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Papessa [141]
1 year ago
8

when a unit load is secured to a pallet, it is more difficult for pilferage to take place. true false

Engineering
2 answers:
igor_vitrenko [27]1 year ago
8 0

Answer: True

Explanation:

When a unit load is secured to a pallet, it is more difficult for pilferage to take place. This is because the unit load is more securely attached to the pallet, making it more difficult for someone to remove items without being noticed. Additionally, securing a unit load to a pallet can also make it easier to transport and handle, which can further reduce the risk of pilferage.

skelet666 [1.2K]1 year ago
5 0

Answer:

It is TRUE.

Explanation:

When a unit load is secured to a pallet, it is more difficult for pilferage to take place.

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3 years ago
*6–24. The beam is used to support a dead load of 400 lb>ft, a live load of 2 k>ft, and a concentrated live load of 8 k. D
lisabon 2012 [21]

Answer:

(a) maximum positive reaction at A = 64.0 k

(b) maximum positive shear at A = 32.0 k

(c) maximum negative moment at C = -540 k·ft

Explanation:

Given;

dead load  Gk = 400 lb/ft

live load Qk = 2 k/ft

concentrated live load Pk =8 k

(a) from the influence line for vertical reaction at A, the maximum positive reaction is

A_{ymax} = 2*(8) +(1/2(20 - 0)* (2))*(2 + 0.4) = 64 k

See attachment for the calculations of (b) & (c) including the influence line

3 0
3 years ago
Which of the following color schemes is composed of hues next to eachother on the color wheel ?
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3 years ago
A flywheel made of Grade 30 cast iron (UTS = 217 MPa, UCS = 763 MPa, E = 100 GPa, density = 7100 Kg/m, Poisson's ratio = 0.26) h
hram777 [196]

Answer:

N = 38546.82 rpm

Explanation:

D_{1} = 150 mm

A_{1}= \frac{\pi }{4}\times 150^{2}

              = 17671.45 mm^{2}

D_{2} = 250 mm

A_{2}= \frac{\pi }{4}\times 250^{2}

              = 49087.78 mm^{2}

The centrifugal force acting on the flywheel is fiven by

F = M ( R_{2} - R_{1} ) x w^{2} ------------(1)

Here F = ( -UTS x A_{1} + UCS x A_{2} )

Since density, \rho = \frac{M}{V}

                        \rho = \frac{M}{A\times t}

                        M = \rho \times A\times tM = 7100 \times \frac{\pi }{4}\left ( D_{2}^{2}-D_{1}^{2} \right )\times t

                        M = 7100 \times \frac{\pi }{4}\left ( 250^{2}-150^{2} \right )\times 37

                        M = 8252963901

∴ R_{2} - R_{1} = 50 mm

∴ F = 763\times \frac{\pi }{4}\times 250^{2}-217\times \frac{\pi }{4}\times 150^{2}

  F = 33618968.38 N --------(2)

Now comparing (1) and (2)

33618968.38 = 8252963901\times 50\times \omega ^{2}

∴ ω = 4036.61

We know

\omega = \frac{2\pi N}{60}

4036.61 = \frac{2\pi N}{60}

∴ N = 38546.82 rpm

7 0
3 years ago
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