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2 years ago

Designing a quiet generator to reduce the amount of hazardous noise is an example of what type of risk control

Physics

1 answer:

kirza4 [7]2 years ago

4 0

Designing a quiet generator to reduce the amount of hazardous noise is an

example of Engineering risk control.

<h3>What is Engineering risk control?</h3>

Engineering risk control involves the process of designing or modification

of a product in order to reduce the risks or hazards associated with it. The

methods involved include:

Process control

Enclosure

Isolation

Ventilation

In this scenario, a quiet generator was designed to reduce the amount of

hazardous noise which correctly represents an Engineering risk control.

Read more about Engineering risk control here brainly.com/question/21213968

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4 years ago

One recently discovered extrasolar planet, or exoplanet, orbits a star whose mass is 0.70 times the mass of our sun. This planet

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Answer: Option B

<u>Explanation:</u>

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Where,

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$r_{e}$ = earth’s radius

Now,

$T_{e}=\sqrt{\frac{r_{e}^{3}}{G M_{s}}}$

As, planet mass is equal to 0.7 times the sun mass, so

$T_{p}=\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}$

Taking the ratios of both equation, we get,

$\frac{T_{e}}{T_{p}}=\frac{\sqrt{\frac{r_{e}^{3}}{G M_{s}}}}{\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}}$

$\frac{T_{e}}{T_{p}}=\sqrt{\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}}$

$\left(\frac{T_{e}}{T_{p}}\right)^{2}=\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}$

$\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}=\frac{r_{e}^{3}}{r_{p}^{3}}$

$\frac{r_{e}}{r_{p}}=\left(\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}$

Given $T_{p}=9.5 \text { days }$ and $T_{e}=365 \text { days }$

$\frac{r_{e}}{r_{p}}=\left(\left(\frac{365}{9.5}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=\left(\frac{133225}{90.25} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=(2108.82)^{\frac{1}{3}}$

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7 0

3 years ago

The latent heat of vaporization for water at room temperature is 2430 J/g.

alukav5142 [94]

Answer:

1) $kinectic energy=7.26*10^-^2^0J$

2) $V= 2.0m/s$

3) $T=3.5*10^3K$

4) The Molecules do not burn because of the presences of hydrogen bond in place

Explanation:

From the question we are told that

latent heat of vaporization for water at room temperature is 2430 J/g.

1)Generally in determining the molar mass of water evaporated we have that

-One mole (6.02 x 10. 23 molecules)

-Molar mass of water is 18.02 g/mol

Mathematically the mass of water is give as

$M=\frac{18.02}{6.02*10^-^2^6}$

$M=3*10^-^2^3g$

Therefore

$kinectic energy=2430J/g*3*10^-^2^3g$

$kinectic energy=7.26*10^-^2^0J$

b)Generally the evaporation speed V is given as $V= \sqrt{\frac{K.E*2}{m} }$

Mathematically derived from the equation

$\frac{1}{2} mv^2 =K.E$

To Give

$V= \sqrt{\frac{K.E*2}{m} }$

$V= \sqrt{\frac{7.26*10^-^2^0J*2}{3*10^-^2^3g} }$

$V= 2.0m/s$

c)Generally the equation for velocity $Vrms=\sqrt{\frac{3RT}{M} }$

Therefore

Effective temperature T is given by

$T=\frac{\sqrt{v}*m}{R}$

where

$T=\frac{\sqrt{2.0m/s}*6.02*10^-^2^6}{0.082057 L atm mol-1K-1}$

$T=3.5*10^3K$

4) The Molecules do not burn because of the presences of hydrogen bond in place

3 0

3 years ago