All categories

3 years ago

Whats the best used for cable -stayed bridge

Engineering

1 answer:

nalin [4]3 years ago

7 0

Answer:

a cable -stayed bridge has, one or more towers,from which cable support the bridge deck.

You might be interested in

g Two Standard 1/2" B18.8.2 dowel pins are to be installed in part B with an LN1 fit. The thickness of plate A is .750 +/- .005"

allochka39001 [22]

Answer:

nmuda mudaf A certain vehicle loses 3.5% of its value each year. If the vehicle has an initial value of $11,168, construct a model that represents the value of the vehicle after a certain number of years. Use your model to compute the value of the vehicle at the end of 6 years.

Explanation:

5 0

3 years ago

Fresh cut potato strips with a moisture content of 50% (w/w) on wet basis are fried in peanut oil to produce French fries.

Rufina [12.5K]

Answer:

Percentage of oil uptake by the potato = 5.57% on a wet basis (including the moisture content)

Percentage of oil uptake by the potato = 7.24% on a dry basis (excluding the moisture content)

Explanation:

Starting with 1000kg of fresh potatoes,

It is given that this consists of 50% (w/w) of water.

Meaning that, 1000 kg of fresh potatoes consists of (1000 × 50%) of water = 500 kg of water.

If 1000 kg of potatoes consist of 500 kg of water, then the remaining 500 kg is pure potatoes (since it is stated that freshly cut potatoes have no oil content)

After frying, the weight of the French fries is 700 kg now.

But of this 700 kg, 23 % of its weight is moisture content, i.e. water,

23% of 700 kg = 161 kg.

This means that the amount of potatoes and oil in the French fries is 700 - 161 = 539 kg

But, recall, that the amount of potatoes in this process from the start is 500 kg. This amount doesn't change even on frying the fresh potatoes into French fries.

So, amount of oil in the French fries = 539 - 500 = 39 kg

Percentage of oil uptake by the potato = 100% × (amount of oil intake)/(total mass of potatoes) = 100% × 39/700 = 5.57% on a wet basis (including the moisture content)

And 100% × 39/539 = 7.24% on a dry basis (excluding the moisture content)

4 0

3 years ago

Air at 25 C and 1 atm is flowing over a long flat plate with a velocity of 8 m/s. Determine the distance from the leading edge o

DIA [1.3K]

Answer:

Explanation:

Given data

Temprature of air= $25^{\circ}$

pressure =1 atm

velocity(V)=8m/s

From the table for air at $25^{\circ}$ and 1 atm

kinematic viscosity $\left ( \nu\right )=1.562\times 10^{-5} m^{2}/s$

Reynolds number for turbulent flow $=5\times 10^{5}$

and Re.no.= $\frac{V \times X}{\nu }$

therefore length where turbulent flows start is

X= $\frac{\left ( Re.no.\right )\times \nu }{V}$

X=0.976 m

Thickness of boundary layer is given by

$\delta _x=\frac{5X}{\sqrt{Re_x}}$

$\delta _x=\frac{5\times 0.976}{\sqrt{5\times 10^5}}$

$\delta _x=6.901mm$

for water

kinematic viscosity $\left ( \nu\right )=8.91\times 10^{-7} m^{2}/s$

Reynolds number for turbulent flow $=5\times 10^{5}$

and Re.no.= $\frac{V \times X}{\nu }$

therefore length where turbulent flows start is

X= $\frac{\left ( Re.no.\right )\times \nu }{V}$

X=0.055m

Thickness of boundary layer is given by

$\delta _x=\frac{5X}{\sqrt{Re_x}}$

$\delta _x=\frac{5\times 0.055}{\sqrt{5\times 10^5}}$

$\delta _x=0.3889mm$

5 0

3 years ago

Read 2 more answers

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

4. Calculate the square root of potato using the sqrt() function. Print out the value of potato again to verify that the value o

N76 [4]

Answer:

potato<-100

print(potato)

sqrt(potato)

potato<-potato*2

print(potato)

Explanation:

The complete question is as follows

Create a variable called potato whose value corresponds to the number of potatoes you’ve eaten in the last week. Or something equally ridiculous. Print out the value of potato.

Calculate the square root of potato using the sqrt() function. Print out the value of potato again to verify that the value of potato hasn’t changed.

Reassign the value of potato to potato * 2.

Print out the new value of potato to verify that it has changed

The question was answered using R programming language.

At line 1, I assumed that I ate 100 potatoes in the previous week.

So, potato = 100

At line 2, the value of potato is printed as 100.

At line 3, the square root of potato is calculated using sqrt function: Square for of 100 = 10

At line 4,the initial value of potato is doubled and stored in potato variable. 100 * 2 = 200

At line 5, the new value of potato is printed: 200.

8 0

4 years ago