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Lynna [10]
3 years ago
8

Given asphalt content test data:

Engineering
1 answer:
VMariaS [17]3 years ago
5 0

Answer:

hello your question is incomplete attached below is the complete question

A) overall mean = 5.535,  standard deviation ≈ 0.3239

B ) upper limit = 5.85, lower limit = 5.0

C) Not all the samples meet the contract specifications

D) fluctuation ( unstable Asphalt content )

Explanation:

B) The daily average asphalt content has to obtained in order to determine the upper and lower control limits using an average asphalt content of 5.5% +/- 0.5% everyday

The upper limit : 14 may = ( 5.8 + 5.1 ) / 2 = 5.85

The lower limit : 16 may = ( 5.2 + 4.8 ) / 2 = 5.0

attached below is the required plot

C ) Not all the samples meet the contract specifications and the samples that do not meet up are samples from :

15 may and 16 may . this is because their Asphalt contents are 6.2 and 4.8 respectively and sample number 18 and 20

D ) what can be observed is that the ASPHALT content fluctuates between the dates while the contract specification is fixed

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2 years ago
A plate clutch is used to connect a motor shaft running at 1500rpm to shaft 1. The motor is rated at 4 hp. Using a service facto
vazorg [7]

Answer:

(M_t)_{rated}=61.11lb-in

Explanation:

speed of motor (N)=1500 rpm

power=4 hp = 4 \times 0.7457 =2.9828 KW

service factor(k)= 2.75

now,

KW=\frac{2\pi n M_t}{60 \times 10^6} \\2.9828=\frac{2\pi \times 1500 M_t}{60 \times 10^6}\\M_t=\frac{2.9828\times 60 \times 10^6}{2\pi \times 1500 }

M_t= 18,989.09 \ N-mm= 168.06 lb-in

torque rating

(M_t)_{design}=k_s\times (M_t)_{rated}\\168.06= 2.75\times (M_t)_{rated}\\(M_t)_{rated}=\frac{168.06}{2.75} \\(M_t)_{rated}=61.11lb-in

4 0
3 years ago
The 10 foot wide circle quarter gate AB is articulated at A. Determine the contact force between the gate and the smooth surface
slamgirl [31]

Answer:

F = 641,771.52 \dfrac{lb-ft}{s^2}

Explanation:

Given that

R=8 ft

Width= 10 ft

We know that hydro statics force given as

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ρ is the density of fluid

A projected area on vertical plane

X is distance of center mass of projected plane from free surface of water.

Here

X=8/2  ⇒X=4 ft

A=8 x 10=80  ft^2

So now putting the values

F=ρ g A X

F=62.4(32.14)(80)(4)

F = 641,771.52 \dfrac{lb-ft}{s^2}

   

4 0
3 years ago
Calculate the electroosmotic velocity of an aqueous solution through a glass capillary 5 cm long with a 0.5 mm internal diameter
natita [175]

Answer:

Electroosmotic velocity will be equal to 1.6\times 10^{-4}m/sec

Explanation:

We have given applied voltage v = 100 volt

Length of capillary L = 5 mm = 0.005 m

Zeta potential of the capillary surface \xi =80mV=0.08volt

Dielectric constant of glass is between 5 to 10 here we are taking dielectric constant as \epsilon =10

Viscosity of glass is \eta =10^8

Electroosmotic velocity is given as v_{eo}=\frac{\epsilon \xi }{\eta }\times \frac{v}{L}

v_{eo}=\frac{10\times 0.08 }{10^8 }\times \frac{100}{0.005}=1.6\times 10^{-4}m/sec

So Electroosmotic velocity will be equal to 1.6\times 10^{-4}m/sec

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B is the answer I believe so
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