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RUDIKE [14]
2 years ago
14

Describe how you would control employee exposure to excessive noise in a mining environment

Engineering
1 answer:
Alexandra [31]2 years ago
7 0

Answer:

1. Buy Quiet – select and purchase low-noise tools and machinery

2. Maintain tools and equipment routinely (such 3. as lubricate gears)

3. Reduce vibration where possible

4. Isolate the noise source in an insulated room or enclosure

5. Place a barrier between the noise source and the employee

6. Isolate the employee from the source in a room or booth (such as sound wall or window

Explanation:

Hope my answer will help u.

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Can somebody help me
masha68 [24]
I think the answer is D .
6 0
3 years ago
Alberta Einstein teaches a business class at Podunk University. To evaluate the students in this class, she has given three test
madreJ [45]

Answer:

See explaination

Explanation:

Please kindly check attachment for step by step solution of the given problem represented as a flow chart.

7 0
4 years ago
Coils of various dimensions designed to introduce specified amounts of inductance into a circuit are called (a) semiconductors (
Delicious77 [7]

Answer: d) Inductors

Explanation:  Inductors are the electric component commonly known by the name of coils as well. The work done by inductors is when the electric current flows through the coils of the wire then there is production of the magnetic field. They are the component that have the inductance in particular amount according to the circuit.So, coils of various dimensions are designed to introduce specified amount of inductance  into a circuit are called inductors.

4 0
3 years ago
Two resistors, with resistances R1 and R2, are connected in series. R1 is normally distributed with mean 65 and standard deviati
Sedbober [7]

Answer:

n this question, we are asked to find the probability that  

R1 is normally distributed with mean 65  and standard deviation 10

R2 is normally distributed with mean 75  and standard deviation 5

Both resistor are connected in series.

We need to find P(R2>R1)

the we can re write as,

P(R2>R1) = P(R2-R1>R1-R1)

P(R2>R1) = P(R2-R1>0)

P(R2>R1) = P(R>0)

Where;

R = R2 - R1

Since both and are independent random variable and normally distributed, we can do the linear combinations of mean and standard deviations.

u = u2-u1

u = 75 - 65 = 10ohm

sd = √sd1² + sd2²

sd = √10²+5²

sd = √100+25 = 11.18ohm

Now we will calculate the z-score, to find  P( R>0 )

Z = ( X -u)/sd

the z score of 0 is

z = 0 - 10/11.18

z= - 0.89

4 0
4 years ago
The atmospheric pressure at the top and the bottom of a building is read by a barometer to be 98.5 kPa and 100 kPa, respectively
Fynjy0 [20]

Answer:

127.42m

Explanation:

The air pressure can be understood as the weight exerted by the air column on a body, for this case we must remember that the pressure is calculated by the formula  P=αgh, Where P=pressure, h=gravity, h= height,α=density

So what we must do to solve this problem is to find the length of the air column above and below the building and then subtract them to find the height of the building, taking into account the above the following equation is inferred

h2-h1= building height=H

H=\frac{P1-P2}{g\alpha }

P1=100kPa=100.000Pa

P2=98.5kPa=98.500Pa

α=1.2 kg/m^3

g=9.81m/s^2

H=\frac{100000-98500}{(9.81)(1.2) }=127.42m

4 0
4 years ago
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