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3 years ago

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A four-lane northbound section of interstate has rigid pavement and was designed with an 8-inch slab, 90% reliability, a 700 lb/

in2 concrete modulus of rupture, a 5 million lb/in2 modulus of elasticity, a 3.0 load transfer coefficient, and an overall standard deviation of 0.3. The initial PSI is 4.6 and the TSI is 2.5. The pavement was conservatively designed (assuming the upper limit of the W18 design lane load) to last 20 years, and the CBR is 25 with a drainage coefficient of 1.0. A design mistake was made that ignored 1000 total northbound (daily) passes of trucks with 22-kip single and 30-kip tandem axles. What slab thickness should have been used?

1 answer:

CaHeK987 [17]3 years ago

6 0

Answer:

Detailed solution is given below:

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Draw a radial power circuit arrangement containing 3 single fused sockets and 2 double unfused sockets showing the live, earth a

sweet [91]

Answer:

Attached below is the Radial power circuit arrangement

Explanation:

Radial power circuit arrangement is done in a way that a single cable starts from the fuse box and connects to all the outlet socket contained in the circuit also the cable contains wires ( live , neutral and earth )

The advantage of a radial power circuit arrangement is that it enables easy identification of electrical faults on the circuit.

4 0

3 years ago

With a very precise volumetric measuring device, the volume of a liquid sample is determined to be 6.321 L (liters). Three stude

zheka24 [161]

Answer:

See explanation

Explanation:

Solution:-

- Three students measure the volume of a liquid sample which is 6.321 L.

- Each student measured the liquid sample 4 times. The data is provided for each measurement taken by each student as follows:

Students

Trial A B C

1 6.35 6.31 6.38

2 6.32 6.31 6.32

3 6.33 6.32 6.36

4 6.36 6.35 6.36

- We will define the two terms stated in the question " precision " and "accuracy"

- Precision refers to how close the values are to the sample mean. The dense cluster of data is termed to be more precise. We will use the knowledge of statistics and determine the sample standard deviation for each student.

- The mean measurement taken by each student would be as follows:

$E ( A ) = \frac{6.35 +6.32+6.33+6.36}{4} \\\\E ( A ) = 6.34\\\\E ( B ) = \frac{6.31 +6.31+6.32+6.35}{4} \\\\E ( B ) = 6.3225\\\\E ( C ) = \frac{6.38 +6.32+6.36+6.36}{4} \\\\E ( C ) = 6.355\\$

- The precision can be quantize in terms of variance or standard deviation of data. Therefore, we will calculate the variance of each data:

$Var ( A ) = \frac{6.35^2+6.32^2+6.33^2+6.36^2}{4} - 6.34^2\\\\Var ( A ) = 0.00025\\\\Var ( B ) = \frac{6.31^2+6.31^2+6.32^2+6.35^2}{4} - 6.3225^2\\\\Var ( B ) = 0.00026875\\\\Var ( C ) = \frac{6.38^2+6.32^2+6.36^2+6.36^2}{4} - 6.355^2\\\\Var ( C ) = 0.000475\\$

- We will rank each student sample data in term sof precision by using the values of variance. The smallest spread or variance corresponds to highest precision. So we have:

Var ( A ) < Var ( B ) < Var ( C )

most precise Least precise

- Accuracy refers to how close the sample mean is to the actual data value. Where the actual volume of the liquid specimen was given to be 6.321 L. We will evaluate the percentage difference of sample values obtained by each student .

$P ( A ) = \frac{6.34-6.321}{6.321}*100= 0.30058\\\\P ( B ) = \frac{6.3225-6.321}{6.321}*100= 0.02373\\\\P ( C ) = \frac{6.355-6.321}{6.321}*100= 0.53788\\$

- Now we will rank the sample means values obtained by each student relative to the actual value of the volume of liquid specimen with the help of percentage difference calculated above. The least percentage difference corresponds to the highest accuracy as follows:

P ( B ) < P ( A ) < P ( C )

most accurate least accurate

7 0

3 years ago

A tensile specimen with a 12mm initial diameter and 50mm gage length reaches maximum load at 90KN and fractures at 70KN

Aleksandr-060686 [28]

Answer:

i) 796.18 N/mm^2

ii) 1111.11 N/mm^2

Explanation:

Initial diameter ( D ) = 12 mm

Gage Length = 50 mm

maximum load ( P ) = 90 KN

Fractures at = 70 KN

minimum diameter at fracture = 10mm

<u>Calculate the engineering stress at Maximum load and the True fracture stress</u>

<em>i) Engineering stress at maximum load = P/ A </em>

= P / $\pi \frac{D^2}{4}$ = 90 * 10^3 / ( 3.14 * 12^2 ) / 4

= 90,000 / 113.04 = 796.18 N/mm^2

<em>ii) True Fracture stress = P/A </em>

= 90 * 10^3 / ( 3.24 * 10^2) / 4

= 90000 / 81 = 1111.11 N/mm^2

3 0

3 years ago

A rectangular car-top carrier of 1.7-ft height, 5.0-ft length (front to back), and 4.2-ft width is attached to the top of a car.

Nataliya [291]

Answer:

$\Delta P =1.2 \frac{1.3}{2}(26.822m/s)^2 (4.2*1.7*(0.3048)^2)=13.88 hp$

Explanation:

We can assume that the general formula for the drag force is given by:

$D= C_D \frac{\rho}{2}V^2 A$

And we can see that is proportional to the area. On this case we can calculate the area with the product of the width and the height. And we can express the grad force like this:

$D_1 = C_{D1} \frac{\rho}{2}V^2 (wh)$

Where w is the width and h the height.

