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FrozenT [24]
3 years ago
6

What is a risk assessment

Physics
2 answers:
xxMikexx [17]3 years ago
7 0
A risk assessment<span> is a process to identify potential hazards and analyze what could happen if a hazard occurs. </span>
CaHeK987 [17]3 years ago
3 0

Answer:

Explanation:

Risk assessment is a process of identifying risks and hazards associated with some activity, things, situations and processes which can cause harm to public. After the identification of the risks, the severity and potential of the risk can be analyzed. Further preventing measures can be taken to prevent the risk.

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Which of the following statements are TRUE about nuclear energy?
san4es73 [151]

Answer:

d

Explanation:

4 0
2 years ago
Two sound waves have equal displacement amplitudes, but wave 1 has two-thirds the frequency of wave 2. What is the ratio of the
zlopas [31]

Answer:

\dfrac{I_1}{I_2}=\dfrac{4}{9}

Explanation:

c = Speed of wave

\rho = Density of medium

A = Area

\nu = Frequency

\nu_1=\dfrac{2}{3}\nu_2

Intensity of sound is given by

I=\dfrac{1}{2}\rho c(A\omega)^2\\\Rightarrow I=\dfrac{1}{2}\rho c(A2\pi \nu)^2

So,

I\propto \nu^2

We get

\dfrac{I_1}{I_2}=\dfrac{\nu_1^2}{\nu_2^2}\\\Rightarrow \dfrac{I_1}{I_2}=\dfrac{\dfrac{2}{3}^2\nu_2^2}{\nu_2^2}\\\Rightarrow \dfrac{I_1}{I_2}=\dfrac{4}{9}

The ratio is \dfrac{I_1}{I_2}=\dfrac{4}{9}

8 0
2 years ago
Two astronauts are 2.00 m apart in their spaceship. One speaks to the other. The conversation is transmitted to earth via electr
Eduardwww [97]

Answer:

D=1693742.7m

Explanation:

For sound waves we have v=d/t where v is the speed of sound and d the distance between the astronauts, while for electromagnetic waves we have c=D/t where c is the speed of light and D the distance between the spaceship and Earth. <em>We have written both times as the same</em> because is what is imposed by the problem, so we have t=d/v=D/c, which means:

D=\frac{dc}{v}

And for our values:

D=\frac{(2m)(299792458m/s)}{354m/s}=1693742.7m

5 0
3 years ago
An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a
S_A_V [24]

Answer:

Given:

Thermal Kinetic Energy of an electron, KE_{t} = \frac{3}{2}k_{b}T

k_{b} = 1.38\times 10^{- 23} J/k = Boltzmann's constant

Temperature, T = 1800 K

Solution:

Now, to calculate the de-Broglie wavelength of the electron, \lambda_{e}:

\lambda_{e} = \frac{h}{p_{e}}

\lambda_{e} = \frac{h}{m_{e}{v_{e}}              (1)

where

h = Planck's constant = 6.626\times 10^{- 34}m^{2}kg/s

p_{e} = momentum of an electron

v_{e} = velocity of an electron

m_{e} = 9.1\times 10_{- 31} kg = mass of electon

Now,

Kinetic energy of an electron = thermal kinetic energy

\frac{1}{2}m_{e}v_{e}^{2} = \frac{3}{2}k_{b}T

}v_{e} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{e}}}

}v_{e} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{9.1\times 10_{- 31}}}

v_{e} = 2.86\times 10^{5} m/s                    (2)

Using eqn (2) in (1):

\lambda_{e} = \frac{6.626\times 10^{- 34}}{9.1\times 10_{- 31}\times 2.86\times 10^{5}} = 2.55 nm

Now, to calculate the de-Broglie wavelength of proton, \lambda_{e}:

\lambda_{p} = \frac{h}{p_{p}}

\lambda_{p} = \frac{h}{m_{p}{v_{p}}                             (3)

where

m_{p} = 1.6726\times 10_{- 27} kg = mass of proton

v_{p} = velocity of an proton

Now,

Kinetic energy of a proton = thermal kinetic energy

\frac{1}{2}m_{p}v_{p}^{2} = \frac{3}{2}k_{b}T

}v_{p} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{p}}}

}v_{p} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{1.6726\times 10_{- 27}}}

v_{p} = 6.674\times 10^{3} m/s                               (4)                    

Using eqn (4) in (3):

\lambda_{p} = \frac{6.626\times 10^{- 34}}{1.6726\times 10_{- 27}\times 6.674\times 10^{3}} = 5.94\times 10^{- 11} m = 0.0594 nm

7 0
3 years ago
kristine speeds past a parked police car at 32 m/s. The police car starts from rest with a uniform acceleration of 2.5 m/s^2. Ho
Digiron [165]

32 = 0 \times t +  \frac{1}{2} \times 2.5 \times t^{2}     \\  32 = 0 + 1.25 \times t {}^{2}  \\ 32 = 1.25t {}^{2}   \\   \frac{32}{1.25}  =  \frac{1.25t {}^{2} }{1.25}  \\ t {}^{2}  = 25.6 \\  \sqrt{t {}^{2} }  =  \sqrt{25.6}  \\ t = 5.1seconds \\

7 0
2 years ago
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