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2 years ago

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A disk shaped part is cast out of aluminum. Diameter of the disk = 650 mm and thickness = 16 mm. If the mold constant = 3.2 sec/

mm2 in Chvorinov's rule, how long will it take the casting to solidify

1 answer:

Natasha_Volkova [10]2 years ago

3 0

Answer:

Time for solidification of the casting is 186 sec.

Explanation:

Chvorinov's rule is used to find the solidification time for any casting shape.

Step1

Given:

Diameter of the disk is 650 mm.

Thickness of the disk is 16 mm.

Chvorinov's rule constant is 3.2 sec/mm².

Step2

Calculation:

Total volume of the disk is calculated as follows:

$V=\frac{\pi}{4}d^{2}t$

$V=\frac{\pi}{4}({650})^{2}\times({16})$

V= 5309291.585 mm³.

Step3

Total surface area of the disk is calculated as follows:

$A=\pi dt+2\times (\frac{\pi }{4}d^{2})$

$A=\pi \times 650\times 16+2\times (\frac{\pi }{4}\times 650^{2})$

A= 696334.0117 mm².

Step4

Chvorinov's rule for time calculation is expressed as follows:

$t=K(\frac{V}{A})^{2}$

$t=K(\frac{5309291.585}{696334.0117})^{2}$

$t=3.2(\frac{5309291.585}{696334.0117})^{2}$

t=186 sec.

Thus, the time for solidification of the casting is 186 sec.

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A spring-loaded toy gun is used to shoot a ball of mass m = 1.50 kg straight up in the air. The spring has spring constant k = 6

adell [148]

Answer:

1) a) Mechanical energy is conserved because no dissipative forces perform work on the ball.

2) The muzzle velocity of the ball is approximately 5.272 meters per second.

3) The maximum height of the ball is 1.417 meters.

Explanation:

1) Which of the following statements are true?

a) Mechanical energy is conserved because no dissipative forces perform work on the ball.

True, statement indicates that there is no air resistence and no friction between ball and the inside of the gun because the first never touches the latter one.

b) The forces of gravity and the spring have potential energies associated with them.

False, force of gravity do work on the ball and spring receives a potential energy at being deformated by the ball.

c) No conservative forces act in this problem after the ball is released from the spring gun.

False, the absence of no conservative forces is guaranteed for the entire system according to the statement of the problem.

2) According to the statement, we understand that spring is deformed and once released and just after reaching its equilibrium position, the muzzle velocity is reached. As spring deformation is too small in comparison with height, we can neglect changes in gravitational potential energy. By Principle of Energy Conservation, we describe the motion of the ball by the following expression:

$U_{k, 1}+K_{1}=U_{k,2}+K_{2}$ (Eq. 1)

Where:

$U_{k,1}$ , $U_{k,2}$ - Initial and final elastic potential energies of spring, measured in joules.

$K_{1}$ , $K_{2}$ - Initial and final translational kinetic energies of the ball, measured in joules.

After using definitions of elastic potential and translational kinetic energies, we expand the equation above as:

$\frac{1}{2}\cdot m\cdot (v_{2}^{2}-v_{1}^{2}) = \frac{1}{2}\cdot k\cdot (x_{1}^{2}-x_{2}^{2})$

And the final velocity is cleared:

$m\cdot (v_{2}^{2}-v_{1}^{2}) = k\cdot (x_{1}^{2}-x_{2}^{2})$

$v_{2}^{2}-v_{1}^{2} =\frac{k}{m}\cdot (x_{1}^{2}-x_{2}^{2})$

$v_{2}^{2} =v_{1}^{2}+\frac{k}{m}\cdot (x_{1}^{2}-x_{2}^{2})$

$v_{2} = \sqrt{v_{1}^{2}+\frac{k}{m}\cdot (x_{1}^{2}-x_{2}^{2}) }$ (Eq. 2)

Where:

$v_{1}$ , $v_{2}$ - Initial and final velocities of the ball, measured in meters per second.

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass of the ball, measured in kilograms.

$x_{1}$ , $x_{2}$ - Initial and final position of spring, measured in meters.

If we know that $v_{1} = 0\,\frac{m}{s}$ , $k = 667\,\frac{N}{m}$ , $m = 1.50\,kg$ , $x_{1} = -0.25\,m$ and $x_{2} = 0\,cm$ , the muzzle velocity of the ball is:

$v_{2} =\sqrt{\left(0\,\frac{m}{s} \right)^{2}+\left(\frac{667\,\frac{N}{m} }{1.50\,kg} \right)\cdot [(-0.25\,m)^{2}-(0\,m)^{2}]}$

$v_{2}\approx 5.272\,\frac{m}{s}$

The muzzle velocity of the ball is approximately 5.272 meters per second.

3) After leaving the toy gun, the ball is solely decelerated by gravity. We construct this model by Principle of Energy Conservation:

$U_{g,2}+K_{2} = U_{g,3}+K_{3}$ (Eq. 3)

Where:

$U_{g,2}$ , $U_{g,3}$ - Initial and gravitational potential energies of the ball, measured in joules.

$K_{2}$ , $K_{3}$ - Initial and final translational kinetic energies of the ball, measured in joules.

