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Anni [7]
3 years ago
9

A surface grinding operation is used to finish a flat plate that is 5.50 in wide and 12.500 in long. The starting thickness is 1

.085 in. After grinding the surface, the thickness is 1.000 in. The grinding wheel had a starting diameter of 6.013 in and a width of 0.50 in. After the operation, the diameter of the grinding wheel is 5.997 in. Determine the grinding ratio in this operation.
Engineering
1 answer:
valina [46]3 years ago
5 0

Answer:

77.40

Explanation:

Initial width = 5.5 Inches

Initial length = 12.5 inches

Initial thickness = 1.085 inches

After grinding

Thickness of flat plate  = 1 inch

Grinding wheel

starting diameter( di )= 6.013 inches

width of grinding wheel = 0.5 inch

After operation

Diameter of grinding wheel ( df ) = 5.997

<u>Calculate the grinding ratio in this operation</u>

First step : determine the volume of material removed from flat plate

= Length of flat plate * width of flat plate * change in thickness

= 12.5 * 5.5 * ( 1.085 - 1 )

= 5.8437 in^3

Volume of material from grinding wheel

= π / 4 * ( di^2 - df^2 ) * width  

= π / 4  * ( 6.013^2 - 5.997^2 ) * 0.5

= 0.0755 in^3

<u>Finally the Grinding ratio</u>

= 5.8437  / 0.0755

= 77.4

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kogti [31]

Answer

given,

Speed of vehicle = 65 mi/hr

                            = 65 x 1.4667 = 95.33 ft/s

e = 0.07 ft/ft

f is the lateral friction, f = 0.11

central angle,Δ = 38°

The PI station is

PI = 250 + 50

   = 25050 ft

using super elevation formula

e + f = \dfrac{v^2}{rg}

0.07 + 0.11 =\dfrac{95.33^2}{r\times 32.2}

r = \dfrac{95.33^2}{32.2\times 0.18}

  r = 1568 ft

As the road is two lane with width 12 ft

R = 1568 + 12/2

R = 1574 ft

Length of the curve

L = \dfrac{\piR\Delta}{180}

L = \dfrac{\pi\times 1574\times 38}{180}

L = 1044 ft

Tangent of the curve calculation

  T = R tan(\dfrac{\Delta}{2})

  T = 1574 tan(\dfrac{38}{2})

      T = 542 ft

The station PC and PT are

 PC = PI - T

 PC = 25050 - 542

       = 24508 ft

       = 245 + 8 ft

PT = PC + L

     = 24508 + 1044

     =25552

     = 255 + 52 ft

the middle ordinate calculation

MO = R(1-cos\dfrac{\Delta}{2})

MO = 1574\times (1-cos\dfrac{38}{2})

     MO = 85.75 ft

degree of the curvature

D = \dfrac{5729.578}{R}

D = \dfrac{5729.578}{1574}

D = 3.64°

8 0
4 years ago
A spherical gas container made of steel has a(n) 17-ft outer diameter and a wall thickness of 0.375 in. Knowing that the interna
Arte-miy333 [17]

Answer:

Maximum Normal Stress σ = 8.16 Ksi

Maximum Shearing Stress τ = 4.08 Ksi

Explanation:

Outer diameter of spherical container D = 17 ft

Convert feet to inches D = 17 x 12 in = 204 inches

Wall thickness t = 0.375 in

Internal Pressure P = 60 Psi

Maximum Normal Stress σ = PD / 4t

σ = PD / 4t

σ = (60 psi x 204 in) / (4 x 0.375 in)

σ = 12,240 / 1.5

σ = 8,160 P/in

σ = 8.16 Ksi

Maximum Shearing Stress τ = PD / 8t

τ = PD / 8t

τ = (60 psi x 204 in) / (8 x 0.375 in)

τ = 12,240 / 3

τ = 4,080 P/in

τ = 4.08 Ksi

7 0
3 years ago
Consider a 20-cm X 20-cm X 20-cm cubical body at 477°C suspended in the air. Assuming the body closely approximates a blackbody,
Olin [163]

Answer:

a) The rate at which the cube emits radiation energy is 704.48 W

b) The spectral blackbody emissive power is 194.27 W/m²μm

Explanation:

Given data:

a = side of the cube = 0.2 m

T = temperature = 477°C

Wavelength = 4 µm

a) The surface area is:

A_{s} =6a^{2} =6(0.2)^{2} =0.24m^{2}

According Stefan-Boltzman law, the rate of emission is:

E=\sigma T^{4} A_{s} =5.67x10^{-8} *(477)^{4} *0.24=704.48W

b) Using Plank´s distribution law to get the spectral blackbody emissive power.

E=\frac{C_{1} }{\lambda ^{5}(exp(\frac{C_{2} }{\lambda T}) -1 )} =\frac{3.743x10^{8} }{4^{5}(exp(\frac{1.4387x10x^{4} }{4*477})-1)  } =194.27W/m^{2} \mu m

8 0
4 years ago
Consider the incompressible flow of water through a divergent duct. The inlet velocity and area are 5 ft/s and 10 ft^2, respecti
kenny6666 [7]

Answer:

v₂ = 1.25 m/s

Explanation:

given,

inlet velocity, v₁ = 5 ft/s

inlet Area = A = 10 ft²

outlet velocity, v₂ = ?

outlet Area = 4 A

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Using continuity equation

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     10 v₁ = 40 v₂

  v_2 = \dfrac{v_1}{4}

  v_2 = \dfrac{5}{4}

       v₂ = 1.25 m/s

Hence, the exit velocity of the water through the duct is equal to 1.25 m/s.

6 0
3 years ago
Sun of first 1 nayural numbers​
marin [14]

Answer:

the answer is

n(n + 1)  \div 2

n(n+1)/2

4 0
3 years ago
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