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3 years ago

Which of the following answer options are your employer's responsibility?

Engineering

1 answer:

tino4ka555 [31]3 years ago

3 0

Answer:

Develop a written hazard communication program

Implement a hazard communication program

Maintain a written hazard communication program

Explanation:

To find - Which of the following answer options are your employer's responsibility? Select all that apply.

Develop a written hazard communication program

Implement a hazard communication program

Maintain a written hazard communication program

Solution -

The correct options are -

Develop a written hazard communication program

Implement a hazard communication program

Maintain a written hazard communication program

All are the Responsibilities of an employer

Reason -

The most important duty of the employer is to stay alert and implement a correctly and efficiently written communication program related to hazards of the substances in the workplace.

He also has to maintain the program so that employees do not get affected.

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Tesla Is the best ELECTRIC car brand, Change my mind

pochemuha

Answer:You are correct, no need to change.

Explanation:

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3 years ago

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What i s the value of a capacitor with 250 V applied and has 500 pC of charge? (a) 200 uF (b) 0.5 pF (c) 500 uF (d) 2 pF

exis [7]

Answer:

(d) 2 pF

Explanation: the charge on capacitor is given by the expression

Q=CV

where Q=charge

C=capacitance

V=voltage across the plate of the capacitor

here we have given Q=500 pF, V=250 volt

using this formula C= $\frac{Q}{V}$

=500× $10^{-12}$ × $\frac{1}{250}$

=2× $10^{-12}$

=2 pF

3 0

3 years ago

List two ways you can make an informal survey

solniwko [45]

The two ways you can use to make an informal survey are:

make field observations
interview people using informal unstructured techniques

<h3>What are informal surveys?</h3>

In informal surveys can be regarded as a type of survey that can be made by the researcher by going to the field themselves and this can be done by using different methods or ways.

For instance, the researcher can go out to interview people that can give the data that is needed about the research such as informally asking them questions, unstructured techniques can also be used to solve critical issues.

learn more about survey at: brainly.com/question/6947486

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8 0

1 year ago

Read 2 more answers

Engine oil flows through a 25‐mm‐diameter tube at a rate of 0.5 kg/s. The oil enters the tube at a temperature of 25°C, while th

Elodia [21]

Answer:

a) the log mean temperature difference (Approx. 64.5 deg C)

b) the rate of heat addition into the oil.

The above have been solved for in the below workings

Explanation:

3 0

3 years ago

The water in a large lake is to be used to generate electricity by the installation of a hydraulic turbine-generator. The elevat

ankoles [38]

Answer:

a) 75%

b) 82%

Explanation:

Assumptions:

$\text{The mechanical energy for water at turbine exit is negligible.} \\ \\ \text{The elevation of the lake remains constant.}$

Properties: The density of water $\delta = 1000 kg/m^3$

Conversions:

$165 \ ft \ to \ meters = 50 m \\ \\7000 \ lbm/s \ to \ kilogram/sec = 3175 kg/s \\ \\1564 \ hp \ to \ kilowatt = 1166 kw \\ \\$

Analysis:

Note that the bottom of the lake is the reference level. The potential energy of water at the surface becomes gh. Consider that kinetic energy of water at the lake surface & the turbine exit is negligible and the pressure at both locations is the atmospheric pressure and change in the mechanical energy of water between lake surface & turbine exit are:

$e_{mech_{in}} - e_{mech_{out}} = gh - 0$

Then;

$gh = (9.8 m/s^2) (50 m) \times \dfrac{1 \ kJ/kg}{1000 m^2/s^2}$

gh = 0.491 kJ/kg

$\Delta E_{mech \ fluid} = m(e_{mech_{in}} - e_{mech_{out}} ) \\ \\ = 3175 kg/s \times 0.491 kJ/kg$

= 1559 kW

Therefore; the overall efficiency is:

$\eta _{overall} = \eta_{turbine- generator} = \dfrac{W_{elect\ out}}{\Delta E_{mech \fluid}}$

$= \dfrac{1166 \ kW}{1559 \ kW}$

= 0.75

= 75%

b) mechanical efficiency of the turbine:

$\eta_{turbine- generator} = \eta_{turbine}\times \eta_{generator}$

thus;

$\eta_{turbine} = \dfrac{\eta_{[turbine- generator]} }{\eta_{generator}} \\ \\ \eta_{turbine} = \dfrac{0.75}{0.92} \\ \\ \eta_{turbine} = 0.82 \\ \\ \eta_{turbine} = 82\%$

6 0

3 years ago