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3 years ago

What is the criteria for a guard having to be used on a machine?

Engineering

1 answer:

evablogger [386]3 years ago

8 0

Answer:

Machine Safeguards must meet these minimum general requirements: Prevent contact: The safeguard must prevent hands, arms, or any other part of a worker's body from making contact with dangerous moving parts. Be secure: Workers should not be able to easily remove or tamper with the safeguard.

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- Scrap tire management is primarily regulated at the

kompoz [17]

Scrap tire management is primarily regulated at the state level.

3 0

3 years ago

Read 2 more answers

A Si sample contains 1016 cm-3 In acceptor atoms and a certain number of shallow donors, the In acceptor level is 0.16 eV above

creativ13 [48]

Answer:

6.5 × 10¹⁵/ cm³

Explanation:

Thinking process:

The relation $N_{o} = N_{i} * \frac{E_{f}-E_{i} }{KT}$

With the expression Ef - Ei = 0.36 × 1.6 × 10⁻¹⁹

and ni = 1.5 × 10¹⁰

Temperature, T = 300 K

K = 1.38 × 10⁻²³

This generates N₀ = 1.654 × 10¹⁶ per cube

Now, there are 10¹⁶ per cubic centimeter

Hence, $N_{d} = 1.65*10^{16} - 10^{16} \\ = 6.5 * 10^{15} per cm cube$

5 0

3 years ago

Read 2 more answers

A steel bolt has a modulus of 207 GPa. It holds two rigid plates together at a high temperature under conditions where the creep

VikaD [51]

Answer:

14.36((14MPa) approximately

Explanation:

In this question, we are asked to calculate the stress tightened in a bolt to a stress of 69MPa.

Please check attachment for complete solution and step by step explanation

7 0

3 years ago

Find E[x] when x is sum of two fair dice?

Ksenya-84 [330]

Answer:

When two fair dice are rolled, 6×6=36 observations are obtained.

P(X=2)=P(1,1)=

P(X=3)=P(1,2)+P(2,1)=

P(X=4)=P(1,3)+P(2,2)+P(3,1)=

P(X=5)=P(1,4)+P(2,3)+P(3,2)+P(4,1)=

P(X=6)=P(1,5)+P(2,4)+P(3,3)+P(4,2)+P(5,1)=

P(X=7)=P(1,6)+P(2,5)+P(3,4)+P(4,3)+P(5,2)+P(6,1)=

P(X=8)=P(2,6)+P(3,5)+P(4,4)+P(5,3)+P(6,2)=

P(X=9)=P(3,6)+P(4,5)+P(5,4)+P(6,3)=

P(X=10)=P(4,6)+P(5,5)+P(6,4)=

P(X=11)=P(5,6)+P(6,5)=

P(X=12)=P(6,6)=

Therefore, the required probability distribution is as follows.

Then, E(X)=∑X

⋅P(X

)

=2×

+3×

+4×

+5×

+6×

+7×

+8×

+9×

+10×

+11×

+12×

+1+

E(X

)=∑X

⋅P(X

)

=4×

+9×

+16×

+25×

+36×

+49×

+64×

+81×

+100×

+121×

+144×

+5+

+9+

121

987

329

=54.833

Then, Var(X)=E(X

)−[E(X)]

=54.833−(7)

=54.833−49

=5.833

∴ Standard deviation =

Var(X)

5.833

=2.415

4 0

3 years ago

From an aerial photograph, one observes that on a level section of a (multilane) highway, 25% of the vehicles are trucks, 75% ar

egoroff_w [7]

Answer:

(a) Flow rate of vehicles = No of vehicles per mile * Speed

=No of cars per mile * Speed +No of trucks per mile * Speed

= 0.75*50*60 + 0.25*50*40

=2750 vehicles / hour

(b) Let Density of vehicles on grade = x

Density on flat * Speed =Density on grade * Speed

So,( 0.75*50) * 60 + (0.25*50) * 40 = (0.75* x) * 55 + (0.25* x) * 25

So, x= 57.89

So, Density is around 58 Vehicles per Mile.

(d)Percentage of truck bystationary observer on the grade= 25*30/60 * 25/55 =22.73 %

4 0

3 years ago