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IceJOKER [234]
3 years ago
9

La extensión de un resumen debe estar en un rango de _________ del texto inicial. a)25 a 35 % b)10 a 20 % c)15 a 20 % d)20 a 25

%
Engineering
1 answer:
mestny [16]3 years ago
3 0

Answer:

b)

Explanation:

because it is correct

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If i build thing a and thing a builds thing b did i build thing b
postnew [5]

Answer:

<h3>Yes</h3>

Explanation:

If you build thing "a" and thing "a" builds thing "b" you <u>indirectly</u> build thing "b".

3 0
3 years ago
Read 2 more answers
The normal stress at gage H calculated in Part 1 includes four components: an axial component due to load P, σaxial, P, a bendin
Degger [83]

Answer:

hello your question has some missing information attached to the answer is the missing component

Answer : αaxial,p = -6.034 ksi ( compressive )

             αbend,p = 19.648 ksi ( tensile )

Explanation:

αaxial, p = \frac{-p}{A}   equation 1

αbend, p = \frac{(P*A)*\frac{d}{2} }{I_{z} } equation 2

P = load = 35 kips

A = area of column = 5.8 in^{2}

d = column cross section depth = 9.5 in

I_{Z} = 55.0 in^{4}

Hence equation 1 becomes

αaxial,p = -35 / 5.8 = - 6.034 ksi ( compressive )

equation 2 becomes

αbend, p = \frac{(35*6.5)(\frac{9.2}{2}) }{55} = + 19.648 ksi ( tensile )

7 0
3 years ago
The two shafts of a Hooke’s coupling have their axes inclined at 20°.The shaft A revolves at a uniform speed of 1000 rpm. The sh
lapo4ka [179]

Answer:

33.429 N-m

Explanation:

Given :

Inclination angle of two shaft, α = 20°

Speed of shaft A, N_{A} = 1000 rpm

Mass of flywheel, m = 30 kg

Radius of Gyration, k =100 mm

                                   = 0.1 m

Now we know that for maximum velocity,

\frac{N_{B}}{N_{A}} = \frac{cos\alpha }{1 - sin^{2}\alpha }

\frac{N_{B}}{1000} = \frac{cos20}{1 - sin^{2}20 }

N_{B} = 1064.1 rpm

Now we know

Mass of flywheel, m = 30 kg

Radius of Gyration, k =100 mm

                                   = 0.1 m

Therefore moment of inertia of flywheel, I = m.k^{2}

                                                                      =30 X 0.1^{2}

                                                                     = 0.3 kg-m^{2}

Now torque on the output shaft

T₂ = I x ω

    = 0.3 X 1064.2 rpm

    = 0.3\times \frac{2\pi \times 1064.1}{60}

     = 33.429 N-m

Torque on the Shaft B is 33.429 N-m

4 0
3 years ago
A Service Schedule is...
VikaD [51]

Answer:

option c

Explanation:

8 0
3 years ago
Read 2 more answers
Consider two different versions of algorithm for finding gcd of two numbers (as given below), Estimate how many times faster it
juin [17]

Answer:

Explanation:

Step 1:

a) The formula for compute greatest advisor is

     gcd(m,n) = gcd (n,m mod n)

the gcd(31415,14142) by applying Euclid's algorithm is

    gcd(31,415,14,142) =gcd(14,142,3,131)

                                  =gcd=(3,131, 1,618)

                                   =gcd(1,618, 1,513)

                                   =gcd(1,513, 105)

                                   =gcd(105, 43)

                                    =gcd(43, 19)

                                     =gcd(19, 5)

                                      =gcd(5, 4)

                                      =gcd(4, 1)

                                      =gcd(1, 0)

                                      =1

STEP 2

b)  The number of comparison of given input with the algorithm based on  checking consecutive integers and Euclid's algorithm is

     The number of division using Euclid's algorithm =10 from part (a)

      The consecutive integer checking algorithm:

      The number of iterations =14,142 and 1 or 2 division of iteration.

        14,142 ∠= number of division∠ = 2*14,142

         Euclid's algorithm is faster by at least 14,142/10 =1400 times

          At most 2*14,142/10 =2800 times.

5 0
3 years ago
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