After applying definitions of gravitational potential and translational kinetic energies, we expand the equation above and solve the resulting for the final height:

$m\cdot g \cdot (h_{3}-h_{2}) = \frac{1}{2}\cdot m \cdot (v_{2}^{2}-v_{3}^{2})$

$h_{3}-h_{2}=\frac{v_{2}^{2}-v_{3}^{2}}{2\cdot g}$

$h_{3} = h_{2} +\frac{v_{2}^{2}-v_{3}^{2}}{2\cdot g}$ (Eq. 4)

$h_{2}$ , $h_{3}$ - Initial and final heights of the ball, measured in meters.

$v_{2}$ , $v_{3}$ - Initial and final velocities of the ball, measured in meters per second.

$g$ - Gravitational acceleration, measured in meters per square second.

If we get that $v_{2} = 5.272\,\frac{m}{s}$ , $v_{3} = 0\,\frac{m}{s}$ , $h_{2} = 0\,m$ and $g = 9.807\,\frac{m}{s^{2}}$ , the maximum height of the ball is:

$h_{3} = 0\,m+\frac{\left(5.272\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$h_{3} = 1.417\,m$

The maximum height of the ball is 1.417 meters.

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2 years ago

Let Deterministic Quicksort be the non-randomized Quicksort which takes the first element as a pivot, using the partition routin

juin [17]

Answer:

Answer for the question:

Let Deterministic Quicksort be the non-randomized Quicksort which takes the first element as a pivot, using the partition routine that we covered in class on the quicksort slides. Consider another almost-best case for quicksort, in which the pivot always splits the arrays 1/3: 2/3, i.e., one third is on the left, and two thirds are on the right, for all recursive calls of Deterministic Quicksort. (a) Give the runtime recurrence for this almost-best case. (b) Use the recursion tree to argue why the runtime recurrence solves to Theta (n log n). You do not need to do big-Oh induction. (c) Give a sequence of 4 distinct numbers and a sequence of 13 distinct numbers that cause this almost-best case behavior. (Assume that for 4 numbers the array is split into 1 element on the left side, the pivot, and two elements on the right side. Similarly, for 13 numbers it is split with 4 elements on the left, the pivot, and 8 elements on the right side.)

is given in the attachment.

Explanation:

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2 years ago

Write multiple if statements: If carYear is before 1967, print "Probably has few safety features." (without quotes). If after 19

Free_Kalibri [48]

Answer:

The solution code is written in Python 3.

carYear = 1995
if(carYear < 1967):
print("Probably has few safety features.\n")
if(carYear > 1970):
print("Probably has head rests. \n")
if(carYear > 1991):
print("Probably has electronic stability control.\n")
if(carYear > 2002):
print("Probably has airbags. \n")

Explanation:

Firstly, create a variable, <em>carYear</em> to hold the value of year of the car make. (Line 1)

Next, create multiple if statements as required by the question (Line 3-13). The operator "<" denotes "smaller" and therefore <em>carYear < 1967</em> means any year before 1967. On another hand, the operator ">" denotes "bigger" and therefore <em>carYear > 1970 </em>means any year after 1970.

The print statement in each of the if statements is done using the Python built-in function <em>print()</em>. The "\n" is an escape sequence that create a new line at the end of each printed phrase.

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2 years ago

The emissivity of galvanized steel sheet, a common roofing material, is ε = 0.13 at temperatures around 300 K, while its absorpt

Step2247 [10]

Answer:

759.99W/m²

Explanation:

Question: If the temperature of the sheet is 77C,what is the incident solar radiation on aday with Tinf= Tsurr= 16°C?

Given

Energy Equation of the Gas

αs * Gs * A + h * A * (T inf - Tg) + εσA (Tsurr⁴- Tg⁴) = 0

Where σ= 5.67 *10^-8 W/m²K⁴ (Stefan-Boltzmann constant)

ε = 0.13 (Emisivity)

αs = 0.65 (Absorptivity for solar radiation)

h = 7W/m²K⁴

Tg = 77 + 273.15K = 350.15K

T inf = 16 + 273.15 = 288.15K

T surr= T inf = 288.15

Substitute the above values in the Gas Equation, we have

0.65 * Gs * A + 7 * A * (288.15 - 350.15) + 0.13 * 5.67 * 10^-8 * A * (288.15⁴ - 350.15⁴) = 0

0.65 * Gs * A = - 7 * A * (288.15 - 350.15) - 0.13 * 5.67 * 10^-8 * A * (288.15⁴ - 350.15⁴)

A cancels out, so we are left with

0.65 * Gs = - 7 * (288.15 - 350.15) - 0.13 * 5.67 * 10^-8 * (288.15⁴ - 350.15⁴)

0.65Gs = 434 - 0.7372 * 10^-8(−8,137,940,481.697)

0.65Gs = 434 + 0.7372 * 81.37940481697

0.65Gs = 493.992897231070284

Gs = 493.992897231070284/0.65

Gs = 759.9890726631850

Gs = 759.99W/m² ------- Approximated

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A shipment of rebar that weighs 745 kg would weigh roughly how much in pounds

Andre45 [30]

Answer:

Dont no but will check

Explanation:

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