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3 years ago

What are the main causes of injuries when using forklifts?

Engineering

2 answers:

astra-53 [7]3 years ago

6 0

Answer:

overturns

Explanation:

Elis [28]3 years ago

5 0

Collisions with other forklifts

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LUNES MARTES MIÉRCOLES JUEVES VIERNES SÁBADO DOMINGO

scZoUnD [109]

Answer:

Explanation:

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3 years ago

(TCO 4) A system samples a sinusoid of frequency 190 Hz at a rate of 120 Hz and writes the sampled signal to its output without

steposvetlana [31]

Answer:

The frequency that the sampling system will generate in its output is 70 Hz

Explanation:

Given;

F = 190 Hz

Fs = 120 Hz

Output Frequency = F - nFs

When n = 1

Output Frequency = 190 - 120 = 70 Hz

Therefore, if a system samples a sinusoid of frequency 190 Hz at a rate of 120 Hz and writes the sampled signal to its output without further modification, the frequency that the sampling system will generate in its output is 70 Hz

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3 years ago

Technician A says that 5W-30 would be better to use than 20W-50 in most vehicles in

shtirl [24]

Technician is correct sorry if im wronghg

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3 years ago

Read 2 more answers

Imagine you are a process safety consultant and you have been tasked to make a metal refinery site DSEAR compliant. What are the

masya89 [10]

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Control measures. ...

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Places where explosive atmospheres may occur ('ATEX' requirements)

hse uk

4 0

2 years ago

A steel pipe of 400-mm outer diameter is fabricated from 10-mm-thick plate by welding along a helix that forms an angle of 20° w

Verdich [7]

Explanation:

Outer di ameter $d_{0}=400 \mathrm{mm}[tex] Thickness of the cylinder [tex]t=10 \mathrm{mm}$

$\therefore[tex] Inner diam eter [tex]d_{i}=d_{0}-2 t=400-2 \times 10$

$d_{1}=380 \mathrm{mm}$

Given loading on the cylinder $P=300 \mathrm{kN}$ Helix an gle of the weld form $\theta=20^{\circ}$

(i) Normal stress on the plane at angle $\theta=20^{\circ}$ is

$\sigma=\frac{P \cos ^{2} \theta}{A_{0}}$

$\text { Where } A_{0}=\frac{\pi}{4}\left(d_{0}^{2}-d_{1}^{2}\right)$

$\quad=\frac{\pi}{4}\left(400^{2}-380^{2}\right)$

$=12252.21 \mathrm{mm}^{2}$

$=12.25221 \times 10^{-9} \mathrm{m}^{2}$

$\sigma=\frac{-300 \times 10^{2} \times \cos ^{2} 20}{12.25221 \times 10^{-1}}$

$=-21.6 \mathrm{MPa}$

(ii) Shear stress along an angle of $\theta=20^{\circ}$ is $\tau=\frac{P}{A_{0}} \cos \theta \sin \theta$

$=\frac{-300 \times 10^{-1} \times \cos 20 \times \sin 20}{12.25221 \times 10^{-3}}$

$=-7.86 \mathrm{MPa}$

3 0

3 years ago