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nata0808 [166]
2 years ago
15

List five pieces of personal safety equipment which must be in everyday use in the workshop​

Engineering
1 answer:
tatyana61 [14]2 years ago
3 0

Answer:

safety glasses, hearing protection, respirator, work gloves, and work boots.

Explanation:

please give brainliest!! <3

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A 650-kN column load is supported on a 1.5 m square, 0.5 m deep spread footing. The soil below is a well-graded, normally consol
insens350 [35]

<u>Explanation:</u>

Determine the weight of footing

W_{f}=\gamma(L)(B)(D)

Where W_{f} is the weight of footing, γ is the unit weight of concrete,  L is the length of footing is the width of footing, and D is the depth of footing

Substitute 2 m \text { for } L, 1.5 m \text { for } B, 0.5 m \text { for } D \text { and } 23.6 kN / m ^{3} for γ in the equation

\begin{aligned}W_{f} &=\left(23.6 kN / m ^{3}\right)(2 m )(1.5 m )(0.5 m ) \\&=35.4 kN\end{aligned}

Therefore, the weight of the footing is 35.4 kN

Determine the initial vertical effective stress.

\sigma_{z p}^{\prime}=\gamma(D+B)-u

Here,   \sigma_{z^{p}}^{\prime} is initial vertical stress at a depth below ground surface  γ is the unit weight of soil, D is depth and u is pore water pressure.

Substitute 18 kN / m ^{3} \text { for } \gamma, 1.5 m \text { for } B, 0.5 m \text { for } D \text { and } 0 for u in the equation

\begin{aligned}\sigma_{z p}^{\prime} &=\left(18 kN / m ^{3}\right)(1.5+0.5) m -0 \\&=36 kPa\end{aligned}

Therefore, the initial vertical stress is 36 kPa

Determine the vertical effective stress.

\sigma_{z D}^{\prime}=\gamma D

Here,   \sigma_{z^{p}}^{\prime} is initial vertical stress at a depth below ground surface  γ is the unit weight of soil, D is depth and u is pore water pressure.

Substitute \(18 kN / m ^{3}\) for \(\gamma, 0.5 m\) for \(D\) and 0 for \(u\) in the equation.

\begin{aligned}\sigma_{z b}^{\prime} &=\left(18 kN / m ^{3}\right)(0.5 m )-0 \\&=9 kPa\end{aligned}

Therefore, the vertical stress at a depth below the ground surface is

9 kPa

Determine the influence factor at the midpoint of soil layer,

I_{e p}=0.5+0.1 \sqrt{\frac{q-\sigma_{s 0}^{\prime}}{\sigma_{z p}^{\prime}}}

Here I_{e p} is the influence factor at the midpoint of soil layer  \sigma_{z^{p}}^{\prime} is initial vertical stress, \sigma_{z^{p}}^{\prime} is vertical effective stress, and Q is bearing pressure

Substitute 36 kPa for \(\sigma_{z p}^{\prime}, 228.47\) kPa for \(q,\) and 9 kPa for \(\sigma_{z D}^{\prime}\) in the equation\begin{aligned}I_{\epsilon P} &=0.5+0.1 \sqrt{\frac{228.47 kPa -9 kPa }{36 kPa }} \\&=0.75\end{aligned}

Therefore the influence factor at the midpoint of the soil layer is 0.693

6 0
3 years ago
(a) Design a lag compensation to meet the following specifications: The step response settling time is to be less than 5 sec, th
Pavlova-9 [17]

Answer:

Please see the attached file for the complete answer.

Explanation:

Download pdf
4 0
2 years ago
Some extremely hazardous materials used in welding operations include.
enyata [817]

Many of the substances in welding smoke, such as chromium, nickel, arsenic, asbestos, manganese, silica, beryllium, cadmium, nitrogen oxides, phosgene, acrolein, fluorine compounds, carbon monoxide, cobalt, copper, lead, ozone, selenium, and zinc, can be extremely toxic.

3 0
2 years ago
Calculate the maximum load that a 7075 series aluminum alloy bar (with a T6 temper heat treatment) can support without permanent
Aleksandr [31]

Answer:

The maximum load the bar can withstand = 35.43 KN

Explanation:

Ultimate tensile strength of the given aluminium bar \sigma = 540 M pa

Cross section area of the bar = 8.1^{2}  = 65.61 mm^{2}

We know that the ultimate strength of the bar is calculated from

\sigma = \frac{P_{max} }{A}

540 = \frac{P_{max} }{65.61}

P_{max} = 540 × 65.61

P_{max} =  35.43 KN

Therefore the maximum load the bar can withstand = 35.43 KN

6 0
3 years ago
A spherical hot air balloon is initially filled with air having 120 kPa pressure and 24 °C temperature. Initial diameter of the
tino4ka555 [31]

Answer:

v = 1.076 m /s

Explanation:

Initial volume of balloon = 4/3 x  3.14 x (9.905/2)³

=508.56 m³

Final volume of balloon = 4/3 x 3.14 x (16.502/2)³

= 2351.73 m³

Increase in volume = 1843.17 m³

Cross sectional area of inlet  A  = 3.14 x( 1.458/2)²

A = 1.6687 m²

Volume rate of flow of air = cross sectional area x velocity of inflow

= 1 .6687 V [ V is velocity of inflow ]

Total time taken = Increase in volume / rate of flow of air

17.108 X 60 = 1843.17 / 1.6687 V

V = \frac{1843.17}{1.6687\times17.108\times60}

v = 1.076 m /s

8 0
3 years ago
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