Question 6 of 10

In thermodynamics, a closed system is a system where _____

Nady [450]

In a closed system, energy in form of heat (work) can be exchanged but not matter.

The answer to your question is C.

Hope it helped!

3 0

3 years ago

The first ionization energy of a hydrogen atom is 2.18 aj (attojoules). what is the frequency and wavelength, in nanometers, of

jasenka [17]

1) Frequency: $3.29\cdot 10^{15}Hz$

the energy of the photon absorbed must be equal to the ionization enegy of the atom, which is

$E=2.18 aJ=2.18\cdot 10^{-18} J$

The energy of a photon is given by

$E=hf$

where $h=6.63\cdot 10^{-34}Js$ is the Planck's constant. By using the energy written above and by re-arranging thsi formula, we can calculate the frequency of the photon:

$f=\frac{E}{h}=\frac{2.18\cdot 10^{-18} J}{6.63\cdot 10^{-34} Js}=3.29\cdot 10^{15} Hz$

2) Wavelength: 91.2 nm

The wavelength of the photon can be found from its frequency, by using the following relationship:

$\lambda=\frac{c}{f}$

where $c=3\cdot 10^8 m/s$ is the speed of light and f is the frequency. Substituting the frequency, we find

$\lambda=\frac{3\cdot 10^8 m/s}{3.29\cdot 10^{15}Hz}=9.12\cdot 10^{-8} m=91.2 nm$

5 0

2 years ago

The aurora is caused when electrons and protons, moving in the earth’s magnetic field of ≈5.0×10−5T, collide with molecules of t

ollegr [7]

Answer:

8.79*10^6 rad/s

Explanation:

To find the frequency of the circular orbit for an electron you use the following expression, for the radius of the trajectory of an electron, that travels trough a constant magnetic field:

$r=\frac{mv}{qB}$ (1)

r: radius of the trajectory

m: mass of the electron = 9.1*10^-31 kg

v: speed of the electron = 1.0*10^6 m/s

q: charge of the electron = 1.6*10^-19 C

B: magnitude of the magnetic field = 5.0*10^-5 T

You use the fact that the angular frequency in a circular motion is given by:

$\omega=\frac{v}{r}$

Then, you solve the equation (1) in order to obtain v/r:

$\frac{v}{r}=\omega=\frac{qB}{m}$

Finally, you replace the values of the parameters:

$\omega=\frac{(1.6*10^{-19}C)(5.0*10^{-5}T)}{9.1*10^{-31}kg}\\\\\omega=8.79*10^6\frac{rad}{s}$

hence, the angular frequency is 8.79*10^6 rad/s

The frequency is:

$f=2\pi \omega=5.5*10^7Hz$

5 0

3 years ago

In an economy, the demand for labor is given by the equation W = 15 - (1/200) L and the supply of labor is given by the equation

mr_godi [17]

Answer:

the equilibrium wage rate is 10 and the equilibrium quantity of labor is 1000 workers

Explanation:

The equilibrium wage rate and the equilibrium quantity of labor are found as the point where the equation of demand intercepts the equation of supply, so the equilibrium quantity of labor is:

$W_{Demand} = W_{Supply}$

15 - (1/200) L = 5 + (1/200) L

15 - 5 = (1/200) L + (1/200) L

10 = (2/200) L

(10*200)/2 = L

1000 = L

Then, the equilibrium wage rate is calculated using either the equation of demand for labor or the equation of supply of labor. If we use the equation of demand for labor, we get:

W = 15 - (1/200) L

W = 15 - (1/200) 1000

W = 10

Finally, the equilibrium wage rate is 10 and the equilibrium quantity of labor is 1000 workers

7 0

3 years ago

HELP DUE TODAY

olga2289 [7]

The answer is B I think sorry if it’s wrong

3 0

3 years ago