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3 years ago

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one airline limits the size of carry-on luggage to a volume of 40,000 cm. A passenger has a carry-on that has an area of 1,960 c

m and is 23 cm high. Is the passenger's luggeage ok to carry onto the airplane?

1 answer:

PolarNik [594]3 years ago

7 0

<u>No, since the volume of the passenger's luggage ( 45,080 cm³) exceeds the allotted volume for carry-on luggages (40,000 cm³).</u>

Explanation:

<h2>Given:</h2>

Acceptable volume = 40,000 cm³

Area of luggage = 1,960 cm²

Height of luggage = 23 cm

<h2>Question:</h2>

Is the passenger's luggeage ok to carry onto the airplane

<h2>Equation:</h2>

V = l x w x h

or we can use

V = A x h

Since A = l x w

where: V - volume

A - area

l - length

w - width

h - height

<h2>Solution:</h2>

V = A x h

V = ( 1,960 cm²)(23 cm)

V = 45,080 cm³

45,080 cm³ is greater than the acceptable volume 40,000 cm³

<h2>Final Answer:</h2><h3><u>No, since the volume of the passenger's luggage ( 45,080 cm³) exceeds the allotted volume for carry-on luggages (40,000 cm³).</u></h3><h3 />

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Explanation:

The complete statement of this question is "Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320 MPa, and a factor of safety of 4.2 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of P = 450 kN"

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The minimum allowable cross section area is cleared in the shearing stress equation:

$A = \frac{4}{3}\cdot \frac{V}{\tau_{max}}$

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The diameter is now cleared and computed:

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$D =\sqrt{\frac{4}{\pi}\cdot (1.640\times 10^{-3}\,m^{2})$

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