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2 years ago

A statement of the second law of thermodynamics is that:__________.

Physics

1 answer:

adoni [48]2 years ago

7 0

Answer:

in a spontaneous process, the entropy of the universe increases.

Explanation:

Entropy is a measure of of the degree of randomness or disorderliness in a system.

The second law of thermodynamics can be stated as follows; "in any spontaneous process, the entropy of the universe increases."

The universe here refers to the system's disorder and the disorder of the surroundings. Therefore, a spontaneous process can occur, in which the entropy of the system decreases, only if the entropy increases in the surroundings.

For instance, when ice freezes, the entropy of liquid water decreases, that is, the entropy of the system decreases. However, heat is given off to the surroundings and the entropy of the surroundings increases. This is an obvious expression of this law.

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NASA scientists suggest using rotating cylindrical spacecraft to replicate gravity while in a weightless environment. Consider s

dusya [7]

Answer:

Explanation:

Given

diameter of spacecraft $d=148\ m$

radius $r=74\ m$

Force of gravity $F_g$ =mg

where m =mass of object

g=acceleration due to gravity on earth

Suppose v is the speed at which spacecraft is rotating so a net centripetal acceleration is acting on spacecraft which is given by

$F_c=\frac{mv^2}{r}$

$F_c=F_g$

$\frac{mv^2}{r}=mg$

$\frac{v^2}{r}=g$

$v=\sqrt{gr}$

$v=\sqrt{1450.4}$

$v=38.08\ m/s$

8 0

2 years ago

A boat moves up and down as water waves pass under the boat. If the amplitude of the wave gets bigger, then

xeze [42]

The anwser is A or D

7 0

2 years ago

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X-rays with an energy of 300 keV undergo Compton scattering from a target. If the scattered rays are detected at 30 relative to

lys-0071 [83]

Answer:

a) $\Delta \lambda = \lambda' -\lambda_o = \frac{h}{m_e c} (1-cos \theta)$

For this case we know the following values:

$h = 6.63 x10^{-34} Js$

$m_e = 9.109 x10^{-31} kg$

$c = 3x10^8 m/s$

$\theta = 37$

So then if we replace we got:

$\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm$

b) $\lambda_0 = \frac{hc}{E_0}$

With $E_0 = 300 k eV= 300000 eV$

And replacing we have:

$\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm$

And then the scattered wavelength is given by:

$\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm$

And the energy of the scattered photon is given by:

$E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV$

c) $E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV$

Explanation

Part a

For this case we can use the Compton shift equation given by:

$\Delta \lambda = \lambda' -\lambda_0 = \frac{h}{m_e c} (1-cos \theta)$

For this case we know the following values:

$h = 6.63 x10^{-34} Js$

$m_e = 9.109 x10^{-31} kg$

$c = 3x10^8 m/s$

$\theta = 37$

So then if we replace we got:

$\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm$

Part b

For this cas we can calculate the wavelength of the phton with this formula:

$\lambda_0 = \frac{hc}{E_0}$

With $E_0 = 300 k eV= 300000 eV$

And replacing we have:

$\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm$

And then the scattered wavelength is given by:

$\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm$

And the energy of the scattered photon is given by:

$E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV$

Part c

For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:

$E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV$

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2 years ago

Read 2 more answers

What is black body radiation? Explain in detail.

tangare [24]

An object that absorbs all radiation falling on it, at all wavelengths, is called a black body. When a black body is at a uniform temperature, its emission has a characteristic frequency distribution that depends on the temperature. Its emission is called black-body radiation

hope it helps

3 0

2 years ago

Natalie accelerate her skateboard along a straight path from 0 m/s to 4.0 m/s in 2.5 s. find her average acceleration

Vikentia [17]

Acceleration=(speed end - speed start)/ time
Data:
speed end=4 m/s
speed start=0 m/s
time=2.5 s

acceleration=(4 m/s - 0 m/s)/2.5 s=1.6 m/s²

Answer: the acceleration would be 1.6 m/s²

8 0

2 years ago