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Blababa [14]
3 years ago
15

Under the right conditions, it is possible, due to surface tension, to have metal objects float on water. Consider placing a sho

rt length of a small diameter steel (sp. wt.=500 lb/ft3) rod on a surface of water. What is the maximum diameter, Dmax⁡, that the rod can have before it will sink? Assume that the surface tension forces act vertically upward. Note: A standard paper clip has a diameter of 0.036 in. Partially unfold a paper clip and see if you can get it to float on water. Do the results of this experiment support your analysis?
Engineering
1 answer:
stiv31 [10]3 years ago
4 0

Answer:

D = 0.060732 in

Explanation:

given data

sp. wt. = 500 lb/ft³

diameter = 0.036 in

solution

we get here maximum diameter of rod that is express as

D = \sqrt{\frac{8 \sigma }{\pi y}}   ......................1

here \sigma surface tension of water at 60⁰f  = 5.03 × 10^{-3}  lb/ft and y = 500 lb/ft³

so put here value and we will get

D = \sqrt{\frac{8 \times 5.03 \times 10^{-3} }{\pi \times 500}}

D = 0.005061 ft

D = 0.060732 in

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Answer:

water based and solvent based

Explanation:

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5 0
4 years ago
Solve using Matlab the problems:
Firlakuza [10]

Answer:

Explanation:

% Clears variables and screen

clear; clc

% Asks user for input

n = input('Total number of objects: ');

r = input('Size of subgroup: ');

% Computes and displays permutation according to basic formulas

p = 1;

for i = n - r + 1 : n

   p = p*i;

end

str1 = [num2str(p) ' permutations'];

disp(str1)

% Computes and displays combinations according to basic formulas

str2 = [num2str(p/factorial(r)) ' combinations'];

disp(str2)

=================================================================================

Example: check

How many permutations and combinations can be made of the 15 alphabets, taking four at a time?

The answer is:

32760 permutations

1365 combinations

==================================================================================

7 0
3 years ago
A glass plate is subjected to a tensile stress of 40 MPa. If the specific surface energy is 0.3 J/m2 and the modulus of elastici
Alika [10]

Answer:

The maximum length of a surface flaw is 8.24 μm

8.24 μm

Explanation:

Given that:

The modulus of elasticity E = 69 GPa

The specific surface energy \delta_s = 0.3 J/m²

The length of the surface flaw "a" = ??

From the theory of the brittle fracture;

\sigma _c = \bigg (  \dfrac{2E \delta_s}{\pi a}  \bigg )^{1/2}

Making a the subject of the formula; we have:

a = \bigg (  \dfrac{2 \times E \times \delta_s}{\pi \sigma _c ^2}  \bigg )

a= \bigg (  \dfrac{2 \times 69*10^9 \times 0.3}{\pi (40*10^6)^2}  \bigg )

a = 8.24 × 10⁻⁶ m

a = 8.24 μm

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3 0
3 years ago
g Part 2: The features arrived the grammar Splitting categories and non-terminals gets out of hand fast. An alternative to the p
koban [17]

Answer:

The split is given by including spaces in both tabs

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The bracket notation can be used to indicate the split. Here is an example:

String [ ] parts = s. split ( "[/]")

3 0
3 years ago
Write a new ARMv8 assembly file called "lab04b.S" which is called by your main function. It should have the following specificat
Len [333]

Answer:

my_mul:

.globl my_mul

my_mul:

   //Multiply X0 and X1

   //   Does not handle negative X1!

   //   Note : This is an in efficient way to multipy!

   SUB SP, SP, 16       //make room for X19 on the stack

   STUR X19, [SP, 0]    //push X19

   ADD X19, X1, XZR     //set X19 equal to X1

   ADD X9 , XZR , XZR //set X9 to 0

mult_loop:

   CBZ X19, mult_eol

   ADD X9, X9, X0

   SUB X19, X19, 1

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mult_eol:

   LDUR X19, [SP, 0]

   ADD X0, X9, XZR      // Move X9 to X0 to return

   ADD SP, SP, 16       // reset the stack

   BR X30

Explanation:

6 0
4 years ago
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