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Tatiana [17]
2 years ago
10

Consider a vortex filament of strength in the shape of a closed circular loop of radius R. Obtain an expression for the velocity

induced at the center of the loop in terms of T and R
Engineering
1 answer:
zysi [14]2 years ago
3 0

Answer:

<em>v</em><em> </em>= T/(2R)

Explanation:

Given

R = radius

T = strength

From Biot - Savart Law

d<em>v</em> = (T/4π)* (d<em>l</em> x <em>r</em>)/r³

Velocity induced at center

<em>v </em>= ∫  (T/4π)* (d<em>l</em> x <em>r</em>)/r³

⇒   <em>v </em>= ∫  (T/4π)* (d<em>l</em> x <em>R</em>)/R³  (<em>k</em>)        <em>k</em><em>:</em> unit vector perpendicular to plane of loop

⇒   <em>v </em>= (T/4π)(1/R²) ∫ dl

If l ∈  (0, 2πR)

⇒   <em>v </em>= (T/4π)(1/R²)(2πR)  (<em>k</em>)    ⇒   <em>v </em>= T/(2R)  (<em>k</em>)  

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Theorems of Pappus and Guldinus are used to find: a. The surface area and volume of a body of rotation b. The surface area and v
s2008m [1.1K]

Answer:

option (a). The surface area and volume of a body of rotation

is the correct option

Explanation:

Theorems of Pappus and Guldinus are used to find the surface area and volume of a revolving body. It is neither applicable for surface areas and volumes of a symmetric body nor it helps to find the overall mass of any body. Thus, it can help to calculate the surface area and volume of any body rotated in 2-D frame(or any 2-D curve).

It is given or calculated as the product of area, perpendicular distance from the axis and length of the 2-D curve.

6 0
2 years ago
If a server takes precisely 15 seconds to serve a customer and customers arrive exactly every 20 seconds, what is the average wa
labwork [276]
The average waiting time is 10 seconds
7 0
2 years ago
Outline an algorithm in **pseudo code** for checking whether an array H[1..n] is a heap and determine its time efficiency.
svlad2 [7]

Answer:

Condition to break: H[j] \geq max {H[2j] , H[2j+1]}

Efficiency: O(n).

Explanation:

Previous concepts

Heap algorithm is used to create all the possible permutations with K possible objects. Was created by B. R Heap in 1963.

Parental dominance condition represent a condition that is satisfied when the parent element is greater than his children.

Solution to the problem

We assume that we have an array H of size n for the algorithm.

It's important on this case analyze the parental dominance condition in order to the algorithm can work and construc a heap.

For this case we can set a counter j =1,2,... [n/2] (We just check until n/2 since in order to create a heap we need to satisfy minimum n/2 possible comparisionsand we need to check this:Break condition: [tex]H[j] \geq max {H[2j] , H[2j+1]}

And we just need to check on the array the last condition and if is not satisfied for any value of the counter j we need to stop the algorithm and the array would not a heap. Otherwise if we satisfy the condition for each j =1,2,.....,[n/2]p then we will have a heap.

On this case this algorithm needs to compare 2*(n/2) times the values and the efficiency is given by O(n).

3 0
2 years ago
A shaft consisting of a steel tube of 50-mm outer diameter is to transmit 100 kW of power while rotating at a frequency of 34 Hz
nikitadnepr [17]

Answer:

25 - \sqrt[4]{26.66*10^{-8} }  mm

Explanation:

Given data

steel tube : outer diameter = 50-mm

power transmitted = 100 KW

frequency(f) = 34 Hz

shearing stress ≤ 60 MPa

Determine tube thickness

firstly we calculate the ; power, angular velocity and torque of the tube

power = T(torque) * w (angular velocity)

angular velocity ( w ) = 2\pif = 2 * \pi * 34 = 213.71

Torque (T) = power / angular velocity = 100000 / 213.71 = 467.92 N.m/s

next we calculate the inner diameter  using the relation

  \frac{J}{c_{2}  } = \frac{T}{t_{max} }  = 467.92 / (60 * 10^6) =  7.8 * 10^-6 m^3

also

c2 = (50/2) = 25 mm

\frac{J}{c_{2} } = \frac{\pi }{2c_{2} } ( c^{4} _{2} - c^{4} _{1} ) =  \frac{\pi }{0.050} [ ( 0.025^{4} - c^{4} _{1}  ) ]

therefore; 0.025^4 - c^{4} _{1} = 0.050 / \pi (7.8 *10^-6)

c^{4} _{1} = 39.06 * 10 ^-8 - ( 1.59*10^-2 * 7.8*10^-6)

    39.06 * 10^-8 - 12.402 * 10^-8 =26.66 *10^-8

c_{1} = \sqrt[4]{26.66 * 10^{-8} }  =

THE TUBE THICKNESS

c_{2} - c_{1} = 25 - \sqrt[4]{26.66*10^{-8} }  mm

4 0
2 years ago
______process in sheet metal is used for producing fluid tight joints. A - Hemming B- Seaming C-Beading D-Roll forming
SOVA2 [1]

Answer:

option B

Explanation:

The correct answer is option B

Seaming is an operation in which the edges are folded over another part to achieve the tight fit.

Seaming is generally used to join other parts together.

So, seaming is generally used for producing fluid-tight joints.

This process is used in the food industry on canned goods, metal roofing, and in the automotive industry.  

5 0
2 years ago
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