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3 years ago

Why is it important to use PPE in doing the task of upgrading a computer server?

1 answer:

3 0

PPE is equipment that will protect workers against health or safety risks on the job. Explanation: The purpose is to reduce employee exposure to hazards when engineering and administrative controls are not feasible or effective to reduce these risks to acceptable levels.

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A metal plate of 400 mm in length, 200mm in width and 30 mm in depth is to be machined by orthogonal cutting using a tool of wid

dmitriy555 [2]

Complete Question:

A metal plate of 400 mm in length, 200mm in width and 30 mm in depth is to be machined by orthogonal cutting using a tool of width 5mm and a depth of cut of 0.5 mm. Estimate the minimum time required to reduce the depth of the plate by 20 mm if the tool moves at 400 mm per second.

Answer:

$T_{min} =$ 26 mins 40 secs

Explanation:

Reduction in depth, Δd = 20 mm

Depth of cut, $d_c = 0.5 mm$

Number of passes necessary for this reduction, $n = \frac{\triangle d}{d_c}$

n = 20/0.5

n = 40 passes

Tool width, w = 5 mm

Width of metal plate, W = 200 mm

For a reduction in the depth per pass, tool will travel W/w = 200/5 = 40 times

Speed of tool, v = 100 mm/s

$Time/pass = \frac{40*400}{400} \\Time/pass = 40 sec$

minimum time required to reduce the depth of the plate by 20 mm:

$T_{min} =$ number of passes * Time/pass

$T_{min} =$ n * Time/pass

$T_{min} =$ 40 * 40

$T_{min} =$ 1600 = 26 mins 40 secs

3 0

4 years ago

Read 2 more answers

The _______ is a tendency to assume that people with one positive attribute (e.g., who are physically attractive) also have othe

morpeh [17]

Answer:

Halo effect

Explanation:

In psychology, halo effect is defined as the tendency to assume that people with one positive attribute (e.g. physical attractiveness) also have other positive traits. Mere Exposure is the tendency to feel more positively toward a stimulus as a result of repeated exposure to it.

7 0

3 years ago

Extra points!!!!! List the general types of housing (Example: condo)

Nadusha1986 [10]

Answer:

co-op, apartment, townhome, manor etc

6 0

3 years ago

Read 2 more answers

Determine the resolution of a manometer required to measure the velocity of air at 50 m/s using a pitot-static tube and a manome

oksano4ka [1.4K]

Answer:

a) Δh = 2 cm, b) Δh = 0.4 cm

Explanation:

Let's start by using Bernoulli's equation for the Pitot tube, we define two points 1 for the small entry point and point 2 for the larger diameter entry point.

P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

Point 1 is called the stagnation point where the fluid velocity is reduced to zero (v₁ = 0), in general pitot tubes are used in such a way that the height of point 2 of is the same of point 1

y₁ = y₂

subtitute

P₁ = P₂ + ½ ρ v₂²

P₁ -P₂ = ½ ρ v²

where ρ is the density of fluid

now we measure the pressure on the included beforehand as a pair of communicating tubes filled with mercury, we set our reference system at the point of the mercury bottom surface

ΔP =ρ_{Hg} g h - ρ g h

ΔP = (ρ_{Hg} - ρ) g h

as the static pressure we can equalize the equations

ΔP = P₁ - P₂

(ρ_{Hg} - ρ) g h = ½ ρ v²

v = $\sqrt{\frac{2 (\rho_{Hg} - \rho) g}{\rho } } \ \sqrt{h}$

in this expression the densities are constant

v = A √h

A = $\sqrt{\frac{2(\rho_{Hg} - \rho ) g}{\rho } }$

They indicate the density of mercury rhohg = 13600 kg / m³, the density of dry air at 20ºC is rho air = 1.29 kg/m³

we look for the constant

A = $\sqrt{\frac{2( 13600 - 1.29) \ 9.8}{1.29} }$

A = 454.55

we substitute

v = 454.55 √h

to calculate the uncertainty or error of the velocity

h = $\frac{1}{454.55^2} \ v^2$

Δh = $\frac{dh}{dv}$ Δv

$\frac{\Delta h}{h } = 2 \ \frac{\Delta v}{v}$

Suppose we have a height reading of h = 20 cm = 0.20 m

a) uncertainty 2.5 m / s ( 0.05)

$\frac{\delta v}{v} = 0.05$

$\frac{\Delta h}{h}$ = 2 0.05

Δh = 0.1 h

Δh = 0.1 20 cm

Δh = 2 cm

b) uncertainty 0.5 m / s ( Δv/v= 0.01)

$\frac{\Delta h}{h}$ = 2 0.01

Δh = 0.02 h

Δh = 0.02 20

Δh = 0.1 20 cm

Δh = 0.4 cm = 4 mm

5 0

3 years ago

The Environmental Protection Agency (EPA) has standards and regulations that says that the lead level in soil cannot exceed the

DENIUS [597]

Answer:

See below

Explanation:

Check One-Sample T-Interval Conditions

Random Sample? √

Sample Size ≥30? √

Independent? √

Population Standard Deviation Unknown? √

One-Sample T-Interval Information

Formula --> $CI=\bar{x}\pm t^*(\frac{S_x}{\sqrt{n}})$
Sample Mean --> $\bar{x}=390.25$
Critical Value --> $t^*=2.0096$ (given $df=n-1=50-1=49$ degrees of freedom at a 95% confidence level)
Sample Size --> $n=50$
Sample Standard Deviation --> $S_x=30.5$

Problem 1

The critical t-value, as mentioned previously, would be $t^*=2.0096$ , making the 95% confidence interval equal to $CI=\bar{x}\pm t^*(\frac{S_x}{\sqrt{n}})=390.25\pm2.0096(\frac{30.5}{\sqrt{50}})\approx\{381.5819,398.9181\}$

This interval suggests that we are 95% confident that the true mean levels of lead in soil are between 381.5819 and 398.9181 parts per million (ppm), which satisfies the EPA's regulated maximum of 400 ppm.

3 0

2 years ago