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Andre45 [30]
3 years ago
12

A hole is to be drilled in the plate at A. The diameters of the bits available to drill the hole range from 12 to 24 mm in 3-mm.

increments. Determine: (a) the diameter d of the largest bit that can be used if the allowable load at the hole is not to exceed that at the fillets.

Physics
1 answer:
blondinia [14]3 years ago
3 0

Answer:

The maximum load is when the hole size is 12 mm so the largest bit such that the load is not exceeding the P value is 12 mm.

Explanation:

As the question is not complete, the complete question is found online and is attached herewith.

From the given data

Allowable stress is given  as \sigma_{all}=145 MPa

Diameters of the bits are 12 mm, 15 mm, 18 mm, 21 mm, 24 mm

For the fillet, the dimensions are given in the attached question which are as

radius rf=9 mm

Small diameter d=75 mm

Large Diameter D=112.5 mm

The ratios of the diameters are calculated as

D/d=1.5

rf/d=0.12

So for these values and from the table, the value of K is found as 2.1

The area is given as

A=d*t

where d=75 mm, t=12 mm

A=75*12=900x10^-6 m2

Now the value of the average stress is given as

\sigma_{av}=\dfrac{\sigma_{max}}{K}\\P=\sigma_{av}A\\P=\dfrac{\sigma_{max} A}{K}\\P=\dfrac{145 \times 10^6 *900\times 10^{-6}}{2.1}\\P=62.1 kN

So for the various holes the value of r is the radius of the hole so

D=112.5 mm as given above whereas

d=D-2r

so the areas, value of k and respective values of P and σ are calculated for each radii and are attached with the solution.

As seen from the attached values, it is observed that the maximum load is when the hole size is 12 mm so the largest bit such that the load is not exceeding the P value is 12 mm.

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A 12 oz can of soda is left in a car on a hot day. In the morning the soda temperature was 60oF with a gauge pressure of 40 psi.
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In this case, volume of the can remains constant. The relationship between pressure and temperature at constant volume is given by:

P/T = Constant

Then
\frac{ P_{1} }{ T_{1} } = \frac{ P_{2} }{ T_{2} }

Where
P1 = 40 psi
P2 = ?
T1 = 60°F ≈ 289 K
T2 = 90°F ≈ 305 K (note, 363 K is not right)

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3 0
3 years ago
A rock is thrown at a window that is located 18.0 m above the ground. The rock is thrown at an angle of 40.0° above horizontal.
Korvikt [17]

Answer:

B) 27.3 m

Explanation:

The rock describes a parabolic path.

The parabolic movement results from the composition of a uniform rectilinear motion (horizontal ) and a uniformly accelerated rectilinear motion of upward or downward motion (vertical ).

The equation of uniform rectilinear motion (horizontal ) for the x axis is :

x =  vx*t   Equation (1)

Where:  

x: horizontal position in meters (m)

t : time (s)

vx: horizontal velocity  in m/s  

The equations of uniformly accelerated rectilinear motion of upward (vertical ) for the y axis  are:

(vfy)² = (v₀y)² - 2g(y- y₀)    Equation (2)

vfy = v₀y -gt    Equation (3)

Where:  

y: vertical position in meters (m)  

y₀ : initial vertical position in meters (m)  

t : time in seconds (s)

v₀y: initial  vertical velocity  in m/s  

vfy: final  vertical velocity  in m/s  

g: acceleration due to gravity in m/s²

Data

v₀ = 30 m/s , at an angle  α=40.0° above the horizontal

v₀x = vx = 30*cos40° = 22.98 m/s

v₀y = 30*sin40° = 19.28 m/s

y₀ = 2m

y =  18.0 m

g = 9.8 m/s²

Calculation of the time (t) it takes for the rock to reach at  18 m above the ground

We replace data in the equation (2)

(vfy)² = (v₀y)² - 2g(y- y₀)    

(vfy)² = (19.28)² - 2(9.8)(18- 2)

(vfy)² = 371.86 - 313.6

(vfy)² = 58.26

v_{f} = \sqrt{58.26}

vfy = 7.63 m/s

We replace vfy = 7.63 m/s in the equation (2)

vfy = v₀y - gt

7.63 = 19.28 - (9.8)(t)

(9.8)(t) = 11.65

t = 11.65 / (9.8)

t = 1.19 s

Horizontal distance from where the rock was thrown to the window

We replace t = 1.19 s , in the equation (1)

x =  vx*t  

x = (22.98)* ( 1.19 )

x = 27.3 m

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Answer:

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