The last formula is without consider the area of the carrier, but if we use the area for the carrier we got:

$D_2 = C_{D2} \frac{\rho}{2}V^2 (wh+ A_{carrier})$

If we want to find the additional power added with the carrier we just need to take the difference between the multiplication of drag force by the velocity (assuming equal velocities for both cases) of the two cases, and we got:

$\Delta P = C_{D2} \frac{\rho}{2}V^2 (wh+ A_{carrier}) V- C_{D1} \frac{\rho}{2}V^2 (wh) V$

We can assume the same drag coeeficient $C_{D1}=C_{D2}=C_{D}$ and we got:

$\Delta P = C_{D} \frac{\rho}{2}V^2 (wh+ A_{carrier}) V- C_{D} \frac{\rho}{2}V^2 (wh) V$

$\Delta P = C_{D} \frac{\rho}{2}V^3 (A_{carrier})$

1.7 ft =0.518 m

60 mph = 26.822 m/s

In order to find the drag coeffcient we ned to estimate the Reynolds number first like this:

$R_E= \frac{Vl}{v}= \frac{26.822m/s*0.518 m}{1.58x10^{-4} Pa s}= 8.79 x10^{4}$

And the value for the kinematic vicosity was obtained from the table of physical properties of the air under standard conditions.

Now we can find the aspect ratio like this:

$\frac{l}{h}=\frac{5}{1.7}2.941$

And we can estimate the calue of $C_D = 1.2$ from a figure.

And we can calculate the power difference like this:

$\Delta P =1.2 \frac{1.3}{2}(26.822m/s)^2 (4.2*1.7*(0.3048)^2)=13.88 hp$

8 0

3 years ago

The accompanying specific gravity values describe various wood types used in construction. 0.320.350.360.360.370.380.400.400.40

marysya [2.9K]

Answer:

$\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ \\ {0.3} & {\vert} & {2\ 5\ 6\ 6\ 7\ 8} \ \\ \\{0.4} & {\vert} & {0\ 0\ 0\ 1\ 1\ 2\ 2\ 2\ 2\ 2\ 3\ 4\ 5\ 6\ 6\ 7\ 8\ 8\ 9} \ \\ \ \\ {0.5} & {\vert} & {1\ 4\ 4\ 5\ 8} \ \\ \ \\ {0.6} & {\vert} & {3\ 6\ 6\ 7\ 8} \ \\ \ \\ {0.7} & {\vert} & {8} \ \ \end{array}$

Explanation:

Given

$0.32,\ 0.35,\ 0.36,\ 0.36,\ 0.37,\ 0.38,\ 0.40,\ 0.40,\ 0.40,\ 0.41,$

$0.41,\ 0.42,\ 0.42,\ 0.42,\ 0.42,\ 0.42,\ 0.43,\ 0.44,\ 0.45,\ 0.46,$

$0.46,\ 0.47,\ 0.48,\ 0.48,\ 0.49,\ 0.51,\ 0.54,\ 0.54,\ 0.55,$

$0.58,\ 0.63,\ 0.66,\ 0.66,\ 0.67,\ 0.68,\ 0.78.$

Required

Plot a steam and leaf display for the given data

Start by categorizing the data by their tenth values:

$0.32,\ 0.35,\ 0.36,\ 0.36,\ 0.37,\ 0.38.$

$0.40,\ 0.40,\ 0.40,\ 0.41,\ 0.41,\ 0.42,\ 0.42,\ 0.42,\ 0.42,\ 0.42,$

$0.43,\ 0.44,\ 0.45,\ 0.46,\ 0.46,\ 0.47,\ 0.48,\ 0.48,\ 0.49.$

$0.51,\ 0.54,\ 0.54,\ 0.55,\ 0.58.$

$0.63,\ 0.66,\ 0.66,\ 0.67,\ 0.68.$

$0.78.$

The 0.3's is will be plotted as thus:

$\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.3} & {\vert} & {2\ 5\ 6\ 6\ 7\ 8} \ \ \end{array}$

The 0.4's is as follows:

$\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.4} & {\vert} & {0\ 0\ 0\ 1\ 1\ 2\ 2\ 2\ 2\ 2\ 3\ 4\ 5\ 6\ 6\ 7\ 8\ 8\ 9} \ \ \end{array}$

The 0.5's is as follows:

$\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.5} & {\vert} & {1\ 4\ 4\ 5\ 8} \ \ \end{array}$

The 0.6's is as thus:

$\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.6} & {\vert} & {3\ 6\ 6\ 7\ 8} \ \ \end{array}$

Lastly, the 0.7's is as thus:

$\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.7} & {\vert} & {8} \ \ \end{array}$

The combined steam and leaf plot is:

$\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ \\ {0.3} & {\vert} & {2\ 5\ 6\ 6\ 7\ 8} \ \\ \\{0.4} & {\vert} & {0\ 0\ 0\ 1\ 1\ 2\ 2\ 2\ 2\ 2\ 3\ 4\ 5\ 6\ 6\ 7\ 8\ 8\ 9} \ \\ \ \\ {0.5} & {\vert} & {1\ 4\ 4\ 5\ 8} \ \\ \ \\ {0.6} & {\vert} & {3\ 6\ 6\ 7\ 8} \ \\ \ \\ {0.7} & {\vert} & {8} \ \ \end{array}$

8 0

3 years ago