What happens to the volume of a gas when pressure is applied? A. increases B. decreases C. Stays the same D. evaporates

olga nikolaevna [1]

Answer:

Professor Hawking had just turned 21 when he was diagnosed with a very rare slow-progressing form of ALS, a form of motor neurone disease (MND). He was at the end of his time at Oxford when he started to notice early signs of his disease. He was getting more clumsy and fell over several times without knowing why.

Explanation:

none

6 0

2 years ago

g You shine orange laser light that has a wavelength of 600 nm through a narrow slit. The slit forms a diffraction pattern on a

zimovet [89]

Answer:

λ = 3 10⁻⁷ m, UV laser

Explanation:

The diffraction phenomenon is described by the expression

a sin θ = m λ

let's use trigonometry

tan θ = y / L

as in this phenomenon the angles are small

tan θ = $\frac{sin \ \theta}{cos \ \theta}$ = sin θ

sin θ = y / L

we substitute

a y / L = m λ

let's apply this equation to the initial data

a 0.04 / L = 1 600 10⁻⁹

a / L = 1.5 10⁻⁵

now they tell us that we change the laser and we have y = 0.04 m for m = 2

a 0.04 / L = 2 λ

a / L = 50 λ

we solve the two expression is

1.5 10⁻⁵ = 50 λ

λ = 1.5 10⁻⁵ / 50

λ = 3 10⁻⁷ m

UV laser

3 0

2 years ago

After rubbing a balloon on your shirt, your hair sticks up when the balloon is near your head. Explain how tiny particles on you

skad [1K]

Almost right. protons are positive and electrons are negative. so when you run the balloon on your hair, electrons are transferred between them (i’m not sure which direction) and now one is positively charged as it lost negative particles and one is negative as it gained negative particles (electrons). opposite charges attract

3 0

1 year ago

What change in entropy occurs when a 0.15 kg ice cube at -18 °C is transformed into steam at 120 °c 4.

Studentka2010 [4]

Answer: The change in entropy of the given process is 1324.8 J/K

Explanation:

The processes involved in the given problem are:

$1.)H_2O(s)(-18^oC,255K)\rightarrow H_2O(s)(0^oC,273K)\\2.)H_2O(s)(0^oC,273K)\rightarrow H_2O(l)(0^oC,273K)\\3.)H_2O(l)(0^oC,273K)\rightarrow H_2O(l)(100^oC,373K)\\4.)H_2O(l)(100^oC,373K)\rightarrow H_2O(g)(100^oC,373K)\\5.)H_2O(g)(100^oC,373K)\rightarrow H_2O(g)(120^oC,393K)$

Pressure is taken as constant.

To calculate the entropy change for same phase at different temperature, we use the equation:

$\Delta S=m\times C_{p,m}\times \ln (\frac{T_2}{T_1})$ .......(1)

where,

$\Delta S$ = Entropy change

$C_{p,m}$ = specific heat capacity of medium

m = mass of ice = 0.15 kg = 150 g (Conversion factor: 1 kg = 1000 g)

$T_2$ = final temperature

$T_1$ = initial temperature

To calculate the entropy change for different phase at same temperature, we use the equation:

$\Delta S=m\times \frac{\Delta H_{f,v}}{T}$ .......(2)

where,

$\Delta S$ = Entropy change

m = mass of ice

$\Delta H_{f,v}$ = enthalpy of fusion of vaporization

T = temperature of the system

Calculating the entropy change for each process:

For process 1:

We are given:

$m=150g\\C_{p,s}=2.06J/gK\\T_1=255K\\T_2=273K$

Putting values in equation 1, we get:

$\Delta S_1=150g\times 2.06J/g.K\times \ln(\frac{273K}{255K})\\\\\Delta S_1=21.1J/K$

For process 2:

We are given:

$m=150g\\\Delta H_{fusion}=334.16J/g\\T=273K$

Putting values in equation 2, we get:

$\Delta S_2=\frac{150g\times 334.16J/g}{273K}\\\\\Delta S_2=183.6J/K$

For process 3:

We are given:

$m=150g\\C_{p,l}=4.184J/gK\\T_1=273K\\T_2=373K$

Putting values in equation 1, we get:

$\Delta S_3=150g\times 4.184J/g.K\times \ln(\frac{373K}{273K})\\\\\Delta S_3=195.9J/K$

For process 4:

We are given:

$m=150g\\\Delta H_{vaporization}=2259J/g\\T=373K$

Putting values in equation 2, we get:

$\Delta S_2=\frac{150g\times 2259J/g}{373K}\\\\\Delta S_2=908.4J/K$

For process 5:

We are given:

$m=150g\\C_{p,g}=2.02J/gK\\T_1=373K\\T_2=393K$

Putting values in equation 1, we get:

$\Delta S_5=150g\times 2.02J/g.K\times \ln(\frac{393K}{373K})\\\\\Delta S_5=15.8J/K$

Total entropy change for the process = $\Delta S_1+\Delta S_2+\Delta S_3+\Delta S_4+\Delta S_5$

Total entropy change for the process = $[21.1+183.6+195.9+908.4+15.8]J/K=1324.8J/K$

Hence, the change in entropy of the given process is 1324.8 J/K

4 0

3 years ago

;-; please help me....

sineoko [7]

The answer is 24N. Since the body is moving with constant velocity all the forces must balance (equal & opposite)

5 0

2 years ago