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3 years ago

A suitcase has a width of 55 cm. Which unit conversation fraction should you use to find the width in inches?

Physics

1 answer:

Crank3 years ago

6 0

Answer:

2,54 cm are equal to 1 inch

Explanation:

Doing the conversion:

$55[cm]*\frac{1[inch]}{2,54[cm]} =21,65[inch]$

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4 years ago

A cantilever beam with a width b=100 mm and depth h=150 mm has a length L=2 m and is subjected to a point load P =500 N at B. Ca

DerKrebs [107]

Answer:

Explanation:

Given that:

width b=100mm

depth h=150 mm

length L=2 m =200mm

point load P =500 N

Calculate moment of inertia

$I=\frac{bh^3}{12} \\\\=\frac{100 \times 150^3}{12} \\\\=28125000\ m m^4$

Point C is subjected to bending moment

Calculate the bending moment of point C

M = P x 1.5

= 500 x 1.5

= 750 N.m

M = 750 × 10³ N.mm

Calculate bending stress at point C

$\sigma=\frac{M.y}{I} \\\\=\frac{(750\times10^3)(25)}{28125000} \\\\=0.0667 \ MPa\\\\ \sigma =666.67\ kPa$

Calculate the first moment of area below point C

$Q=A \bar y\\\\=(50 \times 100)(25 +\frac{50}{2} )\\\\Q=250000\ mm$

Now calculate shear stress at point C

$=\frac{FQ}{It}$

$=\frac{500*250000}{28125000*100} \\\\=0.0444\ MPa\\\\=44.4\ KPa$

Calculate the principal stress at point C

$\sigma_{1,2}=\frac{\sigma_x+\sigma_y}{2} \pm\sqrt{(\frac{\sigma_x-\sigma_y}{2} ) + (\tau)^2} \\\\=\frac{666.67+0}{2} \pm\sqrt{(\frac{666.67-0}{2} )^2 \pm(44.44)^2} \ [ \sigma_y=0]\\\\=333.33\pm336.28\\\\ \sigma_1=333.33+336.28\\=669.61KPa\\\\\sigma_2=333.33-336.28\\=-2.95KPa$

Calculate the maximum shear stress at piont C

$\tau=\frac{\sigma_1-\sigma_2}{2}\\\\=\frac{669.61-(-2.95)}{2} \\\\=336.28KPa$

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3 years ago

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