Ratio equivalent to 12 red beans to 5 total beans

A ramp from an expressway with a design speed of 30 mi/h connects with a local road, forming a T intersection. An additional lan

hram777 [196]

Answer:

the width of the turning roadway = 15 ft

Explanation:

Given that:

A ramp from an expressway with a design speed(u) = 30 mi/h connects with a local road

Using 0.08 for superelevation(e)

The minimum radius of the curve on the road can be determined by using the expression:

$R = \dfrac{u^2}{15(e+f_s)}$

where;

R= radius

$f_s$ = coefficient of friction

From the tables of coefficient of friction for a design speed at 30 mi/h ;

$f_s$ = 0.20

So;

$R = \dfrac{30^2}{15(0.08+0.20)}$

$R = \dfrac{900}{15(0.28)}$

$R = \dfrac{900}{4.2}$

R = 214.29 ft

R ≅ 215 ft

However; given that :

The turning roadway has stabilized shoulders on both sides and will provide for a onelane, one-way operation with no provision for passing a stalled vehicle.

From the tables of "Design widths of pavement for turning roads"

For a One-way operation with no provision for passing a stalled vehicle; this criteria falls under Case 1 operation

Similarly; we are told that the design vehicle is a single-unit truck; so therefore , it falls under traffic condition B.

As such in Case 1 operation that falls under traffic condition B in accordance with the Design widths of pavement for turning roads;

If the radius = 215 ft; the value for the width of the turning roadway for this conditions = 15ft

Hence; the width of the turning roadway = 15 ft

5 0

3 years ago

Write a program that adds the numbers 1 through 5 into the numbers ArrayList and then prints out the first element in the list.

nekit [7.7K]

Answer:

Answer code is given below along with comments to explain each step

Explanation:

Method 1:

// size of the array numbers to store 1 to 5 integers

int Size = 5;

// here we have initialized our array numbers, the type of array is integers and size is 5, right now it is empty.

ArrayList<Integer> numbers = new ArrayList<Integer>(Size);

// lets store values into the array numbers one by one

numbers.add(1)

numbers.add(2)

numbers.add(3)

numbers.add(4)

numbers.add(5)

// here we are finding the first element in the array numbers by get() function, the first element is at the 0th position

int firstElement = numbers.get(0);

// to print the first element in the array numbers

System.out.println(firstElement);

Method 2:

int Size = 5;

ArrayList<Integer> numbers = new ArrayList<Integer>(Size);

// here we are using a for loop instead of adding manually one by one to store the values from 1 to 5

for (int i = 0; i < Size; i++)

{

// adding the values 1 to 5 into array numbers

numbers.add(i);

}

//here we are finding the first element in the array numbers

int firstElement = numbers.get(0);

// to print the first element in the array

System.out.println(firstElement);

6 0

3 years ago

In a production facility, 3 cm thick large brass plates (k = 110 W/mC, α = 33.9 × 10-6 m2 /s) that are initially at a uniform

zysi [14]

Answer:

Explanation:

Given.

Thickness of brass plate $t = 3 cm$

Thermal conductivity of brass $k = 110 W/m.°C$

Density of brass $\rho = 8530 kg/m^3$

Specific heat of brass $C_p =380J/kg.°C$

Thermal diffusivity of brass $\alpha = 33.9\times 10^{-6} m^2/s$

Temperature of oven $T_{\infty} = 700°C$

The initial temperature $T_i= 25°C$

Plate remain in the oven $t =10 min$

Heat conduction in the plate is one-dimensional since the plate is large relative to its thickness and there is thermal symmetry about the center plane.

The thermal properties of the plate are constant.

The heat transfer coefficient is constant and uniform over the entire surface.

The Fourier number is > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified).

The Biot number for this process $Bi = \frac{hL}{k}\\\\Bi=\frac{(80 W/m^2.°C)(0.015 m)}{(110 W/m.°C)}\\=Bi =0.0109$

The constants $\lambda_1$ and $A_1$ corresponding to this Biot are, from 11-2 tables.

The interpolation method used to find the

$\lambda_1=0.1039$ and $A_1=1.0018
$

The Fourier number $\tau=\frac{\alpha t}{L^2}\\\\\tau=\frac{(33.9\times 10^{-6} m^2/s)(10 min \times 60 s/min)}{(0.015m)^2}
\\\\\tau=90.4>0.2$

Therefore, the one-term approximate solution (or the transient temperature charts) is applicable.

Then the temperature at the surface of the plates becomes

$\theta(L,t)_{wall}=\frac{T(x,t)-T_{\infty}}{(T_i-T_{\infty})}\\\\\theta(L,t)_{wall}=A_1e^{-\lambda_1^2\tau}\cos(\lambda_1L/L)\\\\\theta(L,t)_{wall}=(1.0018)e^{-(0.1039^2(90.4))}\cos(0.1039)\\\\\theta(L,t)_{wall}=0.378\\\\\frac{T(L,t)-700}{25-700}=0.378\\\\T(L,t)=445°C$

3 0

3 years ago

Read 2 more answers

A hot-water stream at 60oC enters a mixing chamber with a mass flow rate of 0.5 kg/s where it is mixed with a stream of cold wat

Nezavi [6.7K]

Answer: 0.306 kg/s

Explanation:

If we assume that the specific heat for all stream is constant, we can then say, the mass flow rate of cold water can be calculated by using the energy balance equation.

The energy balance equation basically states that, the heat gained is equal to the heat lost

m'1h1 + m'2h2 = m'3h3

m'1T1 + m'2T2 = (m'1 + m'2) T3

0.5 (60) + m'2(10) = (0.5 + m'2)41

30 + 10m'2 = 20.5 + 41m'2

30 - 20.5 = 41m'2 - 10m'2

9.5 = 31m'2

m'2 = 9.5 / 31

m'2 = 0.306 kg/s

7 0

3 years ago

Read 2 more answers

A pressurized 2-m-diameter tank of water has a 10-cm-diameter orifice at the bottom where water discharges to the atmosphere. Th

alukav5142 [94]

Answer:

A fluid is defined as a material that deforms continuously and permanently under the

application of a shearing stress.

• The pressure at a point in a fluid is independent of the orientation of the surface

passing through the point; the pressure is isotropic.

• The force due to a pressure p acting on one side of a small element of surface dA

defined by a unit normal vector n is given by −pndA.

• Pressure is transmitted through a fluid at the speed of sound.

• The units we use depend on whatever system we have chosen, and they include quantities

like feet, seconds, newtons and pascals. In contrast, a dimension is a more

abstract notion, and it is the term used to describe concepts such as mass, length and

time.

• The specific gravity (SG) of a solid or liquid is the ratio of its density to that of water

at the same temperature.

• A Newtonian fluid is one where the viscous stress is proportional to the rate of strain

(velocity gradient). The constant of proportionality is the viscosity, µ, which is a

property of the fluid, and depends on temperature.

• At the boundary between a solid and a fluid, the fluid and solid velocities are equal;

this is called the “no-slip condition.” As a consequence, for large Reynolds numbers

(>> 1), boundary layers form close to the solid boundary. In the boundary layer,

large velocity gradients are found, and so viscous effects are important.

• At the interface between two fluids, surface tension may become important. Surface

tension leads to the formation of a meniscus, drops and bubbles, and the capillary rise

observed in small tubes, because surface tension can resist pressure differences across

the interface.

5 0

3 years ago