{L}}{\gamma _{w}}=\frac{300000}{\gamma _{w}}+\frac{3.5^{2}}{2g}-\frac{9.72^{2}}{2g}(\because z_{L}=z_{u})\\\\\frac{P_{L}}{\gamma _{w}}=26.38m\\\\\therefore P_{L}=258885.8Pa\\\\\therefore P_{L}=2.588bars

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3 years ago
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Answer:

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Explanation:

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int a, b, sum;

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Answer:

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