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katrin [286]
3 years ago
9

Give two examples of workplace environments where considerations must be made with respect to the possibility of electric discha

rges. Explain why this is a necessary concern
Physics
1 answer:
katrin2010 [14]3 years ago
4 0

Answer:

1.) Oil and gas environment

2.) Mechanical engineering and electrical environment.

Explanation:

1.) Oil and gas environment.

Electricity and electric discharge are recognized as serious workplace hazard in oil and gas companies, exposing employees to burns, fires, and explosions.

2.) Mechanical engineering and electrical environment.

Electricity or electric discharge has long been recognized as a serious workplace hazard, exposing employees to electric shock, electrocution, burns, fires, and explosions. Workers could die from electrocutions at work. This is

accounting for on-the-job fatalities and what makes these more tragic is that most of these fatalities could have been easily avoided.

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devlian [24]
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8 0
3 years ago
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Which of the following examples illustrates static friction?
vivado [14]

Answer:

A box sits stationary  on a ramp

Explanation:

Static friction is a force which keeps an object at rest as it is in the case of the box. It has to be overcome for the object to be set into motion.

Static force of friction is calculated as follows:

F= μη

F is static force of friction.

μ is the coefficient of static friction.

η is the normal force.

6 0
3 years ago
Find the wavelength in meters for a transverse mechanical wave with an amplitude of 10 cm and a radian frequency of 20π rad/s if
nydimaria [60]

Answer:

The wavelength of the wave is 20 m.

Explanation:

Given that,

Amplitude = 10 cm

Radial frequency \omega = 20\pi\ rad/s

Bulk modulus = 40 MPa

Density = 1000 kg/m³

We need to calculate the velocity of the wave in the medium

Using formula of velocity

v=\sqrt{\dfrac{k}{\rho}}

Put the value into the formula

v=\sqrt{\dfrac{40\times10^{6}}{10^3}}

v=200\ m/s

We need to calculate the wavelength

Using formula of wavelength

\lambda =\dfrac{v}{f}

\lambda=\dfrac{v\times2\pi}{\omega}

Put the value into the formula

\lambda=\dfrac{200\times2\pi}{20\pi}

\lambda=20\ m

Hence, The wavelength of the wave is 20 m.

5 0
3 years ago
If the given wave has a frequency of 100 Hz, what frequency will the sixth harmonic have?
alukav5142 [94]

Answer:

600Hz

Explanation:

In electrical systems of alternating current, the harmonics are, as in acoustics, frequencies multiples of the fundamental working frequency of the system and whose amplitude decreases as the multiple increases. For example, if we have systems fed by the 50 Hz network, harmonics of 100, 150, 200, etc. may appear.

In our case having a fundamental wave of 100Hz, I can have harmonics of 200,300,400, ..., 600Hz

4 0
3 years ago
Suppose you illuminate two thin slits by monochromatic coherent light in air and find that they produce their first interference
strojnjashka [21]

Answer:

1.7323

Explanation:

To develop this problem, it is necessary to apply the concepts related to refractive indices and Snell's law.

From the data given we have to:

n_{air}=1

\theta_{liquid} = 19.38\°

\theta_{air}35.09\°

Where n means the index of refraction.

We need to calculate the index of refraction of the liquid, then applying Snell's law we have:

n_1sin\theta_1 = n_2sin\theta_2

n_{air}sin\theta_{air} = n_{liquid}sin\theta_{liquid}

n_{liquid} = \frac{n_{air}sin\theta_{air}}{sin\theta_{liquid}}

Replacing the values we have:

n_{liquid}=\frac{(1)sin(35.09)}{sin(19.38)}

n_liquid = 1.7323

Therefore the refractive index for the liquid is 1.7323

6 0
4 years ago
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