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marissa [1.9K]
3 years ago
12

Help I will brainliest

Engineering
1 answer:
ValentinkaMS [17]3 years ago
3 0

Answer:

hacking?

Explanation:

jsd

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Vapor lock occurs when the gasoline is cooled and forms a gel, preventing fuel flow and
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True

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How is the art of orgami useful in science and technology?
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A lot of origami used symmetry of paper to create things. And it happens that a lot of things in science can be inferred or easily determined using symmetry. The most recent thing I can think of is exploiting symmetry in Gauss’s Law for electrical flux
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3 years ago
An inverted tee lintel is made of two 8" x 1/2" steel plates. Calculate the maximum bending stress in tension and compression wh
Kamila [148]

Answer:

hello your question lacks some information attached is the complete question

A) (i)maximum bending stress in tension = 0.287 * 10^6 Ib-in

    (ii) maximum bending stress in compression =  0.7413*10^6 Ib-in

B) (i)  The average shear stress at the neutral axis = 0.7904 *10 ^5 psi

    (ii)  Average shear stress at the web = 18.289 * 10^5 psi

    (iii) Average shear stress at the Flange = 1.143 *10^5 psi

Explanation:

First we calculate the centroid of the section,then we calculate the moment of inertia and maximum moment of the beam( find attached the calculation)

A) Calculate the maximum bending stress in tension and compression

lintel load = 10000 Ib

simple span = 6 ft

( (moment of inertia*Y)/ I ) = MAXIMUM BENDING STRESS

I = 53.54

i) The maximum bending stress (fb) in tension=

= \frac{M_{mm}Y }{I}  = \frac{6.48 * 10^6 * 2.375}{53.54} =  0.287 * 10^6 Ib-in

ii) The maximum bending stress (fb) in compression

= \frac{M_{mm}Y }{I} = \frac{6.48 *10^6*(8.5-2.375)}{53.54} = 0.7413*10^6 Ib-in

B) calculate the average shear stress at the neutral axis and the average shear stresses at the web and the flange

i) The average shear stress at the neutral axis

V = \frac{wL}{2} = \frac{1000*6*12}{2} = 3.6*10^5 Ib

Ay = 8 * 0.5 * (2.375 - 0.5 ) + 0.5 * (2.375 - \frac{0.5}{2} ) * \frac{(2.375 - (\frac{0.5}{2} ))}{2}

= 5.878 in^3

t = VQ / Ib  = ( 3.6*10^5 * 5.878 ) / (53.54 8 0.5) = 0.7904 *10 ^5 psi

ii) Average shear stress at the web ( value gotten from the shear stress at the flange )

t = 1.143 * 10^5 * (8 / 0.5 )  psi

  = 18.289 * 10^5 psi

iii) Average shear stress at the Flange

t = VQ / Ib = \frac{3.6*10^5 * 8*0.5*(2.375*(0.5/2))}{53.54 *0.5}

= 1.143 *10^5

4 0
4 years ago
A binary star system consists of two stars of masses m1m1m_1 and m2m2m_2. The stars, which gravitationally attract each other, r
d1i1m1o1n [39]

Answer:

          a_c_2=\dfrac{a_c_1\times m_1}{m_2}

Explanation:

The question is: <em>Find the magnitude of the centripetal acceleration of the star with mass m₂</em>

The <em>centripetal acceleration</em> is the quotient of the centripetal force and the mass.

                a_c=\dfrac{F_c}{m}

Thus, you can write the equations for each star:

     

       a_c_1=\dfrac{F_c_1}{m_1}

       a_c_2=\dfrac{F_c_2}{m_2}

As per Newton's third law, the centripetal forces are equal in magnitude. Then:

       a_c_1\times m_1=a_c_2\times m_2

Now you can clear a_c_2:

          a_c_2=\dfrac{a_c_1\times m_1}{m_2}

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Murrr4er [49]

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but the way is the way but the WAY is not the way

Explanation:

(yoda voice)

5 0
3 years ago
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